TITLE: Finding limits of CDF with indicator function
QUESTION [1 upvotes]: Suppose the continuous random variable, X, has a probability density function (pdf)
defined as $f_x(X) = cx^{-2} \mathbb{1}[1,2](x)$ ; c is a constant, c ∈ $\mathbb{R}$+

(a) Determine c so that the function, $f_x$, is a pdf.
(b) Determine the cumulative distribution function (cdf), $F_X(x)$, and verify
i. $F_x$(−∞) = 0,
ii. $F_X$(+∞) = 1, and
iii. $F_X(x)$ is a non-decreasing monotonic function of x.
So I've started this by assuming that because I'm trying to prove we have a pdf, and because the indicator function attached to my function is on the interval [1,2], I need to solve for c by
$\int_{1}^{2} cx^{-2} dx = 1 \rightarrow c= 2$ and $F_x(x) = -2x^{-1}$
I also am not sure, but I believe it would still only be defined on the interval [1,2].  This is making part b confusing. Do I then take the limit as $x \rightarrow -\infty$ or $x \rightarrow 1$? Same thing for ii, do I take the limit as $x \rightarrow \infty$ or $x \rightarrow 2$? In either case for part ii, I can't get it to equal 1, which makes me think that I did part a wrong.
Can anyone shed some light on where I'm going wrong?

REPLY [1 votes]: We have $f_x(X) = cx^{-2} \mathbb{1}[1,2](x)$, so
$$\int_{-\infty}^\infty cx^{-2} \mathbb{1}[1,2](x)dx=\int_1^2 cx^{-2}dx=-\frac{c}{x}\Bigm|_1^2=\frac{c}{2}=1$$
so $c=2$ as you found.
Next, $F_X(x)=\int_{-\infty}^x f_X(t)dt$.
Consider $x<1$ first
$$
F_X(x)=\int_{-\infty}^x 2t^{-2} \mathbb{1}[1,2](t)dt=\int_{-\infty}^x 0\,dt = 0.
$$
For $1\leq x \leq 2$
$$
F_X(x)=\int_{-\infty}^x 2t^{-2} \mathbb{1}[1,2](t)dt=\int_1^x 2t^{-2}\,dt =\color{red}{ -\frac1t\Bigm|_1^x=1-\frac{1}{x}}.
$$
For $x>2$
$$
F_X(x)=\int_{-\infty}^x 2t^{-2} \mathbb{1}[1,2](t)dt=\int_1^2 2t^{-2}\,dt = 1.
$$
So the CDF is
$$
F_X(x)=\begin{cases}0, & x<1\cr 1-\frac1x, & 1\leq x\leq 2 \cr 1, & x>2 \end{cases}
$$
and you can check all properties:
$$
\lim_{x\to-\infty} F_X(x) = \lim_{x\to-\infty} 0 = 0
$$
and so on.


TITLE: Is the collection $\{x|\;\exists y\;\,\text{such that}\; x\times y = x\}$ a set?
QUESTION [3 upvotes]: My guess is that the collection $A\equiv\{x|\;\exists y\text{ such that } x\times y = x\}$ is a set, moreover, $A=\{\emptyset\}$
Here is my reasoning: clearly, if we consider $x=\emptyset$, then for every set $y$ we have that $\emptyset\times y=\emptyset$, so $\emptyset\in A$.
However, if $x\not=\emptyset$, it shouldn't be true that $x\times y=x$ for any set $y$, for it should contradict some sort of conclusion of the Axiom of Regularity.
Maybe accepting that $A$ is a set different from $\{\emptyset\}$ would lead us to an infinite sequence $(a_{n})_{n\in\omega}$ of the following form:
Suppose that $x\not=\emptyset$ and $x\in A$. Then there exists $a_0\in x$. But then $a_0\in x\times y$, so $a_0=\langle a_1,b\rangle$ for some $a_1\in x$ (note that $a_0\not=a_1$ because we would be contradicting the Axiom of Regularity, since we cannot have $\{a_0\}\in a_0$ nor $\{a_0,b\}\in a_0$) and $b\in y$ (for if $y=\emptyset$, then $x\times y=\emptyset\not=x$, and $x\not\in A$, contrary to our hypothesis).
Note that from the equality $a_0=\langle a_1,b\rangle$ we can conclude that $\{a_1\}\in a_0$.
But since $a_1\in x$, then $a_1\in x\times y$, and $a_1=\langle a_2, b'\rangle$ for some $a_2\in x$ such that $a_1\not=a_2$ (for analogous reasons to $a_0\not=a_1$), and $b'\in y$ that may or may not be equal to $b$ (since it doesn't really matter).
In conclusion we have that: $\{a_2\}\in a_1\in\bigcup a_0$
This way we can construct an infinite-decreasing sequence (renaming the corresponding sets):
$$\dots\in a'_{n+1}\in a'_n\in\dots\;\in a'_1\in a'_0$$
Following the previous reasoning, which cannot occur in ZF.
Is my proof correct? Any suggestions or insights about it?
Thanks in advance for your time.

REPLY [2 votes]: Your proof looks right(with only a need to small modifications depends on your definition of 2-tuple).
Note, that although there is no non empty set with that property, for every set $X$ there is a set $Y$ such that $X×X\subseteq Y×Y\subseteq Y$(and you proved that the last inclusion is equality iff $Y=\emptyset\iff X=\emptyset$):
Let $X$ be a set and define $$Y_0=X;\;\;Y_{k+1}=Y_k\cup(Y_k×Y_k);\;\; Y=\bigcup_{k}Y_k$$
First, clearly $X\subseteq Y$, now let $(a,b)\in Y×Y$, then, be definition, $a\in Y_{n_a},\ b\in Y_{n_b}$, WLOG $n_a<n_b$, because we have $Y_m\subseteq Y_{m+1}$ we have $Y_{n_a}\subseteq Y_{n_b}$ so $a\in Y_{n_b}$ as well, so $(a,b)\in Y_{n_b+1}\subseteq Y$ so $(a,b)\in Y$.


