# convergence in probability of mean of sequence $X_{k+1}=\beta X_k+\epsilon_k$

Let $X_1=0$ and $X_{k+1}=\beta X_k+\epsilon_k$ with $\epsilon_k$ iid normally distributed, $\epsilon_k\sim N(\mu,\sigma^2)$. Let $\mu$ and $\sigma^2$ be fixed.

How can you show the mean of $X_k$ for $k=1,...,n$ converges, e.g. $$\frac1n\sum_{k=1}^nX_k\xrightarrow{P}{} \frac{\mu}{1-\beta} \;?$$

• It is β<1 or not? – Jimmy R. Nov 12 '14 at 20:33
• @Stefanos yes, $-1\leq \beta<1$. – user104143 Nov 12 '14 at 20:35
• @Stefanos where's you answer gone? I've started to check it -.- – user104143 Nov 12 '14 at 20:52
• I had a bad mistake, I am correcting it. Again $|β|<1$ not $|β|\le1$ – Jimmy R. Nov 12 '14 at 20:53

By substituting iteratively you find that \begin{align*}X_{k+1}&=βX_{k}+\epsilon_k=β(βX_{k-1}+\epsilon_{k-1})+\epsilon_k=β^2X_{k-1}+β\epsilon_{k-1}+\epsilon_k=\ldots\\\\&=β^kX_1+\left(\sum_{l=0}^{k-1}β^l\epsilon_{k-l}\right)\\&=\sum_{l=0}^{k-1}β^l\epsilon_{k-l}\end{align*} since $X_1=0$. Thus \begin{array}{rcrcrr}X_2&=&\epsilon_1\\X_3&=&β\epsilon_1&+&\epsilon_2\\X_4&=&β^2\epsilon_1&+&β\epsilon_2&+&\epsilon_3&\\\ldots&&\ldots&&\ldots&&\ldots\\X_n+1&=&β^{n-1}\epsilon_1&+&β^{n-2}\epsilon_2&+&β^{n-3}\epsilon_3&+\ldots+β\epsilon_{n-1}+\epsilon_n\end{array} Summing up both sides, we obtain \begin{align*}\sum_{k=1}^{n}X_{k+1}&=\frac{1-β^n}{1-β}\epsilon_1+\frac{1-β^{n-1}}{1-β}\epsilon_2+\ldots+\frac{1-β^2}{1-β}\epsilon_{n-1}+\frac{1-β}{1-β}\epsilon_n\\&=\sum_{k=1}^{n}\frac{1-β^{n+1-k}}{1-β}\epsilon_k=\ldots\\&=\frac{1}{1-β}\left(\sum_{k=1}^{n}\epsilon_k-β^{n+1}\sum_{k=1}^{n}\frac{1}{β^k}\epsilon_k\right)\end{align*} and finally $$\bar{X}_n=\frac{1}{n(1-β)}\left(\sum_{k=1}^{n}\epsilon_k-β^{n+1}\sum_{k=1}^{n}\frac{1}{β^k}\epsilon_k\right)$$ From this form we can calculate $E[\bar{X}_n]$ and (especially) $Var(\bar{X}_n)$ since the $\epsilon_k$ are independent. Indeed $$E[\bar{X}_n]=\frac{1}{n(1-β)}\left(\sum_{k=1}^{n}μ-β^{n+1}\sum_{k=1}^{n}\frac{1}{β^k}μ\right)=\frac{μ}{1-β}\left(1-\frac{β^{n+1}-1}{n(1-β)}\right) \longrightarrow \frac{μ}{1-β}$$ as $n \to \infty$ and similarly \begin{align*}Var[\bar{X}_n]&=\frac{1}{n^2(1-β)^2}\left(\sum_{k=1}^{n}σ^2-(β^2)^{n+1}\sum_{k=1}^{n}\frac{1}{(β^2)k}σ^2\right)\\\\&=\frac{σ^2}{(1-β)^2}\left(\frac{1}{n}-\frac{(β^2)^{n+1}-1}{n(1-β^2)}\right) \, \longrightarrow \, 0\end{align*} as $n \to \infty$. Hence $$E[\bar{X}_n^2]=Var(\bar{X}_n)+E[\bar{X}_n]^2 \longrightarrow 0+\frac{μ^2}{(1-β)^2}$$ which is enough to obtain $\mathcal L^2$ convergence $$E\left[\left(X_n-\frac{μ}{1-β}\right)^2\right]=E\left[\bar{X}_n^2-2\bar{X}_n\frac{μ}{1-β}+\frac{μ^2}{(1-β)^2}\right]\to \frac{2μ^2}{(1-β)^2}-\frac{2μ^2}{(1-β)^2}=0$$ from which you can conclude that $$\overline{X}_n=\frac{1}{n}\sum_{k=1}^{n}X_k\overset{p}\rightarrow \frac{μ}{1-β}$$ since convergence in $\mathcal L^r$ for $1\le r$ implies convergence in probability (see here).
Summing up both sides of the given relation $$\sum_{k=1}^{n}X_{k+1}=β\sum_{k=1}^{n}X_k+\sum_{k=1}^{n}\epsilon_k$$ gives you after simple calculations that $$\bar{X}_n=\frac{1}{n(1-β)}\left(\sum_{k=1}^{n}\epsilon_k-X_{n+1}\right)$$ and thus obtaining a closed form for $X_{n+1}$ (as above) is enough to obtain the result with less calculations than above.
• Indeed the random variables are not independent, but... You already proved that $E(\bar{X_k})\to\mu$. A similar computation would show that $E(\bar{X_k}^2)\to\mu^2$. Thus $\bar{X_k}\to\mu$ in $L^2$, which is more than the desired result. – Did Nov 12 '14 at 21:12
• ??? Where is the proof of the (asserted) convergence in $L^1$? OP: Surely you can complete the proof since you accepted the answer? – Did Nov 12 '14 at 23:30