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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}) ?

André Goulart
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  • Your question is unclear. Are you trying to ask why (I) holds (of course, the "$n \to \infty$" in (I) is not compatible there) or something else? – Zhanxiong Oct 23 '23 at 19:06
  • Note that for each $n$, I have the integral $\int_{I}x dF_{Y_n}(x)$. Since this integral depends on $n$, as the Bernoulli parameter is $c/n$, I believe it is possible to analyze the convergence of this integral as $n$ tends to infinity. That's why I want to find a specific formula for $\int_{I}x dF_{Y_n}(x)$, in order to analyze its convergence. – André Goulart Oct 23 '23 at 19:48
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    Because you mention "truncation" in the title but these are not truncations of random variables (they are mixtures with atoms at zero), are you sure your mathematical formulation corresponds to what you are thinking of? – whuber Oct 24 '23 at 19:36
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    In order for people that are not so great at reading text to more easily read the question it might be better to start the post with a simple introduction instead of a bombardement with definitions and considerations without having first a background or goal that helps reading trough very long text. – Sextus Empiricus Oct 24 '23 at 21:18
  • "let $\mu_{jn}(dx)$ be the probability distribution of $Y_{jn}= X_j Z_n$" this is unclear. What sort of probability distribution are you refering to? What is $dx$ doing there as input of the function and how do you think that you can integrate it? – Sextus Empiricus Oct 25 '23 at 09:01
  • Your post is full of confusing definitions and equations. Could you express what you want in more simple words instead of verbose and confusing mathematical language? It almost seems as if you are playing around with chatGPT to create a purposefullly confusing question. – Sextus Empiricus Oct 25 '23 at 09:03
  • Dear, in principle $Z$ is a discrete random variable. You may be bothered because I am using the notations as if $Z$ were a continuous random variable, but I do this because in my original problem, $Z$ could be a continuous random variable. But this mathematically shouldn't be a problem, because even if a Bernoulli is discrete, nothing prevents me from calculating $P(Z \leq t)$, for $t$ a real number. In this case, $\mu_{jn}(dx)$ is a discrete probability measure. Thanks for the suggestions. At the moment, my time zone doesn't allow me to improve the question, but I'll do as much as possible. – André Goulart Oct 25 '23 at 09:10
  • Which values does $dx$ take? – Sextus Empiricus Oct 25 '23 at 09:13
  • I don't know if this can help. Suppose $U \sim Bernoulli(p)$ taking two values 0 and 1. The probability distribution of $U$ is discrete: $P(U=1)=p = 1- P(U=0)$. I am using continuous random variables notation. Denote by $\mu$ the probability distribution of $U$. So, the expectation is $\int_{R} x \mu(dx)= 1 P(U=1) + 0 (U=0)$. This is basically what I'm doing, with the difference that my discrete variable $Z$ involves a realization of an AR(1). – André Goulart Oct 25 '23 at 09:28
  • I believe that this is more typically written as $$\int_{R} x d\mu(x)= 1 P(U=1) + 0 (U=0)$$ – Sextus Empiricus Oct 25 '23 at 09:31
  • The use of $dx$, the differential of the variable $x$, as an input to a function, is not a common style of writing equations. – Sextus Empiricus Oct 25 '23 at 09:33

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

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/

Sextus Empiricus
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  • Although, $(X_j){j=0}^{\infty}$ is a fixed realization, I believe we can calculate the probabilities $\mathbb P\left[\frac{-r}{1-\frac{c}{n}} \leq X_j \leq \frac{r}{1-\frac{c}{n}} \right]$ and $P \left[ \frac{r}{-\frac{c}{n}} \leq X_j\leq \frac{-r}{-\frac{c}{n}} \right]$. What I'm saying is that we can study the almost sure convergence of the integral $\int{I} x \mathbb d F_{Y_{n}}(x)$ – André Goulart Oct 25 '23 at 02:50
  • @AndréGoulart what sort of distribution does $F_{Y_n}(x)$ represent according to you? – Sextus Empiricus Oct 25 '23 at 07:58
  • Dear, thank you for your comment. I made a huge confusion. For a moment, I mixed up the definitions of sums of probability measures with their convolution. I have now edited the question correctly, please see the latest update. If you have any further questions, please let me know. Thank you. – André Goulart Oct 25 '23 at 08:56
  • Dear, thank you for your answer. I'm not very familiar with Brownian Motion, but I can conclude that $S_n/n \to E[X_0]=0$ a.s. due to the ergodicity of AR(1). I really learned a lot from your answer. Given that the indicator function is obviously bounded and $S_n/n \to E[X_0]=0$ a.s., I can now conclude that the integral $\int_I x d\mu_{n}(x)$ converges a.s. to zero. Thanks. – André Goulart Oct 26 '23 at 06:40
  • @AndréGoulart I made a little mistake, it is not a Brownian motion, an I(1) process, but instead slightly more complex and an ARIMA process. I guess that the conclusion will remain the same. – Sextus Empiricus Oct 26 '23 at 08:29