TITLE: Existence of a connected neighborhood does not imply local connectedness
QUESTION [1 upvotes]: My Topology teacher told us that the existence of a connected neighborhood of a point $x$ in a Topological space does not imply local connectedness at that point, that is, it is not true in general that there is a neighborhood base of connected neighborhoods of $x$.
He told us that we could get an example of this fact by making some "adjustment" or applying a certain "transformation" on the subspace $E=\{(\frac{1}{n},0)\in \mathbb{R}^2: n \in \mathbb{N}\} \cup \{(0,0)\}$ of the plane $\mathbb{R}^2$ (with the Euclidean subspace topology). I know that $E$ is locally connected at every point of the form $(\frac{1}{n},0)$, since every singleton $\{(\frac{1}{n},0)\}$ is open and connected in $E$, so it generates a  connected neighborhood base at $(\frac{1}{n},0)$ whose only element is $\{(\frac{1}{n},0)\}$. On the other hand, $E$ is not locally connected at $(0,0)$, because every neighborhood of this point is disconnected.
Now, I don't know what to do to generate a space to exemplify the said fact. I suppose I should find a way to get some connected neighborhood of $(0,0)$ such that the collection of the connected neighborhoods don't form a base, but I really don't have good ideas on how to proceed. Any thoughts? Thanks on advance.

REPLY [1 votes]: HINT: $\big(E\times[0,1]\big)\cup\big([0,1]\times\{1\}\big)$ (draw a picture) is connected, so every point of it has a connected nbhd, but it is not locally connected at the origin.


TITLE: Statistics Question? Monte Carlo Simulation? Combinatorics?
QUESTION [0 upvotes]: A term in an NBA team's lease agreement states that it must pay a penalty fee every time it fails to reach $17,000$ attendees in three consecutive home games.
The onerous fee has caused the team to request assistance in assessing its risk. The team estimates the probability it will reach $17,000$ attendees in a home game is $75$%, and each game has an equal likelihood of eclipsing the $17,000$ threshold.
Given the $75\ \%$ likelihood of hitting that number each game, what are the odds that the team will avoid a three-game streak with attendance under $17,000$ for the entire season ?.
Please build a simple simulator to determine your answer ( note that there are $41$ regular season home games; assume each game is independent ).

How do I approach this problem and simulate, perhaps with python script ?.

REPLY [0 votes]: Let an attendance over (or equal to) 17000 be a success and anything fewer a failure.
Let $a_n$ be the probability that after n games so far there hasn't been a run of 3 consecutive failures and that the nth game was a success. Similarly let $b_n$ be the probability that after n games so far there hasn't been a run of 3 consecutive failures, the nth game was a failure but the (n-1)th was a success and let $c_n$ be the probability that after n games so far there hasn't been a run of 3 consecutive failures, the nth game was a failure and the (n-1)th was also a failure.
This gives us the recurrance relationship:
$$a_n = \frac{3}{4}(a_{n-1} + b_{n-1} + c_{n-1})\\b_n = \frac{1}{4}a_{n-1}\\c_n = \frac{1}{4}b_{n-1}\\(a_2=\frac{3}{4}, b_2=\frac{3}{16}, c_2=\frac{1}{16})$$
Then the probability of going the entire season (41 games) without three consecutive failures is $a_{41} + b_{41} + c_{41}$
Then using the following python code I got the probability to be approximately 0.61866:
a = 3/4
b = 3/16
c = 1/16
for x in range(39):
    a, b, c = 3/4*(a+b+c), 1/4*(a), 1/4*(b)
print(a+b+c)


TITLE: Find the conectives of a composed proposition given a a true table
QUESTION [3 upvotes]: How can I construct a proposition P  with the values given in the table using p,q, and r?
\begin{array}{|c3:c|}\hline
p & q & r & P \\\hline
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 1 \\
1 & 0 & 0 & 1\\
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 0 \\
1  & 1 & 1 & 1 \\ \hline\end{array}
My fist idea wasto fill each row and given those values , keep  filling the values of the following rows but I always got some contradiction, I have no idea how to solve this :/ Thanks if anyone can help me with another idea to solve the problem

REPLY [1 votes]: You can also solve this mechanically line by line. For the first line consider formula $\neg p\wedge \neg q \wedge\neg r$ call it $P_1$. Obivously $P_1$ is true exactly when you use the evaluation $0,0,0$. Similarly you construct a formula $P_i$ for every line with value $1$. The resulting formula is then their disjunction $\bigvee P_i$. This is actually a way how to transform a formula into disjunctive normal form.


TITLE: A⋂C ⊆ B and a∈C⇒a∉A\B Prove it
QUESTION [1 upvotes]: A⋂C ⊆ B and a∈C⇒a∉A\B . We can see with Venn diagram more easily. But if I want to prove it , How Can I prove it.
I tried but I dont know how to start.

REPLY [0 votes]: I will give a proof that is probably not the most efficient, but good for understanding.
Firstly, suppose $A\cap B\neq \emptyset$, otherwise it is trivial. A good way to visualize what is happening is to use the fact $$C=C\cap (B\cup B^c)=(C\cap B)\cup (C\cap B^c).$$
It reads $a$ is in $C\cap B$ or in $C\cap B^c$.

If $a\in C\cap B$, then $(C\cap B)\cap(A\cap B^c)=C\cap A\cap (B\cap B^c)=\emptyset$. It means $a\notin A\cap B^c$. Indeed, $a$ cannot be in $B$ and in $B^c$ at the same time.
If $a\in C\cap B^c$, then $(C\cap B^c)\cap(A\cap B^c)=C\cap A\cap B^c\subseteq B\cap B^c=\emptyset$.

We conclude that regardless the case,$a\notin (A\cap B)$.
I hope it gives you conditions to visualize the problem in Venn's Diagram now.


TITLE: Subset $A\in\mathcal{B}(\mathbb{R})$ which contains half of every bounded interval
QUESTION [1 upvotes]: Let $A\in\mathcal{B}(\mathbb{R})$, be a set satisfying that $\forall I\in\ \mathcal{I}_\mathbb{R}\colon \lambda(A\cap I)=\lambda(I)/2$. Can such a set $A$ really exist? The answer is no, see this question. I want to find a proof without using the Lebesgue density theorem.
Here is my attempt: Let $A\subset \mathbb{R}^d$ be open. I can create an open subset $B\subset A$ with measure $\lambda(B)\in (0,\lambda(A))$ such that $B$ is dense in $A$.
This can be done by enumerating $\mathbb{Q}\cap A$ by $\{q_n\}_{n\in\mathbb{N}}$ and setting
$B_\epsilon = \bigcup_{n\in\mathbb{N}}(q_n-\epsilon 2^{-(n+1)},q_n+\epsilon 2^{-(n+1)})$
This set is dense in $A$ and since $\lambda(B)\le \epsilon$ by the continuity in $\epsilon$ I can give it any measure $\lambda(B)=\epsilon$ I want.
Now here is the part, where I somehow implicitly have to use the density theorem:
Since $B$ is dense in $A$ we have that $\lambda(A\cap B_\epsilon) = \lambda(B_\epsilon)=\epsilon$ except on the points on the boundary of $B_\epsilon$, we would then have $\lambda(A\cap B_\epsilon)=0$
which is in some way the density theorem. Then I would obtain a contradiction by assuming that such a set $I$ exists (The density would then be $1/2$ for all $B_\epsilon$).
Is this correct so far?

