Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / M (Probability)

For an exponential distribution with rate $\lambda >0$, direct integration gives the moment-generating functionIt is finite in a neighborhood of zero, so its derivatives there give moments. Differentiating three times gives $M_X^{(3)}(t)=6\lambda /(\lambda -t{)}^4$, hence

Take $Y=a{X}_1+b$. From the definition of a moment-generating function, $\mathbb{E}{e}^{tY}=e^{bt}\mathbb{E}{e}^{at{X}_1}=e^{bt}{M}_1(at)$, for $t$ where the expectation is finite.

Take $Y=X_1+X_2$. Independence makes the joint exponential expectation factor:This is the product rule for the moment-generating function of an independent sum.

Take $Y=X_1+X_1^{\prime }$, where $X_1^{\prime }$ is an independent copy of $X_1$. The preceding factorization gives $\mathbb{E}{e}^{tY}=M_1(t{)}^2$. Using $2X_1$ would instead give $M_1(2t)$, so an independent copy is required.

Take $Y=U{X}_1$, where $U$ is uniform on $(0,1)$ and independent of $X_1$. Conditioning on $U$ givesFor $t$ near zero this is finite because the moment-generating function of $X_1$ is finite on a neighborhood of zero; Tonelli's theorem justifies the conditioning integral for the nonnegative exponential.

Recognize the mixture distributionIt is a half-and-half mixture of exponential distributions with rates one and two. Uniqueness of a moment-generating function finite near zero identifies the law. Therefore

A stationary distribution $\pi =({\pi }_1,{\pi }_2)$ satisfies $\pi P=\pi$ and ${\pi }_1+{\pi }_2=1$. The first equation reduces to the balance relation $a{\pi }_1=b{\pi }_2$. Since $a,b>0$, normalization gives the unique solution

Condition on the full sampled history through time $n$. The Markov property of $X$ implies that its future after $2n$ depends on that history only through $X_{2n}=Y_n$. Summing over the intermediate state givesThus $Y$ is a Markov chain with transition matrix $P^2$. This is a deterministic-time sampled Markov chain.

The chain alternates between the classes $\{1,2\}$ and $\{3,4\}$. It is irreducible and has period two, since no odd return is possible and each state has a positive two-step return probability. Matrix multiplication givesThe even-time sampled chain stays in its initial class and moves between its two states with probability $4/9$ at each sampled step.

Starting at one, state two can first be reached only after an even number of steps. At every even time before that hit, the chain is again at one. By the Markov property and part (i), each successive two-step block hits two with probability $4/9$, while a failure returns to one with probability $5/9$. Therefore, for the first hitting time $H$,Equivalently $H/2$ has the geometric distribution on the positive integers with success probability $4/9$.

Within the class $\{1,2\}$, the two-step matrix has eigenvalues one and $1/9$, with stationary distribution $(1/2,1/2)$. Diagonalizing this symmetric matrix, or solving the recursion $p_{n+1}=4/9+p_n/9$ from $p_0=0$, givesConsequently the limit is $1/2$. Restriction to even times is essential because the original chain has period two.

The definition of convergence in probability isThe variables are defined on a common probability space so that these difference events make sense.

Pointwise, $X\ge a{1}_{\{X\ge a\}}$, since $X$ is nonnegative. Taking expectations and dividing by $a>0$ proves the Markov inequalityIf the expectation is infinite, the inequality remains true in the extended-real sense.

Let ${\overline{X}}_n=n^{-1}{\sum }_i{X}_i$ and write ${\sigma }^2=\mathrm{Var}(X_1)<\infty$. Independence gives $\mathbb{E}{\overline{X}}_n=\mu$ and $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. Apply part (b) to the nonnegative variable $({\overline{X}}_n-\mu {)}^2$, which is the Chebyshev inequality:Thus ${\overline{X}}_n\to \mu$ in probability, proving the finite-variance weak law of large numbers.

Equality forces $\mathbb{E}X<\infty$ because its proposed value is at most $a$. The random variable $Y=X-a{1}_{\{X\ge a\}}$ is nonnegative and has expectation zero, so $Y=0$ almost surely. On $\{X<a\}$ it equals the strictly positive variable $X$, making that event null. On $\{X\ge a\}$ it equals $X-a$, forcing $X=a$ there. HenceThis is the equality in Markov's inequality under strict positivity. The reasoning even works without the discreteness assumption.