# How to compute $\lim_{n\to +\infty}E[Y_{n}Y_{n-1}]$? [closed]

Let $X_{1}, X_{2},\ldots, X_{n},\ldots$ be independent and identically distributed random variables with distribution

$P(X_{n}=0)=P(X_{n}=1)=P(X_{n}=2)=P(X_{n}=3)=\frac{1}{4}$

Assume that the sequence $\{Y_n\}$ is defined as: $Y_{0}=0$ and for all $n\in\mathbb{N}$ we have

$Y_n=\begin{cases} 3 &\text{if } X_n=3,\\\min{\{Y_{n-1},X_n\}} &\text{if } X_n<3. \end{cases}$

Compute $\displaystyle \lim_{n\to +\infty}E[Y_{n}Y_{n-1}]$?

I don't know how to start.

Behavior of $Y_i$ is described by an aperiodic irreducible Markov chain on $4$ states which has a unique limiting distribution $\pi$ (a row vector). Let $f=(0,1,2,3)^T$ (column vector - values of $Y_i$ on $4$ states). Then $$\lim E(Y_{n-1}Y_n)=\pi(f\cdot Pf)$$ where $\cdot$ presents coordinate-wise product. You are left to explicitely contruct $P$ and find $\pi$ s.t. $\pi P=\pi$.

• Thank you for your post but i am not familiar with Markov chain theory, so this problem seems to be too hard for me. Why I recived -1 point for my question? What is wrong with it?
– Leon
Nov 12, 2015 at 6:02
• @Leon Ignore the haters. I don't see a way to approach this problem without Markov Chain theory.
– A.S.
Nov 12, 2015 at 15:56

$$E(Y_{n}Y_{n+1})=E(E(Y_{n}Y_{n+1}|X_{n+1}))\\=\frac{1}{4}[E(Y_{n}\min\{Y_{n},0\})+E(Y_{n}\min\{Y_{n},1\})+E(Y_{n}\min\{Y_{n},2\})+3E(Y_n)]$$ Now, note that for all $n\ge 1,\ Y_n\in \{0,1,2,3\}$. Then, \begin{eqnarray*} E(Y_{n}\min\{Y_{n},0\})&=&0\\ E(Y_{n}\min\{Y_{n},1\})&=&P(Y_{n}=1)+2P(Y_{n}=2)+3P(Y_{n}=3)\\ E(Y_{n}\min\{Y_{n},2\})&=&P(Y_{n}=1)+4P(Y_{n}=2)+6P(Y_{n}=3) \end{eqnarray*} Let $$p_n:=P(Y_n=1),\ q_n:=P(Y_n=2),\ r_n=P(Y_n=3)$$ Then, we have the recursion, $$E(Y_{n+1}Y_{n})=\frac{5p_n+12q_n+18r_n}{4},\ n\ge 1$$ Now, we note that $$p_{n+1}=P(Y_{n+1}=1)=P(Y_n= 1,1\le X_{n+1}<3)+P(1< Y_n, X_{n+1}=1)\\=\frac{2p_n+q_n+r_n}{4}\quad(\because X_{n+1}\perp Y_n)\\ q_{n+1}=\frac{(q_n+r_n)}{4}\\ r_{n+1}=\frac{1}{4}$$ Thus, defining $$v_{n+1}=[p_{n+1}\quad q_{n+1}]^T$$ we get the recursive equation, $$v_{n+1}=Av_n+a$$ where $$A=\begin{bmatrix} 2/4 & 1/4\\ 0 & 1/4 \end{bmatrix}\\ a=[1/16\quad 1/16]^T$$ Also, note the initial conditions $$p_1=0,\ q_1=0$$ Thus, noting that the maximum eigenvalue of matrix $A$ is $1/2$, the sequence $v_n$ converges, and we can find the limit as $$\lim_{n\to \infty}v_n=(I-A)^{-1}a$$ Thus $$\lim_{n\to \infty}E(Y_nY_{n-1})=[5/4\quad 3](I-A)^{-1}a+\frac{9}{8}.$$

• $P(Y_{n+1}=1)=P(Y_n\ge 1,1\le X_{n+1}<3)$ is not quite correct. If $Y_n=X_{n+1}=2$ then $Y_{n+1}=2$, not $Y_{n+1}=1$.
– user940
Nov 11, 2015 at 18:35
• In the first line should be $E(Y_{n}Y_{n-1})=E[E(Y_{n}Y_{n-1}|X_{n})]$?
– Leon
Nov 12, 2015 at 5:58
• @ByronSchmuland, sorry, that was a mistake, I am editing it. Nov 12, 2015 at 6:08
• @ByronSchmuland, I have edited my answer. Nov 12, 2015 at 6:45
• Your equations for $p_{n+1}$ and $q_{n+1}$ are still incorrect...
– user940
Nov 12, 2015 at 16:57