New answers tagged

0

In my answer I will give a proof of: $X_{n}\stackrel{P}{\to}X$ if every subsequence $\left(X_{n_{k}}\right)$ of $\left(X_{n}\right)$ contains a further subsequence $\left(X_{n_{k_{i}}}\right)$ with $X_{n_{k_{i}}}\stackrel{P}{\to}X$. This is a stronger statement because $X_{n_{k_{i}}}\stackrel{\text{a.s}}{\to}X$ implies ...


0

This follows from a generalized dominated convergence theorem. Let $\zeta_n:=X_n^k$ and $\zeta:=X^k$. Since $\mathbb{E}|\zeta_n|\le c_k[\mathbb{E}|X_n-X|^k+\mathbb{E}|\zeta|]$ and $k\le p$, $\zeta_n\in L^1$ ($\zeta\in L^1$ by assumption). In addition, you've already shown that $\mathbb{E}|\zeta_n|\to\mathbb{E}|\zeta|$. Thus, for any subsequence $n_m$, there ...


0

Some hints: (a) If $X_n\to 0$ in probability, what should be $\lim_{n\to \infty}\mathbb P\{X_n\gt 1/2\}$? What is it in this case? (b) Notice that since $X_n$ takes the values $0$ and $1$, $X_n(\omega)\to 0$ means that $X_n(\omega)=0$ for $n$ large enough (depending on $\omega$). The key word to find the necessary and sufficient condition is the ...


1

Yes. $A$ and $A^\top$ have the same eigenvalues, and $A^\top \cdot \vec 1 =\vec 1$, because all row sums of $A^\top$ are equal to $1$.


1

The condition $E(XY)=E(X)E(Y)$ is equivalent to the condition that $X$ and $Y$ are uncorrelated. It is true that if $X$ and $Y$ are independent random variables, then they are uncorrelated. You are asking whether the converse is true, i.e., whether variables that are uncorrelated are necessarily independent. The answer is no. There are many ...


0

Short answer: $P(X=Y)=P(X=0,Y=0)+P(X=1,Y=1).$


1

$X$ lies in the ball around $0$ with radius $c$ and so does the expectation of $X$ by Jensen's inequality, since the norm is a convex function. The distance between $X$ and $\mathbb{E}[X]$ is therefore bounded almost surely by $2c$. This upper bound is tight. To see this, take your probability space to be $[0,1]$ with the uniform distribution and define the ...


1

$P(X=Y)$ means the probability that $X$ is equal to $Y$. Hint: there are four possible pairs $(x,y)$ to consider.


1

The theorem is of the form If $X,Y$ independent, then for all nonnegative functions we have an equality you ask whether this implies about If we have the equality for a specific function (which is not even non-negative), then $X,Y$ independent The fact that the identity is not non-negative is only one of at least three points that prevent this ...


-1

No, consider $X = Y$ with both random variables being a constant $1$ over $(0,1)$ and zero elsewhere. Then, $$ \mathbb{E}[XY] = \mathbb{E}[X^2] = 1 = \mathbb{E}[X]^2 $$ but clearly since $X=Y$ they are not independent.


0

HINT probability to never return to 0 starting at 0 is to always move to the right, i.e. $$\lim_{n \to \infty} (1/2)^n = 0$$


1

If $Y_n^k=\sum_{i=1}^k X_n^i>\epsilon$, one of those summands must be greater than $\epsilon/k$. So $P\{Y_n^k>\epsilon\}\le P\{\cup_{i=1}^k \{X_n^i>\epsilon/k\}\}\le \sum_{i=1}^k P\{X_n^i>\epsilon/k\} \to 0$ as $n\to\infty$.


1

Any rotationally invariant $\alpha$-stable Lévy process can be written as a subordinate process. More precisely, if $(B_t)_{t \geq 0}$ is a ($d$-dimensional) Brownian motion and $(S_t)_{t \geq 0}$ an $\frac{\alpha}{2}$-stable subordinator such that $(B_t)_{t \geq 0}$ and $(S_t)_{t \geq 0}$ are independent, then $$X_t := B_{S_t}, \qquad t \geq 0,$$ is a ...


