Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / M

The independent identically distributed central limit theorem states that for independent identically distributed random variables $U_k$ with finite mean $\mu$ and positive finite variance ${\sigma }^2$,For $0<u<1$, continuity, strict monotonicity and the endpoint limits of the cumulative distribution function give a unique $x=F^{-1}(u)$. Then $P(F(X)\le u)=P(X\le x)=F(x)=u$. Thus $Y=F(X)$ has the uniform distribution on $(0,1)$, by the probability integral transform.

For $W=-\log (1-Y)$, $P(W\le w)=1-e^{-w}$ when $w\ge 0$, and it is zero otherwise. Therefore its probability density function is $e^{-w}{\chi }_{(0,\infty )}(w)$: the exponential distribution of rate one. By integration by parts, $E[W]=1$ and $E[W^2]=2$, so $\mathrm{Var}(W)=1$.

The variables $W_k=-\log (1-F_k(X_k))$ are independent because they are separate measurable functions of independent random variables. The probability integral transform makes all of them identically distributed as $\mathrm{Exp}(1)$, even though the $X_k$ need not be identically distributed. The displayed $Z_n$ is exactly $n^{-1/2}{\sum }_{k=1}^n(W_k-1)$. The central limit theorem therefore gives

A Poisson process of rate $\lambda$ is a counting process starting at zero, with independent stationary increments, such that an increment over an interval of length $t$ has the Poisson distribution of parameter $\lambda t$. It has right-continuous paths with unit jumps almost surely.

Independence of the two processes makes increments of $A+B$ independent on disjoint intervals. The sum of independent Poisson random variables with parameters $\alpha t$ and $\beta t$ is Poisson with parameter $(\alpha +\beta )t$, as follows by convolution:Thus $A+B$ is a Poisson process of rate $\alpha +\beta$. Its first red and green waiting times are independent exponential random variables of rates $\alpha ,\beta$, so

The state-zero stationary distribution equation is ${\pi }_0=q{\pi }_0+q{\pi }_1$, so ${\pi }_1=(p/q){\pi }_0$. The state-one equation is ${\pi }_1=p{\pi }_0+q{\pi }_2$, giving ${\pi }_2=(p/q){\pi }_1$. More generally, the stationary probability flow across the cut $\{0,\dots ,i\}$ is balanced, so $p{\pi }_i=q{\pi }_{i+1}$. This also follows inductively from ${\pi }_i=p{\pi }_{i-1}+q{\pi }_{i+1}$.

Hence ${\pi }_i=(p/q{)}^i{\pi }_0$. Since $p<q$, the geometric series is summable, and normalization givesThese probabilities satisfy detailed balance and therefore all the stationarity equations. The Markov chain is an irreducible Markov chain, so this normalized stationary distribution is unique.