Mean of a Reciprocal of a Normal Distribution

I have a random variable, $\eta$, with a normal distribution with zero mean and unity variance. I process this through the function $$f(x)=\frac{1}{1-x}$$

I thought I should be able to replace this exactly with the Taylor series, giving me $$f(x)=\sum_{i=0}^\infty x^i$$

and I would be able to approximate the function with fewer terms if $x$ is sufficiently small.

I want to see how the infinite sum behaves when I calculate the mean of the random variable $$\nu=f(\eta)$$ using the Taylor series expansion.

So I have the expected value $$E(\nu) = E(f(\eta)) = E(\sum_{i=0}^\infty \eta^i) = \sum_{i=0}^\infty E(\eta^i)$$

Given that $\eta$ is has a normal distribution with zero mean, then all of the terms with even $i$ are zero and all terms with odd $i$ are $$E({\eta^i})=\sigma^i (i-1)!!$$

So I could rewrite the calculation of the mean as $$E(\nu) = 1+\sum_{n=1}^\infty \sigma^{2n} (2n-1)!!$$

To see if this converges, the ratio test is used and the ratio of the $n+1$ term to the $n$ term is $\sigma^2 (2n+1)$, which says this does not converge in any practical sense.

What bothers me about this conclusion is that I am seeing convergence when I do a Monte Carlo simulation of a problem involving this type of ratio distribution.

Where did I go wrong in my assumptions or analysis?

• Indeed, $(1-\eta)^{-1}$ is not integrable since the PDF of $\eta$ is bounded below by a positive constant in a neighbourhood of $1$. To go further, you might want to make "I am seeing convergence when I do a Monte Carlo simulation of a problem involving this type of ratio distribution" much more precise. – Did Jan 8 '16 at 23:16
• @Did Could you elaborate on your first point? In my simulations, I am seeing systematic bias in the sample mean value as I increase the variance. I thought I could explain this from the presence of the even terms, but clearly there is something I don't understand well enough. – Jim Jan 9 '16 at 0:49
• As Did points out, the expectation diverges. However, you can get meaningful results if you were to instead evaluate $E[f(\nu) | \ |\nu|<1-\epsilon)$ for each $\epsilon>0$. – Alex R. Jan 9 '16 at 0:54
• @AlexR. I think I'm beginning to see. The simulation and real values may approximate normal distributions but they aren't landing close to 1, where the expression is undefined. And that would be why I'm not seeing divergence in these cases. – Jim Jan 9 '16 at 1:41
• After giving it some thought, I'd like to replace the normal distribution with something similar, but that doesn't go to where the pdf is zero outside a limited range. I want to model something realistic (this is related to an electronic device that does have a limited range of output), but I'm looking for a model that is amenable to the approach I'm with the Taylor series, where the moments are well understood. Does anyone have any suggestions? – Jim Jan 9 '16 at 3:52

If your simulation is suggesting convergence then it may be a coincidence of that particular run.

Here is a particular simulation with $1$ million cases using R, looking at the cumulative sample mean $\frac1n \sum{f(x_i)}$. Looking at the first half, it might hint at convergence despite having $x_{128031}\approx 1.000002$, but then simulation $x_{780491}$ was about $0.999999$ so $f(x_{780491})\approx 10^6$ and this substantially changed the simulated sample mean.

set.seed(2016)
cases <- 1000000
x <- rnorm(cases, mean=0, sd=1)
f <- 1/(1-x)
m <- cumsum(f)/(1:cases)
plot(m, ylim=c(-2,2), type="l", col="red")
abline(h=0)

In general, values of $x_i$ close to $1$ can happen more often than the smoothing effect of sample averaging.