Past exam of the probability course of the University of Oxford / 2023 / 1 / c

Direct integration of the uniform distribution yields $\mathbb{E}{U}_j=b/2$ and $\mathbb{E}{U}_j^2=b^2/3$. The expectation of a product of independent random variables givesSubtracting the square of the expected value gives the variance:Since $W_n\ge 0$, Markov's inequality implies, for every $\epsilon >0$,Hence $\boxed{W_n\to 0\text{ in probability for }0<b<2.}$ The first moment suffices: the variance need not tend to zero throughout this range.

If $0<b\le 1$, then $0<U_j\le 1$ almost surely. Intersecting these events over all $j$ still gives an event of probability one. On it, $0\le W_{n+1}\le W_n$, so the products have a finite nonnegative limit $W_{\infty }$.

For every $\epsilon >0$, the inclusion $\{W_{\infty }>\epsilon \}\subseteq \{W_n>\epsilon \}$ and part (c)(i) imply $\mathbb{P}(W_{\infty }>\epsilon )=0$. Taking $\epsilon =1/m$ for positive integers $m$ proves $W_{\infty }=0$ almost surely. ThusThe monotonicity is the extra ingredient that upgrades convergence in probability to almost sure convergence here.

For $\alpha >0$, integration and independence of random variables giveBy the fractional moment criterion for convergence in probability,The parenthesized factor is less than one precisely when $\alpha \log b-\log (1+\alpha )<0$. This expression vanishes at $\alpha =0$ and has derivative $\log b-1$ there. If $b<e$, the derivative is negative, so some sufficiently small $\alpha >0$ makes the factor strictly less than one. The bound then tends to zero geometrically, proving