Recursive Variance

Question

What will be the distribution or features about the following $x$?

$x=\mu+\epsilon$

where $\epsilon\sim N(0,x^{-1})$.

It seems interesting in econometrics if we allow $x$ being a time series and $x_t\sim N(\mu,x_{t-1}^{-1})$ which is called autoregressive conditional heteroskedasticity. But what if we cancel $t$ and make it purely recursive?

Answer 1

With trial answers to some of the questions in my own comment, suppose we begin the time series at $t = 1$ with $X_1 = \mu = 50$ and that the SD of the normal distribution is $1/X_{t-1}$.

Then the iteration seems to converge quickly to $Norm(\mu = 50,\; \sigma = 1/\mu = 1/50).$ Similarly for $\mu=5$. A According to a Kolmogorov-Smirnov goodness-of-fit test the simulated data are consistent with a normal distribution with these parameters. The first 5000 simulated observations are also consistent with a normal distribution according to a Shapiro-Wilk test. A plot of autocorrelations shows no significant autocorrelations up to lag 40.

If my choices are at all realistic in your situation, it seems to me that the univariate version of this idea must be simply $X \sim Norm(\mu, 1/\mu).$

Below is a simulation in R (ACF plot not shown):

 mu = 50;  m = 10^4;  x = numeric(m);  x[1] = mu
 for(t in 2:m) x[t] = rnorm(1, mu, 1/abs(x[t-1]))
 mean(x);  sd(x)
 ## 49.99985
 ## 0.02007789

 ks.test(x, pnorm, mu, 1/mu)

 One-sample Kolmogorov-Smirnov test:  data:  x 
 D = 0.0082, p-value = 0.5055      # consistent with NORM(50, 1/50)
 alternative hypothesis: two.sided 

 shapiro.test(x[1:5000])

 Shapiro-Wilk normality test:  data:  x[1:5000] 
 W = 0.9996, p-value = 0.4416      # consistent with normal