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Let $X_0, \ldots$ be i.i.d. $\mathbb{P}\{X_i = -1\} = \mathbb{P}\{X_i = 1\} = 1 / 2$. Given $a, b \in \mathbb{R}, |b| < 1$, consider the stochastic process $W_k$ defined as $$ W_0 = a X_0\\ W_k = b W_{k - 1} + X_k, \; k = 1, 2, \ldots $$ Find $\mathbb{E}[W_K]$ and $\operatorname{Var}(W_k)$.

My attempt: First, note that $\mathbb{E}[X_i] = 0$, $\operatorname{Var}(X_i) = 1$, and $\mathbb{E}[X_i^2] = 1$ for all $i$. Now,

$$ \begin{align*} \mathbb{E}[W_0] &= a \mathbb{E}[X_0] = 0\\ \mathbb{E}[W_1] &= b \mathbb{E}[W_0] + \mathbb{E}[X_1] = 0\\ \mathbb{E}[W_2] &= b \mathbb{E}[W_1] + \mathbb{E}[X_2] = 0\\ &\vdots\\ \mathbb{E}[W_k] &= 0 \end{align*} $$ (which there shouldn't be many problems showing inductively).

For the variance, we have (because the expectation is $0$) $$ \begin{align*} \operatorname{Var}[W_0] &= \mathbb{E}[W_0^2] = a^2 \mathbb{E}[X_0^2] = a^2\\ \operatorname{Var}[W_1] &= \mathbb{E}[(b W_0 + X_1)^2] = b^2 \mathbb{E}[W_0^2] + 2 b \mathbb{E}[W_0 X_1] + \mathbb{E}[X_1^2] = b^2 a^2 + 1\\ \operatorname{Var}[W_2] &= \mathbb{E}[(b W_1 + X_2)^2] = b^2 \mathbb{E}[W_1^2] + 2 b \mathbb{E}[W_1 X_2] + \mathbb{E}[X_2^2] = b^4 a^2 + 2\\ &\vdots\\ \operatorname{Var}[W_k] &= \mathbb{E}[(b W_{k - 1} + X_k)^2] = b^2 \mathbb{E}[W_{k - 1}^2] + 2 b \mathbb{E}[W_{k - 1} X_k] + \mathbb{E}[X_k^2] = b^{2 k} a^2 + k\\ \end{align*} $$

Am I correct? Note that I assume independence between $W_{k - 1}$ and $X_k$ (and I'm not sure if I can do that).

EDIT: Okay, I see a mistake. I believe

$$ \operatorname{Var}[W_2] = \mathbb{E}[(b W_1 + X_2)^2] = b^2 \color{red}{\mathbb{E}[W_1^2]} + 2 b \mathbb{E}[W_1 X_2] + \mathbb{E}[X_2^2] = b^4 a^2 + 2 $$

should in fact be

$$ \operatorname{Var}[W_2] = \mathbb{E}[(b W_1 + X_2)^2] = b^2 \mathbb{E}[W_1^2] + 2 b \mathbb{E}[W_1 X_2] + \mathbb{E}[X_2^2] = b^2 (b^2 a^2 + 1) + 1 $$

leading to a geometric sequence in the end. Can anyone confirm?

d125q
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1 Answers1

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Write : \begin{align*} W_k&=bW_{k-1}+X_k=b\color{red}{(bW_{k-2}+X_{k-1})}+X_k\\ &=b^2W_{k-2}+bX_{k-1}+X_k\\ & \qquad\vdots\\ &=b^{k-1}W_{1}+b^{k-2}X_{2}+\cdots+X_k=b^{k-1}\color{red}{(bW_0+X_1)}+b^{k-2}X_{2}+\cdots+X_k\\ &=ab^{k}X_{0}+b^{k-1}X_{1}+b^{k-2}X_{2}+\cdots+X_k \end{align*}

Now you don't have independence problem !

$$\mathbb{E}(W_k)=\sum_{j=1}^k 0=0$$ $$Var(W_k)=Var(ab^kX_0)+\sum_{j=1}^k Var(b^{k-j}X_j)=a^2(b^{2})^k+(b^{2})^{k-1}+(b^2)^{k-2}+\cdots+1$$

Fardad Pouran
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