Let $X_1,X_2,…,X_n$ be independent random variables, each distributed as $Exp(\lambda)$. I am given 2 estimators of $\lambda$:
$$\hat{\lambda_1}=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{X_i}$$
$$\hat{\lambda_2}=\frac{n}{\sum_{i=1}^{n}X_i}$$
I have to calculate the bias of each estimator.
$$Bias_1=E(\hat{\lambda_1})-\lambda=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{\lambda^{-1}}-\lambda=\frac{1}{n}n\lambda-\lambda=\lambda-\lambda=0$$
$$Bias_2=E(\hat{\lambda_2})-\lambda=\frac{n}{\frac{n}{\lambda}}-\lambda=0$$
Used properties:
- $E(X_i)=\lambda^{-1}$
- $E(\sum_{i=1}^{n}X_i)=\frac{n}{\lambda}$ as the sum of iid exponential variables has a gamma distribution with parameters $n,\lambda$
However, in the next part, I am told that $\hat{\lambda_2}$ is biased and I have to modify it in order to obtain an unbiased estimator. What did I do wrong in bias calculation?
Best Answer
You appear to be assuming that $E[g(Z)]=g(E[Z])$ for any (measurable) function $g$ but this is not true unless $g$ is linear.
For the first estimator the bias is not well defined since $1/X_i$ has an inverse exponential distribution, whose expected value does not exist.
For the second estimator, it helps to note $\sum_{i=1}^n X_i\sim$ Gamma$(n,\lambda)$ (using the $(\alpha,\beta)$ shape-rate parametrization), so $1/\sum_{i=1}^n X_i\sim $Inv-Gamma$(n,\lambda).$