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Yes, the mean is infinite. (If it was finite, since $W$ is integrable, $XY/Z$ would be integrable. Since $XY$ and $Z$ are independent and $XY$ is not identically $0$, $1/Z$ would be integrable. But …

This is to answer a question raised by unknown: in the asymmetric case where the coin moves clockwise with probability $p$ and counterclockwise with probability $1-p$, the new head is located $k$ seat …

Yes it can: the probability that $|(XY/Z)-W|\ge t$ is bounded by $c/t+o(1/t)$ with $c^2=2/\pi^3$, when $t\to\infty$. (By the way, Yemon Choi DID NOT suggest that the tail estimate might be interesting …

First, $Z$ is distributed like $bX$, with $b>0$ and $b^2=a^2+(1−a)^2$. Second, for every $x$, $u(bx)=\sqrt{b}u(x)$. Third, $E(u(X))=-E(\sqrt{X};X>0)$ is negative. Hence, $E(u(Z))=\sqrt{b}E(u(X))$ is a …

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following constructi …

Let $D$ denote a set of probability distributions of positive random variables and $r$ a positive real number. Consider the following properties: For every random variables $X$ and $T$ with probabil …

Let $M_n(T)=\max\{S_1(T),\ldots,S_n(T)\}$ where the $S_j(T)$ are i.i.d. and distributed like $S(T)$. A partial answer to your question is that $E(M_n(T))/T\to\mu$ when $T\to+\infty$ as soon as the f …

In the symmetric case, your second question is easy. Assume that each outcome is equally likely. You need $N_0=1$ trial to get $1$ outcome. Once you got $k$ different outcomes, you need $N_k$ more tri …

In fact $S_d$ does not converge to zero when $d\to\infty$, at least if $a_x=1$ for every $x$. Here is a proof. For every $x$, $G_d(x)=P_0^{(d)}(\mathtt{hit}\ x)G_d(0)\le G_d(0)$ hence $G_d(x)^2/G_d( …

This solution is ugly, sorry. Proceed by contradiction and assume that 0000, 0001 and 0010 all have probability zero. Every event where one specifies two coordinates and one leaves free the two remai …

Assume the distribution of $X_n$ is binomial $(n,p)$. Then, for every real number $t$, $$ \mathrm E(\mathrm e^{\mathrm it(X_n-np)})=\mathrm e^{\mathrm itnp}(1-p+p\mathrm e^{\mathrm it})^n. $$ Assuming …

While you are at it, you could allow the value $+\infty$ as well... The resulting object is a random variable from a probability space $(\Omega,\mathcal{F},P)$ to a bona fide measurable space $(E,\m …

The horror, the horror... :-) Recall that $\Phi(x)=P[X\leqslant x]$ for every $x$, where the random variable $X$ is standard normal, and that, for every suitable function $u$, $$ \int_{-\infty}^{+\in …

Gibbs measures provide a solution. Introduce the energy $H(x)$ of a configuration $x=(x_k)_{1\le k\le N}$ in $S=\{0,1\}^N$ as $$ H(x)=\sum_kx_k+\sum_{k\ne\ell}x_kx_\ell, $$ and, for every parameter $a …

A partial answer is that, when $X$ is not integrable, $\hat X_x$ the residual lifetime of $\min\{X,x\}$ converges to infinity in distribution when $x\to\infty$. Since the residual lifetime $\hat X$ of …