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

The strong law of large numbers states that, for independent and identically distributed random variables $X_1,X_2,\dots$ with $\mathbb{E}|X_1|<\infty$,Thus, outside an event of probability zero, the sample mean converges to the common expected value.

For a Borel set $A\subseteq \mathbb{R}$, the joint probability density givesFor each $x\ne 0$, substitute $z=xy$. The change of variables for probability densities contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional Lebesgue measure zero, so omitting it changes no probability. By the Tonelli theorem,The bracket is nonnegative and its integral over $\mathbb{R}$ is one. This proves the probability density of a product of random variables formula:As usual, a probability density function is determined only almost everywhere; exceptional infinite values are harmless.

For the independent random variables with the specified uniform distribution, $f(x,y)=b^{-2}$ on $(0,b{)}^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. ThereforeThe value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks normalization of a probability distribution: ${\int }_0^1-\log sds=1$. This is the product of independent uniform random variables law.

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

The inverse transformation of a random variable is $u=e^{-v}$, with absolute derivative $e^{-v}$. The condition $0<u<b$ becomes $v>-\log b$. The change of variables for probability densities givesEquivalently, $V_n+\log b$ has the exponential distribution of rate one. Applying a separate measurable transformation of a random variable to each independent $U_n$ preserves independence of random variables, so the $V_n$ are independent and identically distributed.

Taking logarithms of the product givesBy part (d)(i), $V_j=E_j-\log b$ in probability distribution, where $E_j$ has the exponential distribution of rate one. Thus $\mathbb{E}|V_j|\le 1+|\log b|<\infty$ and $\mathbb{E}{V}_j=1-\log b$. The strong law of large numbers givesIf $b<e$, choose $\delta =(1-\log b)/2>0$. Eventually $\log W_n\le -\delta n$, so $W_n\le e^{-\delta n}\to 0$. If $b>e$, choose $\delta =(\log b-1)/2>0$; eventually $W_n\ge e^{\delta n}\to \infty$. ThereforeIn particular, the second range cannot have almost sure convergence to zero. This is the almost-sure growth rate of independent products. At $b=e$ the logarithmic limit is zero, which by itself decides neither behavior; that boundary case is not requested.