# Unbiased estimator of the reciprocal of a truncated normal distribution

I have a random variable $$y$$, which is a function of $$x$$, i.e. $$y = \frac{1}{x} - 1$$, where $$x$$ is a random variable with a truncated normal distribution with mean $$mu$$ and variance $$\sigma^2$$. In particular $$x$$ cannot be larger than 1, and less than or equal to some positive value $$b$$, and would be normalized accordingly. A possible estimator could be $$\frac{1}{\mu} - 1$$ but it is biased. What is the error distribution and variance (or approximate variance) of the estimator?

• Estimator of what? – Dave Oct 1 '20 at 0:23
• I am after an estimator of the value of y. x is the result of another estimation process that has a mean of mu, but has a truncated normal error distribution. I would expect the value of $y$ would be $\frac{1}{\mu} -1$, but if I do a monte carlo simulation of the $x$ errors, the mean of $\frac{1}{x} - 1$ is not equal to $\frac{1}{\mu} -1$. – Obromios Oct 1 '20 at 0:32
• You can always derive an unbiased numerical based mean estimate. Use a Monte Carlo generation formula to simulate a normal, which you appropriately truncate yielding x. Compute -1 + 1/x as your random deviate. Repeat n times and record the mean estimate. – AJKOER Oct 1 '20 at 1:00
• But $y$ isn’t a value; it’s a distribution. Do you mean that you want to estimate the mean of $y$? In that case, the sample mean is an unbiased estimator of the population mean whenever the population mean exists. – Dave Oct 1 '20 at 1:04
• Dave, I do want want to estimate the mean of y, so AKOER's comment is the way to go, and the naive point estimator i.e. $\frac{1}{\mu} -1$ is biased. From the monte carlo generation, I can get the error distribution and variance. I was hoping for a more analytical approach, but this works for me. Do one of you want to do up an answer and I will accept it, or I can add the answer? Thank you both. – Obromios Oct 1 '20 at 1:18