Question

Let Z1, Z2, . . . be a sequence of independent standard normal random variables. Define X0 = 0 and Xn+1 = (nXn + (Zn+1))/ (n + 1) , n = 0, 1, 2, . . . . The stochastic process {Xn, n = 0, 1, 2, } is a Markov chain, but with a continuous state space.

(a) Find E(Xn) and Var(Xn).

(b) Give probability distribution of Xn.

(c) Find limn→∞ P(Xn > epsilon) for any epsilon > 0.

Answer 1

(a)

Consider:

$\begin{align*} X_{n+1} &= \frac{nX_n + Z_{n+1}}{n+1} \\ \Rightarrow (n+1)X_{n+1} &= nX_n + Z_{n+1} \\ &= ((n-1)X_{n-1} + Z_n) + Z_{n+1} \\ &= (n-1)X_{n-1} + (Z_n + Z_{n+1}) \\ &= (n-2)X_{n-2} + (Z_{n-1} + Z_n + Z_{n+1}) \\ \text{Continuing in }&\text{this way, we get:} \\ (n+1)X_{n+1} &= 0*X_0 + (Z_1+ Z_2 + ... + Z_{n+1}) \\ \Rightarrow X_{n+1} &= \frac{1}{n+1}(Z_1 + Z_2 + ... + Z_{n+1}) \\ \Rightarrow X_{n} &= \frac{1}{n}(Z_1 + Z_2 + ... + Z_{n}) \\ \end{align*}$

Now, $\begin{align*} Z_1 , Z_2 , ... , Z_{n} \end{align*}$ are a sequence of independent standard normal random variables. Thus:

$\begin{align*} Z_i &\sim N(0,1) \ \ \ \ \forall i = 1,...,n \\ \Rightarrow Z_1 + Z_2 + ... + Z_n &\sim N(0,n) \\ \Rightarrow \frac{1}{n}(Z_1 + Z_2 + ... + Z_n) &\sim N(0,1/n) \\ \Rightarrow X_n &\sim N(0,1/n) \end{align*}$

Thus,

$\begin{align*} E(X_n) &=0 \\ Var(X_n) &= \frac{1}{n} \end{align*}$

(b)

In part (a), we have already derived the probability distribution of X_n:

Xn N(0, 1/n) 7l

(c)

We have:

$\begin{align*} X_n &\sim N(0,1/n) \\ \Rightarrow \frac{X_n - 0}{\sqrt{1/n}} &\sim N(0,1) \\ \Rightarrow \sqrt{n} X_n &\sim N(0,1) \end{align*}$

Now, consider:

$\begin{align*} \lim_{n \to \infty} P(X_n > \epsilon) &= \lim_{n \to \infty} P(\sqrt{n}X_n > \sqrt{n}\epsilon) \\ &= \lim_{n \to \infty} P(N(0,1) > \sqrt{n}\epsilon) \\ &= \lim_{n \to \infty} (1-P(N(0,1) \le \sqrt{n}\epsilon)) \\ &= \lim_{n \to \infty} (1-\Phi(\sqrt{n}\epsilon)) \\ \text{where }\Phi(.) \text{ is t} &\text{he CDF of N(0,1)} \\ \Rightarrow \lim_{n \to \infty} P(X_n > \epsilon) &= 1 - \lim_{n \to \infty} \Phi(\sqrt{n}\epsilon) \\ &= 1 - \Phi(\infty) \ \ \ \ [\text{Since, }\epsilon > 0] \\ &= 1 - 1 \\ &= \textbf{0 \ \ \ \ \ \ [Answer]} \end{align*}$

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