A question related to the convergence of mathematical expectations restricted to an Interval centered on zero

Question

Let $(X_j)_{j= \mathbb 0}^\infty$ a fixed realization of strictly stationary AR(1) process: $$X_j = 0.9 \,X_{j-1}+ \eta_{j}, \quad (\eta_j) \overset{iid}{\sim} N(0,1)$$ For each $n$, consider $B_n\sim Bernoulli(c/n)$, with $c>0$. Suppose $(X_j)_{j= \mathbb 0}^\infty$ and $(B_n)_{n \in \mathbb N}$ are independent. Define $Y_{jn} = X_j Z_n$, where $Z_n = B_n - c/n$. Note that $E[Z_n]=0$ and $Z_n$ take values $1- c/n$ and $-c/n$ with probability $c/n$ and $1- c/n$, respectively. Now define $$Y_n = \sum_{j=0}^n Y_{jn}$$

For each $jn$, let $\mu_{jn}(dx)$ be the probability distribution of $Y_{jn}= X_j Z_n$. Now, consider $r>0$ and $I = (-r,r)$. I want to find a formula for: $$\int_I x \mu_n(dx)= \sum_{j=0}^n \int x \mathbb 1_{I} \mu_{jn}(dx), \quad \mu_n(dx) := \sum_{j=0}^n \mu_{jn}(dx)\label{I}\tag{I}$$ and investigate the almost sure convergence of $\int_I x \mu_n(dx)$, as $n \to \infty$.

First, I tried to find $\int x \mathbb \mu_{jn}(dx)$. For $n$ large enough, we have: $$ \begin{aligned} \int_{I} x \mathbb \mu_{jn}(dx)&= X_j \left(1-\frac{c}{n}\right) \mathbb P\left[-r \leq X_j \left(1-\frac{c}{n}\right) \leq r\right] + X_j (-\frac{c}{n}) \mathbb P[-r \leq X_j (-\frac{c}{n}) \leq r]\\ &=X_j \left(1-\frac{c}{n}\right) \mathbb P\left[\frac{-r}{1-\frac{c}{n}} \leq X_j \leq \frac{r}{1-\frac{c}{n}} \right] - \frac{X_jc}{n} \mathbb P \left[ \frac{r}{-\frac{c}{n}} \leq X_j\leq \frac{-r}{-\frac{c}{n}} \right] \end{aligned} $$ Given that we are dealing with a Gaussian $AR(1)$, I believe that $X_j$ has a normal distribution. For example, see here to see that the distribution of $X_1$ has distribution $N\left(0, \frac{1}{1- (0.9)^2} \right)$. But I'm having trouble finding a formula for (\ref{I}) that allows me to analyze its convergence. Could you help me to finda a formula for (\ref{I}) ?

Answer 1

$$\begin{array}{} Y_n &= &\sum_{j=0}^n Y_{jn}\\ & = &\sum_{j=0}^n X_j Z_n \\ & =& Z_n \sum_{j=0}^n X_j \\ & =& Z_n S_n \\ \end{array}$$

Where in the end I redefined the sum of the $X_j$ as $S_n$.

This makes $Y_n$ a scaled and shifted Bernoulli variable with a point masses of $c/n$ and $1-c/n$ at $(1-c/n) S_n$ and $-c/n S_n$.

Because $Y_n$ is a discrete variable we can replace your integral from $-r$ to $r$ with a sum and it will be equal to

$$S_n (1-c/n)c/n (I_{|(1-c/n) S_n| < r} - I_{|-c/n S_n| < r }) \approx \frac{c S_n}{n} (I_{|(1-c/n) S_n| < r} - I_{|-c/n S_n| < r }) $$

This term $\frac{c S_n}{n}$ is like a 1,1,0 ARIMA process divided by $n$, I would conjecture that this converges in the same way as the random walk converges https://math.stackexchange.com/questions/1099655/