4

$S_{n}=n\bar{X}_{n}$ where $\bar{X}_{n}:=\frac{1}{n}\left(X_{1}+\cdots+X_{n}\right)$ and $P\left(\bar{X}_{n}\to\mu\right)=1$ according to the strong law of large numbers. If $\bar{X}_{n}\left(\omega\right)\to\mu>0$ then $S_n(\omega)=n\bar{X}_{n}\left(\omega\right)\rightarrow+\infty$. So: $$\left\{ \bar{X}_{n}\to\mu\right\} \subseteq\left\{ ...


2

Let $a = m -1$, $b = n - m - 1$, $d = a + b = n - 2$. $\theta_{ij}$, $1 \le i \ne j \le n$ be the angular separation between point $x_i$ and $x_j$. $\ell_m$ be the expected angular separation among a pair of $m^{th}$ nearest neighbors. $\mathcal{E}_m$ be the event that $x_2$ is the $m^{th}$ nearest neighbor of $x_1$. $\mathbf{1}_m$ be the indicator ...


2

Note that for each fixed $n$, $X_i^n$ are iid. $$E(X_1^n)=\sum_{k:2^k\leq n\log n}2^k\dfrac{1}{2^k}\sim\log_2 n+\log_2\log n$$ therefore $$E(\dfrac{S_n'}{n\log_2n})\sim\dfrac{n\log_2 n+n\log_2\log n}{n\log_2 n}\to1$$ Finally, $$Var(S_n')=nVar(X_1^n)$$ due to iid-ness and $$Var(X_1^n)\leq E(X_1^n)^2=\sum_{k:2^k\leq n\log ...


1

Let $\mu^d$ be the restriction of a measure $\mu\in\mathcal{P}(\Omega)$ to its atoms ($\mu^d(A)$ is the maximum of $\mu(A\cap S)$ over countable $S\subseteq\Omega$). Then a counterexample is given by $$ \phi(\mu)=\mu^d(\Omega)=\max\left\{\mu(S)\colon S\subseteq\Omega\textrm{ is countable}\right\}. $$


3

Basically no, because this is sensitive to the distribution of $X_1$, which must be given initially. However, typically you could do it if you instead had $X_N,X_{N+1},\dots,X_{N+M}$, where $N$ and $M$ are both large. Then under generic (but not universal) circumstances, there is a unique invariant distribution, the chain converges to this distribution, and ...


3

Partial answer: Simple cases: If we generate only two points, this is like fixing one as north pole and generating the other randomly. Here, the expected angular distance is "clearly" $\frac\pi2$ because points at angular distances $\theta$ are just as likely as points at distance $\pi-\theta$. The same argument holds for the $m$th closest neighbour when ...


2

The answer is yes. As a concrete example of (1), suppose $X_1, X_2, X_3$ are exchangeable. We show $X_1, X_3$ are exchangeable by filling in the omitted slots with the 'everything' event: $$\begin{align} P(X_1\in A, X_3\in B)&=P(X_1\in A, X_2\in{\mathbb R}, X_3\in B)\\ &\stackrel{(*)}=P(X_3\in A, X_2\in {\mathbb R}, X_1\in B)\\ &=P(X_3\in A, ...


4

This is a very valid question, and one that I didn't really touch until my second time learning undergraduate probability. I am by no means an expert; this is merely the intuition I've developed. $X$ is a function. What they don't tell you in undergraduate-level probability is that $$X: \Omega \to \mathbb{R}$$ is a "random variable," where $\Omega$ is the ...


4

Change variables $s=tu$: $$\int_0^t e^{B_s} \, ds = \int_0^1 e^{B_{t u} }t \, du=(*)$$ Use scaling: $(B_{tu}:u \ge 0) \overset{\mbox{dist}}{=}\sqrt{t}( B_{u}:i \ge 0)$. Therefore $$ (*) \overset{\text{dist}}{=} \sqrt{t} \int_0^t e^{\sqrt{t} B_u} \, du $$ Take $\sqrt{t}$-th root: $$t^{1/\sqrt{t}} \left(\int_0^1 e^{\sqrt{t} B_u} \, ...


