UMD Probability Qualifying Exams/Aug2006Probability

Solution
$$\begin{align} P(T_n\leq x)&=P(e^{a_n-\log X_{(2),n}}\leq e^x)=P(\frac{e^{a_n}}{X_{(2),n}}\leq e^x)\\ &=P(X_{(2),n}\geq e^{a_n-x})= n(1-e^{a_n-x})^{n-1}e^{a_n-x} +(1-e^{1_n-x})^n \end{align}$$

The two terms on the right hand side look like the limit definition of the exponential function. Can we choose $$a_n$$ appropriately so that it is?

Let $$a_n=-\log n$$. Then $$\begin{align} P(T_n\leq x)&= n(1-\frac{1}{ne^x})^{n-1}\frac{1}{ne^x} +(1-\frac{1}{ne^x})^n\\ &\to e^{-e^x}e^{-x}+e^{-e^x} \end{align}$$

This is the distribution of $$\lim_{n\to\infty}T_n$$.

Solution
(a)

$$\begin{align} P(\zeta\leq x) &= P(\xi\leq x-\eta) =\sum_{n=-\infty}^\infty P(\eta=n)\int_{-\infty}^{x-n} p_\xi (t)\,dt\\ &=\lim_{N\to\infty}\sum_{n=-N}^N P(\eta=n)\int_{-\infty}^{x-n} p_\xi (t)\,dt=\lim_{N\to\infty}\sum_{n=-N}^N P(\eta=n)\int_{-\infty}^{x} p_\xi (t-n)\,dt\\ &=\lim_{N\to\infty} \int_{-\infty}^{x}\sum_{n=-N}^N P(\eta=n) p_\xi (t-n)\,dt= \int_{-\infty}^{x}\sum_{n=-\infty}^\infty P(\eta=n) p_\xi (t-n)\,dt \end{align}$$

where the last equality follows from Monotone Convergence Theorem.

Hence, we have shown explicitly that $$\zeta$$ has a density and it is given by $$q_\zeta(t)=\sum_{n=-\infty}^\infty P(\eta=n) p_\xi (t-n)$$.

(b) When $$\xi \sim$$ Uniform[0,1] and $$\eta\sim$$ Poisson(1), we have $$p_\xi(x)=\Chi_{[0,1]}(x)$$ and $$p_\eta(k)=\frac{1}{k!e}$$ with support on $$k=0,1,2,...$$.

Then from part (a), the density will be

$$q_\zeta(x)=\sum_{n=-\infty}^\infty p_\eta(n)p_\xi(x-n)=\sum_{n=0}^\infty \frac{1}{k! e} \Chi_{[0,1]}(x)$$.

(a)
We know that $$N(\frac{mk}{n})-N(\frac{m(k-1)}{n})$$ is distributed as Poisson with parameter $$m/n$$. So

$$P(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\geq 2)=1-P(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})=1 \text{ or } 2) = 1- e^{-m/n} - \frac{m}{n} e^{-m/n}$$

Then $$\begin{align} E(W_{m,n})&=E[\sum_{k=1}^n I(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\geq 2)\\ &= \sum_{k=1}^n E(I(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\geq 2)) \text{  by linearity}\\ &= \sum_{k=1}^n P(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\geq 2) = n\left(1- e^{-m/n} - \frac{m}{n} e^{-m/n}\right)=n-e^{-m/n}(n+m) \end{align} $$

(b)
If $$\lim_{n\to\infty}W_{n,n^\alpha}=\infty$$ then we must have $$I(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\geq 2)=0$$ only finitely often. The probability of this even (from part a) is $$e^{-n^{\alpha-1}}(1+n^{\alpha-1})$$.

This decays to 0 for $$\alpha>1$$. Then clearly, we see that the probability of $$\lim_{n\to\infty}W_{n,n^\alpha}=\infty$$ is equal to 1.

I don't know how to show the result for $$1/2<\alpha<1$$...

(a)
We want $$Y_n=E[Y_{n+1}|\mathcal{F}_n]$$. We can compute both sides of this equation explicitly.

$$X_n-an=E[X_{n+1}-a(n+1)|X_n]=(X_n-2)\frac{1}{4}+(X_n+1)\frac{3}{4}-a(n+1)=X_n+1/4-an-a$$

Thus if we want this equality to hold we must have $$a=1/4$$.

Similarly, if we want $$Z_n=E[Z_{n+1}|\mathcal{F}_n]$$ then

$$e^{bX_n}=E[e^{bX_{n+1}}|X_n]=\frac{1}{4} e^{b(X_n-2)}+\frac{3}{4} e^{b(X_n+1)}=e^{bX_n}(\frac{1}{4}e^{-2b}+\frac{3}{4}e^b)$$

We can easily check that $$b=0$$ gives a trivial solution to the equation. Using the substitution $$x=e^b$$ we can find another solution for $$b$$. We should get $$b=\log(1+\sqrt{13})-\log(6)<0$$.

(b)
We've just shown that $$Y_n=X_n-\frac{1}{4} n$$ is a martingale. Thus, $$E[Y_n|Y_0]=E[Y_0]=0$$. Then since each $$\xi$$ is i.i.d., we can apply the Strong Law of Large Numbers to say $$1/n(X_n-n/4)\to 0$$ almost surely. In other words, $$X_n\to\infty$$ almost surely and so certainly $$\tau<\infty$$ almost surely.

Now to calculate $$E[\tau]$$. We introduce new notation: let $$\tau_k(x)=\inf\{n\geq 1: X_n=k, X_0=x\}$$. Then $$\begin{align} E[\tau_n(0)]&=\frac{3}{4}(1+E[\tau_n(1)])+\frac{1}{4}(1+E[\tau_n(-2)])\\ &=\frac{3}{4}(1+E[\tau_{n-1}(0)])+\frac{1}{4}(1+E[\tau_{n+2}(0)]) = 1+\frac{3}{4}\tau_{n-1}(0)+\frac{1}{4}\tau_{n+2}(0) \end{align}$$

by a symmetry argument.

So we can write $$E[\tau_1(0)]=1+\frac{3}{4} 0 + \frac{1}{4} \tau_3(0)$$. But I don't know how to calculate $$E[\tau_1(0)]$$....

(c)
Recall from part (a) that the nontrivial solution for $$b$$ must be some negative number. Then $$\lim_{n\to\infty}Z_n\to 1$$ almost surely by part (a) as well.

However, $$Z_n=e^{bX_n}\neq 1 = E[Z_\infty|\mathcal{F}_n]$$. This by the definition, means the martingale is not right closable. A martingale is right-closable iff uniformly integrable. Thus, we're done.