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Let $\left(\xi_n\right)_{n\geqslant1}$ be a sequence of independent identically distributed random variables with distribution $\mathcal{N}(0,1)$. Let $X_0$ be a random variable in $\mathbb{R}$ with distribution $\mu$, independent from the sequence $\left(\xi_n\right)_{n\geqslant1}$. Let us define the sequence $\left(X_n\right)_{n\geqslant1}$ by $$X_n=\alpha X_{n-1}+\xi_n$$ where $\alpha\in(0,1)$.

Show that $\left(X_n\right)_{n\geqslant0}$ is a Markov chain and compute its transition kernel. Find $\sigma^2>0$ such that $\mathcal{N}\left(0,\sigma^2\right)$ is a stationnary distribution for this chain.

My work

I have computed $X_n=\alpha^nX_0+\sum_{j=1}^n \alpha^{n-j}\xi_j$, however when writing $$\mathbb{P}\left(X_0\in C_0,\dots,X_n\in C_n\right)$$ for $C_0,\dots,C_n$ borel sets of $\mathbb{R}$, I do not know how to use independence to make it look like $$\int_{\substack{x_0\in C_0\\\dots\\x_n\in C_n}}\mu(dx_0)P(x_0,dx_1)\dots P(x_{n-1},dx_n)$$

Flewer47
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1 Answers1

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Observe that $X_0,X_1,..,X_{n-1}$ depend only on $\xi_1,\xi_2,...,\xi_{n-1}$. Hence, $P(X_n\leq x|X_0,X_1,..,X_{n-1})=P(\xi_n\leq x-\alpha X_{n-1}|X_{n-1})$. By independence of $\xi_n$ and $X_{n-1}$ you can compute this as $\Phi (x-\alpha X_{n-1})$.

Kavi Rama Murthy
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  • Got it for the independence. However, I do not understand how I can compute the cdf: I do have an integral but it does not seem so nice (if $X_{n-1}$ is written in terms of $\xi_i$) ? I mean, I should be having a nice expression of the transition kernel in order to compute $\sigma^2$ in the end. – Flewer47 Dec 19 '20 at 10:31
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    You don't have to express $X_{n-1}$ in terms of the $\xi_i$'s. The probability that $X_n$ lands in a set $E$ given $X_{n-1}=y$ is what you call the transition kernel. Here the kernel is $K(y,E)=\int_{E-\alpha y} \phi (t) dt$. @Flewer47 – Kavi Rama Murthy Dec 19 '20 at 11:32