REPLY [2 votes]: Assume it exists and assume that has finite positive measure.
(If has zero measure then it is obvious that is not true)
Then $\forall \epsilon>0 $,exists a covering    $\{I_n:n \in \Bbb{N}\}$ of $A$ such that $\sum_nm(I_n) \leq m(A)+\epsilon$
Thus $A=\bigcup_n(A \cap I_n)$ and $$m(A) \leq \sum_nm(A \cap I_n)=\sum_n\frac{m(I_n)}{2} \leq \frac{m(A)+\epsilon}{2}$$
Choose $\epsilon =\frac{m(A)}{4}$ and you have a contradiction.
if $A$ has infinite measure and we assume that has the property then  set $A_m=A \cap[-m,m]$ will have the same property so we can restrict ourselves to every subinterval of $[-m,m]$ so by the previous case you can arrive to a contradiction again


TITLE: Equivalent alternative to delta-epsilon formulation of limit?
QUESTION [1 upvotes]: Is $\lim_{n \to \infty} x_n = L$ same as:

$\forall n \in \mathbb N,
     \exists \ \varepsilon_n > 0 $ so that:  $$  \ |x_n - L| \leq
     \varepsilon_n$$ and $$\varepsilon_n \to
     0, \ \text{when } n \to \infty$$

I know the second assumes and uses the definition of limit again, but my question is if they are equivalent.
Thanks.

REPLY [1 votes]: You can take $\epsilon_n = 2|x_n-L|$, since you assume $\lim_{n \to \infty} x_n = L$ for every $\mu > 0$ there exists $N>0$ such that $$|x_n-L|< \mu / 2 \qquad \forall n \geq N $$ and so we
  have
$$|\epsilon_n| = 2|x_n-L|< \mu \qquad \forall n \geq N$$
this shows that $\lim_{n \to \infty} \epsilon_n = 0$.

REPLY [1 votes]: The better way to formulate something like your description is to first rewrite the definition of the limit as
$$(\exists \delta : \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}_+)(\forall \varepsilon > 0,x \in \mathbb{R}) \, |x-c|<\delta(\varepsilon,x) \Rightarrow |f(x)-L|<\varepsilon.$$
The function whose existence is implied by this statement is called the modulus of convergence. Now suppose $\delta_1$ is a modulus of convergence and $\delta_2 \leq \delta_1$. Then $\delta_2$ is also a modulus of convergence.
Therefore we can assume that the modulus of convergence is strictly increasing and continuous, by taking an arbitrary modulus of convergence and decreasing it. This means the modulus of convergence has a continuous inverse, which I will call $\varepsilon : \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}_+$.
What we have said is just:
$$(\exists \varepsilon : \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}_+)(\forall \delta > 0,x \in \mathbb{R}) \, |x-c|<\delta \Rightarrow |f(x)-L|<\varepsilon(\delta,x).$$
But the original statement also implies that
$$(\forall x \in \mathbb{R}) \, \lim_{\delta \to 0^+} \varepsilon(\delta,x) = 0$$
because $\varepsilon(\delta,x)$ is the inverse of $\delta(\varepsilon,x)$ which has the same property.
Everything here should go through just fine in the opposite direction: if we assume $\lim_{\delta \to 0^+} \varepsilon(\delta,x) = 0$ then we can invert it (by increasing it if necessary) to get a function $\delta$ defined for arbitrarily small $\varepsilon$, which is what is required for convergence.


TITLE: Why pressure standing waves have different nodes location with respect to the corresponding displacement waves?
QUESTION [2 upvotes]: In acoustics the pressure wave has a $\pi/2$ phase difference with the displacement wave. But I do not understand how this leads to a different position of nodes of pressure inside a tube with respects to the nodes of displacement.
In interference, what is important is the phase difference so if I have two waves interfering creating a standing wave, if both of the pressure waves have a $\pi/2$ phase difference with respect to the corrisponding displacemente waves, then the phase difference between the two pressure waves would be the same of the one between the two displacement waves.
So apparently there is no difference in the interference of pressure or displacemente waves, hence I do not see the reason of the different location of nodes, for istances in the followin situation
Consider two speakers at the same frequency one in front of the other that interfere creating a standing wave: the nodes of pressure would be located exactly where the displacement nodes are, because the phase difference is zero at the midpoint between the speakers both for displacement and pressure waves, then at space intervals of $\lambda/4$ from midpoint the nodes of pressure (and displacement too) would be located.
So why the nodes of a pressure standing wave should have a different location with respect to the ones of the corresponding displacement standing wave?

To clarify the question : in the picture there is a possible situation described with the two speakers, which emits sound in phase at the same frequency. I'm sure that the displacement waves will have an antinode at the midpoint, but what about the pressure wave. Is $A$ or $B$ the correct diagram of the corresponding pressure wave?

On the one side there is a pase difference of $\pi/2$ with the displacement that makes me think about $B$ (node in the center instead of an antinode).
On the other side if both of the waves coming from the two different speakers have the same phase difference with the corresponding pressure wave, then at the midpoint there should still be an antinode (as for displacement) because the phase difference of pressure waves calculated at midpoint should be $$\Phi_1-\Phi_2=k(\frac{L}{2})-\frac{\pi}{2}-(k(\frac{L}{2})-\frac{\pi}{2})=0$$ Which means constructive interference at the midpoint ($L$ is the distance between the speakers and $k=\frac{2\pi}{\lambda}$).

REPLY [1 votes]: There is a difference between pressure and displacement: pressure is "absolute", displacement has a direction. With speakers on the left and right, pointing to each other, the pressure wave from the right will be the mirror image of the one from the left, the displacement wave is not only mirrored but also changes sign (mirrored by the x-axis). Taking that phase change into account, it should make more sense.


TITLE: Do these infinite expressions have a meaning?
QUESTION [0 upvotes]: While playing with expressions, I came up to the following "infinite sums". I haven't seen them anywhere else but maybe I didn't look long enough. Find the values of $s$ and $t$.
$$s=\sum\nolimits_{1}^{\sum\nolimits_{1}^{\sum\nolimits_{1}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~}1}1}1$$
$$t=\int_{0}^{\int_{0}^{\int_{0}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~} dx} dx} dx\\$$
Does this mean that notation is inadequate at representing values in a nonambiguous way or are these expressions just an example of notation abuse?