1

Indicating the dependence of $T$ on $c$ explicitly, you have $$\{T_c=\infty\}=\{\sup_n|M_n|\le c\}$$ Therefore, $$\bigcup_{c>0,c\in\Bbb Q}\{T_c=\infty\}=\{\sup_n|M_n|<\infty\}$$


0

Modulo the following claim (which is true for low dimensions, and is quite plausible, but I didn't find a proof), it can be done. Claim: A subspace generated by $k$ linearly independent $(\pm1)$ vectors contains $2^k$ $(\pm1)$ vectors. First note that this implies that if $a$ is orthogonal to a $k$-codimensional subspace generated by $(\pm1)$ vectors, ...


1

Similar to what @user1952009 was saying, we can fix a positive integer M and then define the finite linear transformation: $$(Y_t^M,\dots,Y_{t+h}^M) = (\sum_{|j|\le M}\psi_jX_{t-j},\dots, \sum_{|j|\le M}\psi_jX_{t+h-j})$$ Here we can without problem make the argument: $$(\sum_{|j|\le M}\psi_jX_{t-j},\dots, \sum_{|j|\le M}\psi_jX_{t+h-j}) \stackrel{d}{=} ...


0

I completely overthought this problem, thank you Did for pointing me in the right direction. Actually, $\begin{align}X_n\xrightarrow{\mathrm{a.s.}} 0 &\iff P(\{w, X_n(w)\to 0\}) = 1 \\ &\iff P(\{w,\exists n, \forall k\geq n,X_n(w)=0\})=1 \text { because } X_n(w) \text{ is integer-valued} \\ &\iff P(\operatorname{liminf X_n=0)} =1 \\ &\iff ...


1

We have that for all $n>1$ , $\mathsf P(Y_n) ~=~ \mathsf P(Y_{n-1})\,\mathsf P(Y_n\mid Y_{n-1})+\mathsf P(Y_{n-1}^\complement)\,\mathsf P(Y_n\mid Y_{n-1}^\complement) \\ \qquad ~~~ = ~ 0.9\,\mathsf P(Y_{n-1})+0.3\,(1-\mathsf P(Y_{n-1}))$ Now in the long term, whatever the initial state, we will approach an equilibrium state. ...


1

Here is a good link for books to learn on the probability theory. Some are more in depth while others meant for more casual reading. There are also a bunch of puzzles on the blog along with some that have code on them, that help with understanding the concepts.


3

You meant $\{X_n\}$ are uniformly integrable if $$ \lim_{M\to\infty} \sup_n E[ |X_n|\chi_{\{|X_n|>M\}}]=0$$ (e.g. https://en.wikipedia.org/wiki/Uniform_integrability) Now for your question. Answer is NO. Why ? In general the random variables are not in $L^p$. Easy example to remember $X_n = X$, where $X$ is a random variable in $L^1$ but not in ...


1

Well one place we could start would be to note that we could define 𝔾 on a larger set; namely any real sequence of coefficients for which the generating series is absolutely convergent or even Abel summable (namely the limit of the series exists as s→1 even if the series diverges when one substitutes s=1). If we are willing to consider quasiprobability ...


0

Here is the idea (you need to fill in some details) $R_j = o(|\Delta t_j|^2 + |\Delta X_j|^2)$ means $$R_j \le r(|\Delta t_j|^2 + |\Delta X_j|^2)(|\Delta t_j|^2 + |\Delta X_j|^2)$$, where $r(y)\to 0 \ as \ y\to 0$ (using the condition that $g$ is twice continuously differentiable, you can show such $r$ exists and does not depends on $j$). Now $$|\sum_j R_j| ...


