Sufficient Estimator How to proceed?

Question

Let $X$ be a random variable with exponential density function. Show that the mean of $X$, denoted $\overline{X}$, is the sufficient estimator of $\lambda$ but not an unbiased estimator of $\lambda$

Answer 1

If your sample $X_1,\ldots,X_n$ is i.i.d. Exp($\lambda$), you have that its joint density is $f(x_1,\ldots,x_n) = \lambda^n e^{-\lambda \sum_i x_i} $ on the non-negative orthant. By the Neyman factorization theorem, we can factor the joint density as $(\lambda^n e^{-\lambda \sum_i x_i} ) (1)$ where the first function captures all the dependency on $\lambda$ and the second one is solely a function of $x_1,\ldots,x_n$. Thus, $\sum_i x_i$ is a sufficient statistic for $\lambda$. Since there exists a bijection between $\sum_i x_i $ and $\frac{1}{N} \sum_i x_i = \bar{x}$, $\bar{x}$ is also a sufficient statistic for $\lambda$.

To show that it is a biased estimator of $\lambda$, we simply calculate $E[\bar{X}] - \lambda$ and show it is non-zero. In this case, $E[\bar{X}] = \frac{1}{N} \sum_i E[X_i] = \frac{1}{N} \sum_i \frac{1}{\lambda} = \frac{1}{\lambda}$. And, $frac{1}{\lambda} - \lambda$ is not zero.