[Math] Maximum likelihood estimator of $\lambda$ and verifying if the estimator is unbiased

Question

$(X_1,\ldots,X_n)$ is a random sample extracted from an exponential law of parameter $\lambda$

Calculate the likelihood estimator $\nu$ of $\lambda$.

Then, if $n=2$: establish if $\nu$ is a unbiased estimator

$$L(\lambda: X_1,\ldots,X_n)=\prod_{i=1}^n \lambda \ e^{-\lambda \ x_i} \ \ 1_{(0,+\infty)} \ (x_i)=$$

$$=\prod_{i=1}^n (1_{(0,+\infty)} \ (x_i)) \ \ \lambda^n \ e^{-\lambda \sum_{i=1}^n x_i} \ \ $$

$$\frac{\partial}{\partial \lambda} L(\lambda: X_1,…,X_n)=\prod_{i=1}^n (1_{(0,+\infty)} \ (x_i)) \ \ n \lambda^{n-1} \ e^{-\lambda \sum_{i=1}^n x_i}-\sum_{1=1}^n x_i \ \lambda^n \ \ e^{-\lambda \sum_{i=1}^n x_i}= \ \ $$

$$=\prod_{i=1}^n (1_{(0,+\infty)} \ (x_i)) \ \ \lambda^{n-1} \ e^{-\lambda \sum_{i=1}^n x_i} \ \ (n- \lambda \sum_{i=1}^n x_i) $$

$$\frac{\partial}{\partial \lambda} L(\lambda: X_1,\ldots,X_n) \ge 0 \Longleftrightarrow \lambda \le \frac{1}{\overline{X}}$$

Maximum likelihood estimator of $\lambda$ is $\nu=\frac{1}{\overline{X}}$

If $n=2$, I think that:

$$\nu=\frac{2}{\sum_{i=1}^2 X_i}$$

and

$$\sum_{i=1}^2 X_i \sim \Gamma(2, \lambda)$$

How can I establish if $\nu$ is a unbiased estimator?

Thanks!

Answer 1

\begin{align} & \operatorname E\left( \frac n {X_1+\cdots+X_n} \right) \\[8pt] = {} & \int_0^{+\infty} \frac n x \cdot f_{X_1+\cdots+X_n}(x)\, dx \\[8pt] = {} & \int_0^{+\infty} \frac n x \cdot \frac 1 {\Gamma(n)} \cdot (\lambda x)^{n-1} e^{-\lambda x} (\lambda\, dx) \\[8pt] = {} & \frac{n\lambda}{\Gamma(n)} \int_0^{+\infty} (\lambda x)^{n-2} e^{-\lambda x} (\lambda\,dx) = 2\lambda. \\[8pt] = {} & \frac{n\lambda \Gamma(n-1)}{\Gamma(n)} = \frac{n\lambda}{n-1}. \end{align}