REPLY [1 votes]: Without thinking too much of how they would be defined, you are essentially looking for solutions to
$$
s = \sum_{n=1}^s 1
$$
and
$$
t = \int_0^t dx
$$
which both can be any value at all (although they must be integral in the first case). So they are certainly not unique...


TITLE: Prove hypergeometric distribution given Binomial distribution
QUESTION [2 upvotes]: i'm trying to prove b) here:

The answers are proveded, so from a) X+Y distributes Binomial B(40,p)
I'm stuck at proving b).
As I understand, I have to prove that P(X=k) =  $ {25 \choose k} * {15 \choose 10-k}\div {40 \choose 10}$ - mark as *.
Now i'm not sure why, but when I'm looking at $P(X=k)*P(Y=10-k) \div P(X+Y=10)$
i'm getting the correct expression - *.
But I can't explain the logic behind it.
Is my intuition correct? How can I prove it?

REPLY [1 votes]: Start by using Bayes' theorem on $\mathbb{P}(X=k|X+Y=10)$, i.e.
$$\mathbb{P}(X=k|X+Y=10) = \frac{\mathbb{P}(X+Y=10|X=k)\mathbb{P}(X=k)}{\mathbb{P}(X+Y=10)} \; .$$
Now $\mathbb{P}(X+Y=10|X=k)=\mathbb{P}(Y=10-k|X=k)=\mathbb{P}(Y=10-k)$ because $X$ and $Y$ are independent. Thus,
$$\mathbb{P}(X=k|X+Y=10) = \frac{\mathbb{P}(Y=10-k)\mathbb{P}(X=k)}{\mathbb{P}(X+Y=10)} \; .$$
Now, it's just a matter of filling in the distributions for $X$ and $Y$, which you know are binomial.


text:
\chapter{Outline of Proof}
\label{sec:outline-of-paper}
\section{Continuum homogenization with $G=\Z^d$}
\label{sec:continuum-homogenization-with-G-equals-Zd}
As described in the introduction, we will construct a function $t(x,\alpha,\w):\R^d \times A \to \R$ using the edge-weights $\tau(x,\alpha,\w)$. Hence, $t(x,\alpha,\w)$ will only inherit the stationarity of the edge-weights on the lattice; i.e.,
\[
  t(x + z,\cdot,\w) = t(x,\cdot,V_z\w)  \quad \forall~z \in \Z^d.
\]
Therefore, we first observe that
\begin{prop}
  The homogenization theorem (Theorem $3.1$) from~\citet{lions_homogenization_2005} holds with $G=\Z^d$.
  \label{prop:homogenization-theorem-for-discrete-translation-group}
\end{prop}
The proof of~\Propref{prop:homogenization-theorem-for-discrete-translation-group} can be summarized as follows: as the functions $T(x)$ are scaled by $n$, it's as if the lattice is scaled to have size $1/n$. The optimal-control interpretation gives us a uniform in $n$ Lipschitz continuity estimate, and hence the scaled functions $T_n(x)$ do not fluctuate too much on the scaled lattice. Therefore, the discrete subadditive ergodic theorem is enough to prove the homogenization theorem. To flesh out some of the details, we'll identify where exactly the subadditive ergodic theorem is used in~\citet{lions_homogenization_2005}.

The classical approach to proving homogenization is to find a corrector to the cell-problem; i.e., to find a function $v(y)$ satisfying
\begin{equation}
  \begin{split}
    H(p+Dv(y),y) & =  \overline{H}(p), \\
    \lim_{|y|\to\infty} \frac{v(y)}{|y|} & =  0.
  \end{split}
    \label{eq:cell-problem-with-du}
\end{equation}
However, correctors with sublinear growth at infinity do not exist in general~\citep{lions_correctors_2003}. To get around this problem,~\citet{lions_homogenization_2005} consider the equation:
    \begin{equation}
        u_t(x,t) + H(p + Du,x) = 0,  \quad u(x,0) = 0.
        \label{eq:finite-time-horizon-problem-PDE-for-cell-problem}
    \end{equation}
A (sub)solution of this equation can be written in terms of a subadditive quantity, and it follows from the subadditive ergodic theorem that
\begin{prop}
  for any $R > 0$ and $\e > 0$, there is a (random) $t_0$ large enough so that for all $t \geq t_0$
  \[
    \left|\frac{u(x,t)}{t} + \overline{H}(p)\right| \leq \e \quad \forall x \in  B_{Rt}(0) ~\almostsurely.
  \]
  \label{prop:convergence-of-u-to-H-p}
\end{prop}
The main ingredient needed to prove~\Thmref{thm:lions-homogenization-theorem}, are approximate corrector-like functions. One way to construct them is through the stationary equation for the cell-problem:
\[
  \e v_{\e}(x) + H(p+Dv_{\e}(x)) = 0.
\]
\citet{lions_homogenization_2005} quote a Abelian-Tauberian theorem for the variational problems in~\Eqref{eq:finite-time-horizon-variational-problem} and~\Eqref{eq:continuous-stationary-problem}, which says that
\begin{equation}
  \lim_{t \to \infty} \frac{u(x,t)}{t} = \lim_{\e \to 0} \e v_{\e}(x) = - \overline{H}(p).
  \label{eq:abelian-tauberian-theorem-for-u-and-v}
\end{equation}
Hence, all we need to do is to prove~\Propref{prop:convergence-of-u-to-H-p} when $G = \Z^d$, and this will give the approximate corrector-like functions.

The variational formula for the limiting Hamiltonian (\Propref{prop:lions-variational-characterization-of-h-bar}) also holds with $G = \Z^d$ in~\Eqref{eq:lions-S-set-definition-translation-group-R}. This requires a little work, rather than merely being an observation like the homogenization theorem. We don't need the continuum variational formula in this paper. We present it here because we'll take an analogous route to prove the discrete variational formula. The proof closely follows~\citet{lions_homogenization_2005}.

The sketch proof for~\Propref{prop:convergence-of-u-to-H-p} (and hence~\Propref{prop:homogenization-theorem-for-discrete-translation-group}), and the proof of the continuum variational formula are in Appendix~\ref{sec:proof-of-homogenization}.

\section{Embedding the discrete problem in the continuum}
\label{sec:taking-the-discrete-problem-into-the-continuum}
Now that we have the appropriate version of~\Thmref{thm:lions-homogenization-theorem} on $\R^d$ with $G = \Z^d$, we have to take discrete \FPP~into the continuum. In addition to defining a continuum version of~\FPP~---as we've sketched in the introduction--- we also need a discrete cell-problem. We will first ``reverse engineer'' the discrete cell-problem from the continuum Hamiltonian.