0

If $X$ and $Y$ have a joint density $f_{X,Y}(x,y)$, then $$ \mathbb P(X \in E) = \int_{E} dx\; \int_{\mathcal Y} dy\; f_{X,Y}(x,y)$$ More generally, their joint distribution is a probability measure $\mu_{X,Y}$ on $\mathcal X \times \mathcal Y$, and $\mathbb P(X \in E) = \mu_{X,Y}(E \times \mathcal Y) = \int_{E \times \mathcal Y} d\mu_{X,Y}$.


0

It depends what you mean by $P (Y\in (y,y+dy))$. If you mean the actual probability that $Y$ falls between $y$ and $dy$, then the extra factor of $dy$ would be unnecessary, but this would be practically impossible to calculate. If however you mean the pdf of $Y$ for a given value of $y$, then you would need the extra $dy$ at the end.


1

The density $g$ of the ratio $Z=X/Y$ is given by the following integral: $$\int_{\mathbb R} |y|f(zy,y)\ \mathsf dy.\tag1 $$ Here $f(x,y)=2(x+y)\mathsf 1_{(0,y)}(x)\mathsf 1_{(x,1)}(y)$, so because $0<\frac xy < y$ we have \begin{align} f(zy,y)&=2y(1+z)\mathsf1_{(0,y)}(zy)\mathsf1_{(zy,1)}(y)\\ &=2y(1+z)\mathsf1_{(0,1)}(y)\mathsf1_{(0,1)}(z). ...


0

The dominated convergence theorem holds when the almost convergence is replaced by convergence in probability. Indeed, if $\left(Y_n\right)_{n\geqslant 1}$ to $0$ in probability and $\sup_n\left|Y_n\right|$ is integrable, then for each sub-sequence $\left(Y_{n_k}\right)_{k\geqslant 1}$, we can extract a subsequence $\left(Y_{n'_k}\right)_{k\geqslant 1}$ ...


2

Recall the following two statements: Lemma 1 (subsequence principle): A sequence $(a_n)_{n \in \mathbb{N}}$ converges to a limit $a$ if, and only if, any subsequence of $(a_n)_{n \in \mathbb{N}}$ contains a further subsequence which converges to $a$. Lemma 2: If $Y_n \to Y$ in probability, then there exists a subsequence $(Y_{n(k)})_{k \in ...


1

WARNING: The following argument attempts to make use of conditional expectations which I do not understand entirely rigorously (as @Did can confirm). $P\left[A<Z \left| \right. \mathcal{G} \right]=E\left[1_{\{A<Z\}} \left| \right. \mathcal{G} \right]$ just by definition. Moreover, a regular conditional distribution such that the following makes sense ...


2

No. By your definition, $U_1,U_2$ will be both uniform on the unit interval. Then, by Holder's inequality $E[U_1U_2]\leq (E[U_1^p])^{1/p}(E[U_2^{(1-1/p)^{-1}}])^{1-1/p} = \left( \frac{1}{p+1} \right)^{1/p} \left( \frac{1}{\left(1-1/p \right)^{-1} +1} \right)^{1-1/p}$ for all $p\geq 1 $. Now, plot the upper bound, and see that there exists a $p$ such that ...


0

The random variables $U_i = \Phi(X_i)$ are $U(0,1)$ uniformly distributed since $\Phi$ is one-to-one and $$P(U_i \leqslant x ) = P(\Phi(X_i) \leqslant x ) = P(X_i \leqslant \Phi^{-1}(x)) = x.$$ We have $E(U_i) = 1/2$ and $$cov(U_1,U_2) = E(U_1U_2) - 1/4 = \int_{-\infty}^\infty\int_{-\infty}^\infty\Phi(x_1)\Phi(x_2)n(x_1,x_2,0.5) \, dx_1\, dx_2 - 1/4,$$ ...


