Probability of Level Crossing

Question

I am kind of stuck on how to proceed on this.

$X_n$ is an IID process with $$f_{X_n}(y)= \frac\lambda2 e^{-\lambda |y|}$$

There is a stationary autoregressive process $Y_n$ defined as $$Y_n=\rho Y_{n-1}+X_n$$

The goal is to find $Pr(Y_n > X_n +1/2)$. Any hints?

Edit: Does the fact that the process is first order Markov and stationary help?

Answer 1

First, $$ \mathsf{P}\left(Y_n-X_n>\frac{1}{2}\right)=\mathsf{P}\left(Y_{n-1}>\frac{1}{2\rho}\right) $$

and $Y_n=(1-\rho L)^{-1}X_{n}=\sum_{k=0}^{\infty}\rho^k X_{n-k}$, where $L$ is the lag operator. Then, using the characteristic function of a Laplace r.v.,

$$ \varphi_{\sum_{k=0}^{N}\rho^k X_{n-k}}(t)=\prod_{k=0}^{N}\varphi_{X_{n-k}}\left(\rho^k t\right)=\exp\left\{-\sum_{k=0}^{N}\ln\left(1+\rho^{2k}\frac{t^2}{\lambda^2}\right)\right\}, $$

which does not converge to the characteristic function of a normal r.v. In fact, for $|t|<\lambda$ (using the Taylor series for $\ln$),

$$ \varphi_{Y_n}(t)=\exp\left\{-\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac{(t/\lambda)^{2k}}{(1-\rho^{2k})} \right\}, $$

which is approximately the c.f. of a normal r.v. for small values of $t$. So, in general, it's hard to find the stationary distribution of $Y$-s given Laplace innovations. The pdf of $Y_n$ ($h_Y$) can be found as the solution of

$$h_Y(x)=\int f_X(x-\rho u)h_Y(u)du$$

via the following recursion

$$h_{Y,n}(x)=\int f_X(x-\rho u)h_{Y,n-1}(u)du$$

starting with some arbitrary pdf $h_{Y,0}$. It can be shown that $h_{Y,n}\rightarrow h$ as $n\rightarrow\infty$.

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However, it seems that there is a typo in the question. For example, in part (a) the authors ask to compute the autocorrelation $R_X(k,j)$ which does not make sense because $X$-s are i.i.d. So, probably their intention was to specify the stationary distribution of $Y$-s. Then you can find the corresponding distribution of innovations by noticing that

$$\varphi_{Y_n}(t)=\varphi_{Y_{n-1}}(\rho t)\varphi_{X_n}(t)$$

which yields

$$\varphi_{X_n}(t)=\frac{1+\rho^2(t/\lambda)^2}{1+(t/\lambda)^2}=\rho^2+(1-\rho^2)\frac{1}{1+(t/\lambda)^2}$$

so that $X_n$ is a mixture of a degenerate r.v. ($\delta_0$) and a $\text{Laplace}(0,\lambda^{-1})$ r.v. with weights $\rho^2$ and $1-\rho^2$, respectively.

Answer 2

This is not an answer but maybe it might generate some ideas:

By a recursive argument one can see that $$ Y_n = \rho^k Y_{n-k} + \sum^k_{i=0}\rho^i X_{n-i}. $$ Since $Y_n$ was defined as stationary we have $|\rho|<1$ so taking $k\rightarrow \infty$ we get $$ Y_n = \sum^{\infty}_{i=0}\rho^i X_{n-i}. $$ Now $$ \mathbb P(Y_n> X_n + 1/2) = \mathbb P(\rho Y_{n-1}> 1/2) =\mathbb P\left( \rho\sum^{\infty}_{i=0}\rho^i X_{i}> 1/2\right) $$

Edit: On the use of CLT:

Note that if we define $Z_i = \rho^i X_i$ then $\text{Var}(Z_i) = \rho^{2i}\text{Var}(X_i) =2 \rho^{2i}$ and hence the sum of their variances is a geometric series$$ s_n^2 := \sum^n_{k=1}2 \rho^{2i} = 2\frac{1-\rho^{2(n+1)}}{1-\rho^2} $$

If $\{Z_i \}$ satisfies Lindeberg's condition then CLT applies and $$ \lim_{n\rightarrow \infty} \frac{\sum^{n}_{i=0}Z_{i}}{s_n} = \frac{\sqrt{1-\rho^2}}{\sqrt{2}}\sum^{\infty}_{i=0}\rho^i X_{i} $$ is a standard Normal rv.