For fixed $p \in \R^d$, the shifted Hamiltonian for \FPP~in~\Eqref{eq:H-for-general-continuous-FPP} can be written as
\[
H(p+q,x) = \sup_{\alpha \in A} \frac{-p\cdot \alpha - q\cdot\alpha}{t(x,\alpha)}.
\]
Comparing this formula with the definition of the continuum Hamiltonian in the optimal-control formulation~\Eqref{eq:continuous-hamiltonian-definition}, it follows that we must take the continuous running-cost to be
\begin{equation}
  l(x,\alpha) = \frac{p\cdot\alpha}{t(x,\alpha)}.
  \label{eq:running-costs-for-continuous-cell-problem}
\end{equation}
Since we'll eventually require the continuous \FPP~to mimic discrete \FPP, we'll define $t(x,\alpha) = \tau(x,\alpha)$ along the edges (see~\Secref{chap:proofs-related-to-penalization-theorems}). For such a $t(x,\alpha)$ and $l(x,\alpha)$, let $u(x,t)$ be the finite time-horizon variational problem defined in~\Eqref{eq:finite-time-horizon-variational-problem}. Suppose a path $g_{x,\alpha}$ traverses an edge $(x,x+\alpha)$. Along this edge, we accumulate cost
\[
\int_0^{\tau(x,\alpha)} \frac{p\cdot \alpha}{\tau(x,\alpha)} = p\cdot \alpha.
\]
This indicates that we must consider discrete running costs of the form
  \begin{equation}
    \lambda(x,\alpha) = p \cdot \alpha .
    \label{eq:running-costs-for-discrete-cell-problem}
  \end{equation}

  For the edge-weights $\tau(x,\alpha)$ and the cell-problem running-cost $\lambda(x,\alpha) = p\cdot\alpha$, we'll consider three optimal-control problems: first-passage percolation $\mathcal{T}(x)$, the finite time cell-problem $\mu(x,t)$, and the stationary cell-problem $\nu(x)$. The latter two are defined in~\Eqref{eq:discrete-finite-time-horizon-problem} and~\Eqref{eq:discrete-stationary-control-problem}.

  Now that we've reverse engineered the cell-problem running costs~\Eqref{eq:running-costs-for-discrete-cell-problem}, we turn to constructing continuous approximations of our three discrete optimal-control problems. Using the edge-weights $\tau(x,\alpha)$ and the running-costs $\lambda(x,\alpha)$, we will define (precisely in~\Secref{chap:proofs-related-to-penalization-theorems}) families of functions $t^{\delta}\colon\R^d \times A \to \R$ and $l^{\delta}\colon\R^d \times A \to \R$ parametrized by $\delta$. Let $T^{\delta}(x)$, $u^{\delta}(x,t)$ and $v^{\delta}(x)$ be defined by~\Eqref{eq:continuousFPT-definition},~\Eqref{eq:finite-time-horizon-variational-problem} and~\Eqref{eq:continuous-stationary-problem} (with $\e = 1$) respectively. As $\delta \to 0$, $T^{\delta},~u^{\delta} \AND v^{\delta}$ will approach $\mathcal{T},~\mu \AND \nu$. Hence, we will frequently refer to the continuum problems as ``\deltaapprox s'' to \FPP.

  Next, we define the scaling for the three functions $T^{\delta}(x),u^{\delta}(x,t) \AND v^{\delta}(x)$ and their discrete counterparts $\mathcal{T}(x),~\mu(x) \AND \nu(x)$. The function $\mathcal{T}_n(x)$ has already been defined in~\Eqref{eq:n-scaling-first-passage-time}; $T^{\delta}_n(x)$ is similarly defined in terms of $T^{\delta}(x)$. For the finite-time horizon problems we similarly define
\begin{equation}
\begin{split}
  u^{\delta}_n(x,t) = \frac{u^{\delta}([nx],nt)}{n} ~\AND~ \mu_n(x,t) = \frac{\mu([nx],nt)}{n}.

  \end{split}
  \label{eq:scaling-for-finite-time-horizon-problem}
\end{equation}

For the stationary problem, the scaled versions $v_n^{\delta}(x)$ and $\nu_n(x)$ are obtained by setting $\e = 1/n$ in the variational problems in~\Eqref{eq:continuous-stationary-problem} and~\Eqref{eq:discrete-stationary-control-problem} respectively.

The following interchange of limits (or commutation diagram) is the main ingredient in the proofs of the discrete homogenization theorem and variational formula (\Thmref{thm:-equation-for-mu} and~\Thmref{thm:discrete-variational-formula-for-H-bar}). The theorem compares each of the three sequences of continuum functions
\begin{equation}
  b^{\delta}_n = T_n^{\delta}(x),~u_n^{\delta}(x,t),\OR \nu_n^{\delta}(x)
  \label{eq:b-sequence-of-continuous-problems}
\end{equation}
with the corresponding discrete versions
\begin{equation}
  \beta_n = \mathcal{T}_n(x), \mu_n(x,t),\OR \nu_n(x).
  \label{eq:beta-sequence-of-continuous-problems}
\end{equation}

\begin{theorem}
  For each fixed $x \in \R^d$ and $t \in \R^+$, every pair $(b^{\delta}_n,\beta_n)$ in~\Eqref{eq:b-sequence-of-continuous-problems} and~\Eqref{eq:beta-sequence-of-continuous-problems} homogenizes. That is, $b^{\delta}_n \to \overline{b}^{\delta}$ and $\beta_n \to \overline{b}$. Further,

    \begin{equation*}
      \lim_{\delta \to 0} \lim_{n \to \infty} b^{\delta}_n = \lim_{n \to \infty} \lim_{\delta \to 0} b^{\delta}_n = \lim_{n \to \infty} \beta_n = \overline{b}.
    \end{equation*}

Stated as a commutative diagram, this is

\begin{equation*}
  \begin{CD}
    b^{\delta}_n @>n>> \overline{b}^{\delta} \\
    @V{\delta}VV  @VV{\delta}V \\
    \beta_n          @>>n> \overline{b}
  \end{CD}
\end{equation*}
  \label{thm:commutation-theorem-for-u-and-T}
\end{theorem}

We will prove theorem~\Thmref{thm:commutation-theorem-for-u-and-T} when $(b,\beta) = (u,\mu)$, and when $(b,\beta) = (T,\mathcal{T})$. The proof for $(b,\beta) = (v,\nu)$ is nearly identical, and to avoid repetition of ideas, we will omit it.

