Answer

The likelihood of the \(n\) random variables will be

\[ \begin{aligned} L &= \prod_{i=1}^{n} f_{X}(x_{i})\newline &= \bigg( \frac{2}{\lambda} \bigg)^{n} \prod_{i=1}^{n} X_{i} \exp \bigg( -\frac{1}{\lambda} \sum_{i=1}^{n} X_{i}^{2} \bigg)\newline \ln L &= -n \ln \bigg( \frac{\lambda}{2} \bigg) = \ln \bigg( \prod_{i=1}^{n} X_{i} \bigg) - \frac{1}{\lambda} \sum_{i=1}^{n} X_{i}^{2}\newline \frac{\partial \ln L}{\partial \lambda} &= -\frac{n}{\lambda} + \frac{1}{\lambda^{2}}\sum_{i=1}^{n} X_{i}^{2} = 0\newline \lambda_{MLE} &= \frac{1}{n} \sum_{i=1}^{n} X_{i}^{2} \end{aligned} \]

To prove that it is an unbiased estimator, we need to show \(E[\lambda_{MLE}] = \lambda\)

\[ \begin{aligned} E[X^{2}] &= \int_{0}^{\infty} \frac{2x^{3}}{\lambda} \exp \bigg(-\frac{x^{2}}{\lambda} \bigg) dx\newline &= \int_{0}^{\infty} \lambda t e^{t} dt \: \text{putting $x^{2}/\lambda = t$}\newline &= \lambda\newline E[\lambda_{MLE}] &= \frac{1}{n} \sum_{i=1}^{n} E[X_{i}^{2}] = \lambda \end{aligned} \]

To show that this is a sufficient statistic, we try to factorize the joint density function

\[ \begin{aligned} T(X) &= \frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}\newline L &= \prod_{i=1}^{n} f_{X}(x_{i})\newline &= \bigg( \frac{2}{\lambda} \bigg)^{n} \prod_{i=1}^{n} X_{i} \exp \bigg( -\frac{1}{\lambda} \sum_{i=1}^{n} X_{i}^{2} \bigg)\newline &= \bigg( \prod_{i=1}^{n} X_{i} \bigg) \bigg( \bigg( \frac{2}{\lambda} \bigg)^{n} \exp \bigg( -\frac{n}{\lambda} T(X) \bigg) \bigg)\newline &= h(X) g(T(X), \lambda) \end{aligned} \]

i.e., we were able to factorize the joint density into two functions, one which is only dependent on the sample, and the other dependent on the parameter \(\lambda\) and on the sample through the statistic \(\lambda_{MLE}\). By Fisher Neyman Factorization Theorem, \(\lambda_{MLE}\) is a sufficient statistic.