Past exam of the probability course of the University of Oxford / 2014 / 1

Convergence in distribution means that the cumulative distribution functions satisfyat every $x$ where the cumulative distribution function of $X$ is continuous. Values at its discontinuities need not converge. This definition uses only the probability distributions, so the random variables need not be defined on the same probability space.

Let $\mu =\mathbb{E}[X_1]$ and ${\sigma }^2=\mathrm{Var}(X_1)\in (0,\infty )$. For independent and identically distributed random variables, the central limit theorem statesThe limit is the standard normal distribution. If the variance is zero, then $X_i=\mu$ almost surely; the centered sums are zero, and division by $\sigma$ is not defined.

The implication convergence in probability implies convergence in distribution follows by comparing nearby events. Fix a continuity point $x$ of the cumulative distribution function $F_X$ and let $\epsilon >0$. If $X\le x-\epsilon$ and $|X_n-X|\le \epsilon$, then $X_n\le x$. Conversely, $X_n\le x$ and $|X_n-X|\le \epsilon$ imply $X\le x+\epsilon$. ThusThe error probability tends to zero by convergence in probability. Taking lower and upper limits and then letting $\epsilon \downarrow 0$, continuity at $x$ squeezes both bounds to $F_X(x)$. This is exactly convergence in distribution.

Subtract adjacent values of the cumulative distribution function. Its value at zero is zero, so the same calculation applies also to $k=1$:The probability mass function sums to one by a telescoping series.

For $x>0$, the event $M_n/n\le x$ says that each of the independent random variables is at most $\lfloor nx\rfloor$. Hence, for all sufficiently large $n$,Since $n/(\lfloor nx\rfloor +1)\to 1/x$ and $\log (1-u)=-u+o(u)$ as $u\to 0$, taking logarithms gives the limit $e^{-1/x}$. For $x\le 0$ the probability is zero. ThereforeThis is the Fréchet limit for a reciprocal integer tail, with a Fréchet distribution of shape one. Its cumulative distribution function is continuous even at zero, so the pointwise calculation establishes convergence in distribution.

Write $S_{n+1}=S_n+X_{n+1}$. The new random variable $X_{n+1}$ is independent of the entire history generated by $X_1,\dots ,X_n$, hence of $(S_1,\dots ,S_n)$. This is a Markov chain obtained from independent innovations: conditional on the past, only the current sum enters the next state. Its time-homogeneous transition probabilities areEvery jump is a positive integer, and each row sums to one by the probability mass function from part (b).

Now $M_{n+1}=\max (M_n,X_{n+1})$. Again $X_{n+1}$ is independent of the whole previous history, so the recursion is a Markov chain obtained from independent innovations. To remain at $s$, the new observation must be at most $s$; to jump to $t>s$, it must equal $t$. Thus the transition probabilities areThe upper tail sums to $1/(s+1)$, complementing the holding probability $s/(s+1)$.