Proposition~\ref{prop:convergence-of-u-to-H-p} says that
\[
  \lim_{t \to \infty} u_n^{\delta}(x,t) = -\overline{H}^{\delta}(p).
\]
Let $m^{\delta}$ and $m$ be the time-constants of $T$ and $\mathcal{T}$. The continuum homogenization theorem says that $m^{\delta}$ is the viscosity solution of
\[
  \overline{H}^{\delta}(Dm^{\delta}(x)) = 0, ~m^{\delta}(0) = 0.
\]
\Thmref{thm:commutation-theorem-for-u-and-T} applied to $(u,\mu)$ says that there exists a function $\overline{H}(p)$ such that
\[
  \overline{H}^{\delta}(p)  \to \overline{H}(p).
\]
\Thmref{thm:commutation-theorem-for-u-and-T} applied to $(T,\mathcal{T})$ to says that for each $x$,
\[
  m^{\delta}(x) \to m(x).
\]

We would like to show that $m(x)$ solves the PDE corresponding to $\overline{H}(p)$. For this, a uniform in $\delta$ Lipschitz estimate for both $\overline{H}^{\delta}(p)$ and $m^{\delta}(x)$ is sufficient. We state this as two separate propositions below.
\begin{prop}
  $\overline{H}^{\delta}(p)$ is Lipschitz continuous in $p$ with constant bounded above by $1/a$, where $a$ is defined in~\Eqref{eq:basic-assumption-on-edge-weights}.
  \label{prop:lipschitz-continuity-of-limit-hamiltonian}
\end{prop}

\begin{prop}
  $m^{\delta}(x)$ is Lipschitz in $x$ with constant bounded above by $b$, where $b$ is defined in~\Eqref{eq:basic-assumption-on-edge-weights}.
  \label{prop:lipschitz-continuity-of-time-constant}
\end{prop}

Since $\overline{H}^{\delta}$ and $m^{\delta}$ converge locally uniformly to $\overline{H}$ and $m$ respectively, the standard stability theorem for viscosity solutions~\citep{crandall_users_1992} implies~\Thmref{thm:-equation-for-mu}.

The proofs of the results in this section are in~\Chapref{chap:proofs-related-to-penalization-theorems}.

\section{Discrete variational formula and solution of the limiting PDE}
\label{sec:outline-discrete-variational-formula}
The commutation theorem also transfers the Abelian-Tauberian theorem over from the continuum relating the limits of $u^{\delta}_n(x,t)$ and $v^{\delta}_{\e}(x)$~\Eqref{eq:abelian-tauberian-theorem-for-u-and-v}. That is, both $\mu_n(x,t)$ and $\nu_{\e}(x)$ converge to $\overline{H}(p)$ almost surely as $n \to \infty$ and $\e \to 0$. This is very useful in establishing the discrete version of the variational formula. We have
\begin{prop} for each $R > 0$ and $\eta > 0$, there is a small enough (random) $\e_0$ such that for all $\e \leq \e_0$,
  \[
    \left|\e \nu_{\e}(x) + \overline{H}(p)\right| < \eta \quad \forall x \in B_{\e^{-1}R}(0).
  \]

  \label{prop:statement-of-tauberian-theorem}
\end{prop}

The stationary cell-problem has the discrete (DPP)
\begin{equation}
  \nu_{\e}(x) = \inf_{\alpha \in A} \left( \alpha \cdot p + e^{-\e \tau(x,\alpha)} \nu_{\e}(x + \alpha) \right).
  \label{eq:discrete-dpp-stationary-problem}
\end{equation}
With a little manipulation of the DPP, we can obtain a discrete version of the stationary PDE in~\Eqref{eq:stationary-problem-pde}.
\begin{prop}
  For all $x \in \Z^d$, there is a constant $C > 0$ uniform in $\e$ and $\w$ such that
  \[
  -C \e \leq \e \nu_{\e}(x) + \mathcal{H}(\nu_{\e},p,x) \leq C \e,
  \]
  \label{prop:approximate-discrete-HJB-for-stationary-problem}
  where $\mathcal{H}$ is the discrete Hamiltonian~\Eqref{eq:discrete-hamiltonian-1}.
\end{prop}

The final ingredient for the variational formula is what is usually called a comparison principle for HJB equations. In the continuum, the comparison principle is stated for sub- and supersolutions of the PDE. The discrete comparison principle we prove is a less general version that suffices for our purposes. Consider the discrete problem in~\Eqref{eq:discrete-finite-time-horizon-problem} for any $\phi\colon\Z^d \to \R$ such that $\Norm{\phi}{Lip} < \infty$. Then,
\begin{prop}
  \[
    \mu(x,t) \geq \phi(x) - t \sup_{x \in \Z^d} \mathcal{H}(\phi,p,x) \quad \forall x \in \Z^d \AND t \in \R^+.
  \]
  \label{prop:comparison-principle}
\end{prop}
Using these facts, the discrete variational formula in~\Thmref{thm:discrete-variational-formula-for-H-bar} is easy to prove.

Now that we have a formula for the limiting Hamiltonian, we can solve the PDE in~\Eqref{eq:limiting-HJB} to obtain the time-constant $m(x)$. It follows directly from the variational formula that
\begin{prop}
  $\overline{H}(p)$ is a norm on $\R^d$.
  \label{prop:limiting-hamiltonian-is-a-norm}
\end{prop}

Let $L(x)$ be the dual norm of $\overline{H}(p)$ on $\R^d$. Consider the set of paths
\[
  \mathcal{A} := \left\{g \in C^1([0,\infty),\R^d) : L(g'(s)) = 1 ~\forall~ s \in [0,t)\right\}.
\]
  Let $T(x)$ be the minimum time function defined by~\Eqref{eq:continuousFPT-definition}. From standard optimal-control theory~\citep{bardi_optimal_1997}, it follows that $T(x)$ is the unique viscosity solution of the metric HJB equation~\Eqref{eq:time-constant-satisfies-a-PDE}. A standard Hopf-Lax formula gives
\begin{prop}
  $T(x) = L(x)$.
  \label{prop:Hopf-Lax-formula-for-u}
\end{prop}
The fact that $\overline{H}(p)$ is the dual norm of $m(x)$ follows immediately, and \Corref{cor:H-is-the-dual-norm-of-time-constant} is proved.

\begin{remark}
  It is natural to question the necessity of taking the discrete problem into the continuum; the PDE is unnecessary once the Hamiltonian has been identified as the dual norm of the time-constant\footnote{observation due to S.R.S Varadhan.}.

  There is a direct proof based on the continuum homogenization theorem of~\citet{kosygina_stochastic_2006}. Such a route has been taken to prove variational formulas for the large deviations of random walks in random environments by~\citet{rosenbluth_quenched_2008}. This work was considerably generalized by~\citet{rassoul-agha_quenched_2013} and \citet{rassoul-agha_quenched_2014}. Subsequently,~\citet{georgiou_variational_2013} extended these ideas to prove variational formulas for the directed random polymer, and for last-passage percolation.

