Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 9

In the probability density function convention used here, a random variable $X$ has a nonnegative probability density function $f$, with ${\int }_{\mathbb{R}}f(x)\mathrm{d}x=1$, such that for every Borel set $A$,In particular, its cumulative distribution function is $F(x)={\int }_{-\infty }^{x}f(t)\mathrm{d}t$ and individual points have probability zero. More precisely, this property is the statement that the distribution is an absolutely continuous probability distribution; continuity of the cumulative distribution function alone is a weaker property and need not imply a probability density function.

For a probability density function $f$, the finite expected value isThe absolute integrability condition makes the signed integral unambiguous. One can also allow an extended expectation when exactly one of the positive and negative parts has infinite integral, obtaining $+\infty$ or $-\infty$. If both parts have infinite integral, the expectation is undefined; subtracting two infinities is not a definition.

Multiply the probability density function by $x$ and integrate over its support:The middle equality is the beta function integral with parameters $\alpha +1$ and $\beta$; the last uses the recurrence for the gamma function. Both parameters are positive, so the integral converges. Equivalently, multiplying the beta distribution probability density function by $x$ converts its kernel to that of a beta distribution probability density function with its first parameter increased by one.^[ref]

The same integral works with both powers shifted:The assumptions $s>-\alpha$ and $t>-\beta$ say precisely that the exponents at each endpoint are greater than $-1$, ensuring integrability there. Values of a negative power at the endpoints themselves can be assigned arbitrarily, because those endpoints have probability zero. This establishes the mixed power moments of a beta distribution without requiring $s$ and $t$ to be integers.^[ref]

Write $Z=U^a$. Since $a>0$, the map is increasing on $[0,1]$, and for $0⩽z⩽1$,Differentiation givesand zero outside this interval. Since $B(1/a,1)=a$, this is exactly the beta distribution probability density function with parameters $(1/a,1)$. The power transform of a uniform random variable therefore satisfies $U^a\sim \mathrm{Beta}(1/a,1)$; the second parameter does not depend on $a$.

Let $W_i=U_i^2$. Applying the same measurable function separately to independent and identically distributed random variables preserves those two properties, andIn particular, these bounded random variables are integrable random variables. The weak law of large numbers givesIf $c>1/3$, then $\{{\overline{W}}_n>c\}\subseteq \{|{\overline{W}}_n-1/3|⩾c-1/3\}$, whose probability tends to zero. If $0⩽c<1/3$, then $\{{\overline{W}}_n⩽c\}\subseteq \{|{\overline{W}}_n-1/3|⩾1/3-c\}$, whose probability also tends to zero. Taking a complement in the second case provesThe positive gap from $1/3$ is essential to this argument. Convergence in probability alone does not determine the answer at the excluded threshold $c=1/3$.