<p>If a stationary distribution exists, the time index no longer matters as t goes to infinity. Replace x_{t+1} and x_{t} by x and solve for x. What you obtain is a random variable whose distribution you can infer. In your case you get x = b/(1-a) N(0,1) which has normal distribution with mean 0 and variance (b/(1-a))^2.</p>

CC BY-SA 3.0
2018-01-30T02:21:30.960Z
2018-01-30T02:21:30.960Z
braaterAfrikaaner
<p>How can I find the stationary distribution (as t goes to infinity) of stochastic difference equations in the form:</p>

<p>$x_{t+1} = a*x_t + b*N(0,1)$</p>

<p>where N(0,1) is a standard normal pdf</p>

<p>I have numerical results for a few examples I'm working through from notes, but I'm having trouble following the analytic derivation. There must be a simpler solution...</p>

CC BY-SA 3.0
2018-01-29T14:35:11.633Z
Are you sure it's not "as $t$ goes to infinity"?
Herr K.
CC BY-SA 3.0
2018-01-29T19:26:25.033Z
Yes, that is what I meant. My apologies for the error.
user14631
<econometrics><statistics><stochastic-processes>
Stochastic process difference equation: stationary distribution
user14631
# Stochastic process difference equation: stationary distribution

(Asked by: username_0 on 2018-01-29)

How can I find the stationary distribution (as t goes to infinity) of stochastic difference equations in the form:

$x\_{t+1} = a\*x\_t + b\*N(0,1)$

where N(0,1) is a standard normal pdf

I have numerical results for a few examples I'm working through from notes, but I'm having trouble following the analytic derivation. There must be a simpler solution...

**Tags:** econometrics, statistics, stochastic-processes

> **Comments:**
> **username_1**: Are you sure it's not "as $t$ goes to infinity"?  
> **username_0**: Yes, that is what I meant. My apologies for the error.  

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## 1 answers

### Answer by username_2 on 2018-01-30. Score: 0

If a stationary distribution exists, the time index no longer matters as t goes to infinity. Replace x\_{t+1} and x\_{t} by x and solve for x. What you obtain is a random variable whose distribution you can infer. In your case you get x = b/(1-a) N(0,1) which has normal distribution with mean 0 and variance (b/(1-a))^2.


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