\end{remark}

\begin{remark}
  We chose the metric-form of the HJB equation so that the limiting Hamiltonian $\overline{H}(p)$ could be interpreted as a norm. This allowed us to solve the PDE for the time-constant. We can make the assumption on the edge-weights $\tau(x,\alpha,\w)$ in~\Eqref{eq:basic-assumption-on-edge-weights} less restrictive if the Hamiltonian is written in the more standard form
  \[
    \mathcal{H}(p,x) = \sup_{\alpha \in A} \left\{-p\cdot\alpha - \tau(x,\alpha) \right\}.
  \]
  Here, $\tau(x,\alpha,\w)$ can take the value $0$ without making the Hamiltonian blow-up. However, even though the homogenization theorem and variational formula are still valid, we do not know the Hopf-Lax formula for the limiting PDE.
\end{remark}

The proofs of the results in this section can be found in~\Chapref{chap:proofs-related-to-discrete-variational-formula}.

config: arxiv


TITLE: Is there absolute proof for gravitational waves?
QUESTION [7 upvotes]: As you probably know gravitational waves seem to have been proven unless you've been living under a rock for the past day. The experiment involves the reflecting off mirrors and waves of light and it is best you look because it is a little bit complicated but not too difficult. The effects are very subtle, and I am wondering if they could be a result of any Earth movements or activity above the Earth's surface. However, it seems most likely that if it is not gravitational waves, it is Earth movements because the experiment took place deep underground. Is there any proof that this is not a result of Earth movements and the only possible explanation is gravitational waves?
I don't want to be the stick in the mud here and spoil the party. However, I do like to know that what we are seeing is the truth as the effects require highly advanced equipment to be observed.

After reading all the great responses, the proof that gravitational waves do exist appears rock solid, and it seems Einstein was right again.

REPLY [0 votes]: There is no absolute proof.  I think we can be confident that a speed-of-light pulse from the direction of southern hemisphere hit the Earth.  However, as I noted on another post (and got a lot of heat and dinged), there is only one laboratory doing the measurements, and "science" requires more than one laboratory to provide proof, even that a pulse hit.  The Livingston-Hanford detectors together make up one laboratory.  One without the other is useless.  That's why two were built simultaneously.
What may be true is that other pulses have been detected that have not been announced.  Let's wait to be true believers.  Maybe they were gamma ray bursts.  It may only take weeks.  No hurry, no worry.


TITLE: graph on mobius strip is nonplanar?
QUESTION [0 upvotes]: Consider the closed curve on the mobius strip that covers both sides of the fundamental polygon:
https://thumb9.shutterstock.com/display_pic_with_logo/1663882/462955276/stock-vector-blue-moebius-strip-or-moebius-band-with-centerline-surface-with-only-one-side-and-one-boundary-462955276.jpg
Is this curve viewed as a graph (with one vertex?) non planar since it must intersect itself in the plane? (help me make this more rigorous)
How do I more rigorously prove that certain graph in the mobius strip is non planar? Do I need to use some planarity criterion like Kuratowski's?

REPLY [1 votes]: I think you are confusing some concepts.
A graph can be crossing free in some embeddings and not in others - think of a circle embedded as figure 8. A graph is planar if it has a crossing free embedding.
The bounding curve of your Mobius strip is just a circle. What you've noticed is that any plane projection of that strip will lead to an embedding of that circle that has crossings. But the circle is still a planar graph (when "viewed as a graph").


TITLE: how to compute this definite integral?
QUESTION [2 upvotes]: it is the second moment of logistic distribution, i.e.
$$
\int_{-\infty}^{\infty}\frac{x^2 e^x }{(1+e^x)^2}dx
$$
I've been struggling with it for hours. but i fail.
thanks first!

REPLY [4 votes]: $$\int_{-\infty}^{\infty} \frac{x^{2} e^{x}}{(1+e^{x})^{2}} \ dx = 2 \int_{0}^{\infty} \frac{x^{2} e^{x}}{(1+e^{x})^{2}} = 2 \int_{0}^{\infty} x^{2} \sum_{n=1}^{\infty} (-1)^{n+1} n e^{-nx} \ dx $$
$$ = 2 \sum_{n=1}^{\infty} (-1)^{n+1} n \int_{0}^{\infty} x^{2} e^{-nx} \ dx = 2 \sum_{n=1}^{\infty} (-1)^{n+1} n \Big(\frac{2}{n^{3}} \Big) $$
$$ = 4 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}} =4 \Big(\frac{\pi^{2}}{12} \Big) = \frac{\pi^{2}}{3}$$


TITLE: Smooth projective toric varieties which are quotients of product of spheres and torii by a free torus action?
QUESTION [4 upvotes]: My question is just as in the box. Is every smooth projective toric variety diffeomorphic to a quotient of $\prod_i S^{n_i} \times T^k$ (I know torus is a one-sphere but I just wanted to make clear I allow this) by a free torus action? If not which ones can be realized like this?  Maybe it can be proven using the Geometric Invariant theory construction. Namely, when is the fiber of the corresponding torus action on $\mathbb{C}^l$ diffeomorphic to the one of the above spaces?

REPLY [5 votes]: Let us consider the case of toric varieties of real dimension $4$ and prove they
can not be represented as such a quotient unless they have second Betti
number $1$ or $2$.
Proof.
Let us introduce some notations. Let $B$ be the toric manifold of real dimension $4$, $n=b_2(B)$. Denote by $E$ the product $\Pi_i S^{n_i}\times T^k$. Let $k_2=b_2(E)$, $k_3=b_3(E)$. Finally, denote by $m$ the dimension of the torus that is acting on $E$.
Suppose that $T^m$ is acting freely on $E$ and $B=E/T^m$.
Then we have the following obvious relation on dimensions:
$$4=dim(E)-m\ge k+2k_2+3k_3-m$$
We will explain now that we must have $n\le 2$ (recall $n=b_2(B)$).
For this purpose we will consider the long exact sequence of homotopy groups
for the fibration $E\to B$.
$$0\to \pi_3(E)\to \pi_3(B)\to 0 \to \pi_2(E)\to \pi_2(B)\to \mathbb Z^m\to \mathbb Z^k\to 0$$
Since $\pi_2(B)=\mathbb Z^n$, from the second half of the sequence we get
$$-k_2+n-m+k=0$$
substituting this in the inequality on dimensions we get
$$4\ge 3k_2+3k_3-n$$
To prove finally that $n\le 2$ we use the classical statement that $\pi_3(B)$ contains sub-group $\mathbb Z^{(n^2+n)/2-1}$. It follows that
$$rk (\pi_3(E))=k_3+k_2\ge (n^2+n)/2-1$$
so
$$4\ge 3((n^2+n)/2-1)-n$$


