UMD Probability Qualifying Exams/Jan2006Probability

(a)
Each $$S_n$$ is clearly $$\mathcal F_n$$-measureable and finite a.s. (Hence $$L^1$$). Therefore we only need to verify the martingale property. That is, we want to show $$S_n^2-cn=E[S_{n+1}^2-c(n+1_|\mathcal{F}_n]$$

$$\begin{align} S_n^2-cn=E[S_{n+1}^2-c(n+1_|\mathcal{F}_n]&=S_n^2+E[X_{n+1}^2 +2\sum_{j=1}^n X_{n+1}X_j |\mathcal F_n] -cn -c\\ &= S_n^2-cn+E[X_{n+1}^2]+2\sum_{j=1}^n X_j E[X_{n+1}] \text{   by independence}\\ &= S_n^2-cn + Var[X_{n+1}] = S_n^2-cn + \sigma^2 \end{align}$$

We can assert that $$\sigma^2$$ exists and is finite since each $$|X_n|\leq 1$$ almost surely. Therefore, in order to make $$S_n^2-cn$$ a martingale, we must have $$c=\sigma^2$$.

Solution
First let us find the distribution of $$\tau_r$$:

$$P(\tau_r<x)=P(N_1(x)\geq r)=\sum_{k=r}^\infty \frac{(\lambda x)^k}{k!}e^{-\lambda x}$$

Thus by the chain rule, our random variable $$\tau_r$$ has probability density function

$$p_{\tau_r}(x)=\sum_{k=r}^\infty \frac{k \lambda (\lambda x)^{k-1}}{k!} e^{-\lambda x} + \frac{(\lambda x)^k}{k!} (-\lambda)e^{-\lambda x} = \lambda \frac{(\lambda x)^{r-1}}{(r-1)!}e^{-\lambda x}$$

So then

$$\begin{align} E[N_2(\tau_r^2)]=E[E[N_2(t)|\tau_r^2=t]]&= \int_0^\infty x^2 \lambda^2 \lambda \frac{\lambda^{r-1}}{(r-1)!}x^{r-1} e^{-\lambda x}\, dx\\ &= \frac{\lambda^{r+2}}{(r-1)!} \int_0^\infty x^{r+1} e^{-\lambda x}\, dx \end{align}$$

Now integrate the remaining integral by parts letting $$u=x^{r+1}, dv= e^{-\lambda x}\,dx$$. We get:

$$\begin{align} E[N_2(\tau_r^2)]&=\frac{\lambda^{r+2}}{(r-1)!} [ -\frac{1}{\lambda} e^{-\lambda x} x^{r+1} |_0^\infty +\int_0^\infty \frac{1}{\lambda}^{r+1} x^r e^{-\lambda x}\, dx]\\ &=\frac{\lambda^{r+2}}{(r-1)!} [\int_0^\infty \frac{1}{\lambda}^{r+1} x^r e^{-\lambda x}\, dx]\\ &=\frac{\lambda^{r+1} (r+1)}{(r-1)!} [\int_0^\infty \frac{1}{\lambda}^{r+1} x^r e^{-\lambda x}\, dx]\\ \end{align}$$

Repeat integration by parts another $$r$$ times and we get

$$ E[N_2(\tau_r^2)]=(r+1) r \lambda \int_0^\infty e^{-\lambda x}\, dx = \lambda (r+1) r \frac{-1}{\lambda} e^{-\lambda x} |_0^\infty = (r+1)r $$

(a)
$$\varphi_{X_{kn}}(t)=(1-1/n)+1/n e^{itk^2}$$

Then by independence, we have $$\varphi_{X_n}(t)=\prod_{k=1}^n \varphi_{X_{kn}}=\prod_{k=1}^n (1-1/n)+1/n e^{itk^2} $$