3

$$ P\{|X_n-Y_n|>\epsilon\}\le P\{|X_n-X|+|Y_n-Y|+|X-Y|>\epsilon\} \\ \le P\{|X_n-X|>\epsilon/3\}+P\{|Y_n-Y|>\epsilon/3\}+P\{|X-Y|>\epsilon/3\} $$


0

If your sample $X_1,\ldots,X_n$ is i.i.d. Exp($\lambda$), you have that its joint density is $f(x_1,\ldots,x_n) = \lambda^n e^{-\lambda \sum_i x_i} $ on the non-negative orthant. By the Neyman factorization theorem, we can factor the joint density as $(\lambda^n e^{-\lambda \sum_i x_i} ) (1)$ where the first function captures all the dependency on $\lambda$ ...


1

The proof is correct, but notice that you get something stronger than what you state in the last line of your question, indeed $X_{\infty}$ is finite almost surely, which is the same as saying $-\infty < X_{\infty} < \infty$ (and not only $X_{\infty} < \infty$). To go from the fact that $E(|X_{\infty}|) < \infty$ to the fact that $X_{\infty} \in ...


0

The Gaussian copula (with correlation matrix $R$): $$ C_R(u)=\Phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R). $$ Then, letting $\zeta:=(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n))^{\top}$, $$ c_R(u)=\frac{\phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R)}{\phi(\Phi^{-1}(u_1))\cdots \phi(\Phi^{-1}(u_1))} \\=(2\pi)^{-n/2}\lvert R ...


1

The partial sums of $\sum X_k$ are bounded by constant $c$ iff $|M_r|\le c$ for every $r$. This last occurs iff $T_c=\infty$ (where I write $T_c$ instead of $T$ to make the dependence on $c$ explicit). Therefore $$ \{\text{partial sums of $\sum X_k$ are bounded}\} = \bigcup_{c=1}^\infty \{T_c=\infty\}.\tag1 $$ So if the LHS has positive probability then ...


2

For any $r\ge 0$ $$ \mathbb{E}(M_{\infty}-M_n)^2=\mathbb{E}(M_{\infty}-M_{n+r})^2+\mathbb{E}(M_{n+r}-M_n)^2 \\\overset{(d)}=\mathbb{E}(M_{\infty}-M_{n+r})^2+\sum_{k=n+1}^{n+r}\mathbb{E}(M_k-M_{k-1})^2. $$ Taking $r\to\infty$, the first term converges to $0$ by (f).


1

You've shown that there is a $c$ such that $P(T=\infty)>0$. So:$$E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+A_n P(T=\infty).$$ Both terms terms on the r.h.s are non-negative and you've also shown that $E[A_{T\wedge n}]$ is finite and you can take $n\rightarrow\infty$, so at the very least $A_\infty<\infty$ since $P(T=\infty)>0$


3

Write $\cos(a_j t) = (\exp(i a_j t) + \exp(-i a_j t))/2$, and expand the product. We get $$ 2^{-n} \sum_{x \in \{-1,1\}^n} \exp \left( i \sum_{j=1}^n x_j a_j t\right) $$ Now note that $$\dfrac{1}{2\pi} \int_0^{2\pi} \exp(ikt)\; dt = \cases{0 & if $k$ is a nonzero integer\cr 1 & if $k = 0$\cr}$$ Thus, assuming the $a_j$ are all integers, your right ...


1

Hint: You can write $$ \mathbb{E}(X\mid X\geq t)=\frac{\int_t^\infty xf_X(x)\,dx}{\int_t^\infty f_X(x) \, dx}. $$ Using this to get a differential equation for $f_X$.


1

Let $r>c>0$ and let $B$ be a Borelset with $0<\mathbb P(X\in B)\leq\frac{c}{r}$ If $f(x)=r\cdot1_B(x)$ then $\mathbb Ef(X)=r\mathbb P(X\in B)\leq c$. The essential supremum of $|f(X)|^2$ is $r^2>c^2$.


0

If I am not mistaken, exchangeability means (for two variables) that the law of $(X_1,X_2)$ is the same as the law of $(X_2,X_1)$. Thus, the space of possible values for the couple $(X_1,X_2)$ must be the same as the space of possible values for the couple $(X_2,X_1)$. In a nutshell, exchangeability is stronger than sharing the same support which is a ...



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