TITLE: On the indecomposable decomposition of the reduction of an integral representation
QUESTION [7 upvotes]: Here is a problem I have been grappling with all day. I started out thinking it might be true but am now inclined to believe it is false, and would like to see a counterexample.
Suppose $G$ is a finite group and $p$ a rational prime, such that $G$ has a unique (normal) sylow $p$-subgroup $P.$ If you want, assume that $G/P$ is cyclic of prime power order.
Let $R$ be either the ring $\mathbb{Z}$ or $\mathbb{Z}_{(p)}.$ By an $R$-representation of $G$ I mean a finitely generated $R[G]$-module which is free as an $R$-module.
Suppose $V$ is such an $R$-representation. Let $\bar{\mathbb{F}_p}$ denote the algebraic closure of the finite field with $p$ elements. By the Krull-Schmidt theorem, the $\bar{\mathbb{F}_p}[G]$-module $V \otimes_R \bar{\mathbb{F}_p}$ can be written as a sum of indecomposables, uniquely so up to reordering. Now in general $\bar{\mathbb{F}_p}[G]$ admits many indecomposable modules, but a particularly nice class is obtained by decomposing $\bar{\mathbb{F}_p}[G/P]$ into one-dimensional constituents (which works because $|G/P|$ is coprime to $p$), and indeed these are precisely the one-dimensional indecomposable $\bar{\mathbb{F}_p}[G]$-modules.
Let me write $\chi(V)$ for the tensor products of all the one-dimensional indecomposables occurring in $V \otimes_R \bar{\mathbb{F}_p}$. I shall say $V$ is good if $\chi(V)^{\otimes 2}$ is the trivial one-dimensional module.
Question: Suppose $V$ is a good $R$-representation and $V = W \oplus W'$ (decomposition as $R$-representations). Does it follow that $W$ is good?
Why one might think this is true: the point is that the representation is integral, so its ordinary character is real. In ordinary representation theory, the number of times a constituent occurs is determined by taking the inner product of characters; it follows that (over $\mathbb{C}$) every one-dimensional constituent with a non-real character occurs precisely as many times as its inverse. Thus multiplying together the one-dimensionals only leaves a real character, which has order two. That is to say, if we do decomposition over $\mathbb{C}$ instead of $\bar{\mathbb{F}_p},$ every $R$-representation is good.
Why one might think this is false: I offer two explanations. For one, the crux of the above argument is the complex conjugation action on a suitable ring of integers, which can be used to identify dual representations. However, this action does not (in general) descend to the finite field. Indeed it may happen that $\mathbb{F}_p$ already contains enough roots of unity to split $\mathbb{F}_p[G/P],$ and so it does not seem possible to use Galois actions like this.
A more direct reason why this proof breaks down is as follows. Instead of trying to decompose over $\bar{\mathbb{F}_p}$ one might try to decompose over the completion $\mathbb{Z}_p$ (this ring of characteristic zero also has Krull-Schmidt). But the problem then is that there can be indecomposable $\mathbb{Z}_p$-modules of dimension $\ge 2$ which have the simple one-dimensional modules as decomposition factors. (And somehow the decomposition factors are the only thing I know how to control via character methods.) Also again note that $\mathbb{Q}_p$ may already contain all the roots of unity we need, so it does not seem possible to use Galois actions.

REPLY [3 votes]: For any finite group $G$ and finite-dimensional $\mathbb{F}_p[G]$-module $M$, I'll denote by $\Omega M$ the kernel of an $\mathbb{F}_p[G]$-projective cover $P(M)\to M$.
Let $F(M)$ be a free $\mathbb{Z}[G]$-module with a surjection $F(M)\to M$, and complete to a short exact sequence
$$0\to K(M)\to F(M)\to M\to0$$
of $\mathbb{Z}[G]$-modules. Then $K(M)$ is $\mathbb{Z}$-free.
Since $pK(M)\leq pF(M)\leq K(M)$, we get a short exact sequence
$$0\to pF(M)/pK(M)\to K(M)/pK(M)\to K(M)/pF(M)\to 0$$
of $\mathbb{F}_p[G]$-modules, where it's easy to check that $pF(M)/pK(M)\cong M$ and $K(M)/pF(M)$, since it's the kernel of a surjection $F(M)/pF(M)\to M$ from a projective $\mathbb{F}_p[G]$-module to $M$, is isomorphic to the direct sum $\Omega M\oplus P$ for some projective $\mathbb{F}_p[G]$-module $P$.
This sequence must certainly split if $\operatorname{Ext}^1_{\mathbb{F}_p[G]}(\Omega M, M)=0$ (or equivalently $\operatorname{Ext}^2_{\mathbb{F}_p[G]}(M,M)=0$), and so in this case $K(M)$ is a $\mathbb{Z}$-free $\mathbb{Z}[G]$-module whose reduction mod $p$ is $M\oplus\Omega M\oplus P$.
Alternatively, it's not hard to see that whether the sequence splits is independent of the choice of $F(M)$, and depends only on the restriction of $M$ to a Sylow $p$-subgroup of $G$. Also, if $M$ lifts to a $\mathbb{Z}$-free $\mathbb{Z}[G]$-module $\tilde{M}$ then one can take $F(M)$ to be a free module that maps onto $\tilde{M}$, and if $\tilde{K}(M)$ is the kernel of $F(M)\to\tilde{M}$ then $K(M)/pK(M)=pF(M)/pK(M)\oplus \tilde{K}(M)/p\tilde{K}(M)$, so there is an explicit splitting. So if the restriction of $M$ to a Sylow $p$-subgroup lifts (which will always be the case if $G$ has a normal Sylow $p$-subgroup and $M$ is simple) then the sequence splits.
Now take, for example, $G=C_7\rtimes C_3$, and $p=7$, and let $M$ be any of the three one-dimensional $\mathbb{F}_p[G]$-modules. Then all of the above applies, and so there is a $\mathbb{Z}$-free $\mathbb{Z}[G]$-module $V$ whose reduction mod $p$ has only one one-dimensional summand, which is isomorphic to $M$ (since $\Omega M$ is indecomposable and $6$-dimensional, and the indecomposable projectives are $7$-dimensional). By taking direct sums of such modules, you can construct a $V$ whose reduction mod $p$ has an arbitrary collection of one-dimensional summands, which shows that the answer to your question is "no".