Past exam of the probability course of the University of Oxford / 2017 / 2 / b

The given expected value and variance imply $M_X(0)=1$, $M_X^{\prime }(0)=0$ and $M_X^{\prime \prime }(0)=1$. The Taylor expansion therefore givesSince $\log (1+u)=u+o(u)$ as $u\to 0$, it follows thatThis is the quadratic expansion of a cumulant-generating function for a centered variable of variance one.

Write $S_n=n^{-1/2}{\sum }_{k=1}^n{X}_k$. By independence of random variables, its moment-generating function isFor each fixed real $t$, the expansion in part (i) yieldsFor $t=0$ the equality is immediate. Exponentiation gives $M_{S_n}(t)\to e^{t^2/2}$, the moment-generating function of a standard normal variable. The functions are finite on a common neighborhood of zero, so the convergence theorem for moment-generating functions provesThis proves the central limit theorem under the exponential-moment hypothesis of this part.

Use the deterministic weights $a_{n,k}=\sqrt{k}/n$. Their squared sum and maximum areTo apply the quadratic expansion of a cumulant-generating function rigorously, write $\log M_X(u)=u^2/2+r(u)$, with $r(u)=o(u^2)$. Fix $t\ne 0$ and letAll arguments $t\sqrt{k}/n$ lie in this shrinking interval. ConsequentlyTaking logs of the moment-generating function and using independence of random variables now givesThe case $t=0$ is immediate. Since $e^{t^2/4}$ is the moment-generating function of a normal distribution of variance $1/2$, the convergence theorem for moment-generating functions givesThis is a weighted central limit theorem with negligible maximum weight. Its limit variance also agrees with the exact value $\mathrm{Var}(T_n)=(n+1)/(2n)$ from variance additivity for independent random variables.