Find variance of unbiased estimator L, where $L = \dfrac\pi4\sqrt{X_1X_2}$.
$f(x) = \dfrac1\theta e^{\frac{-x}{\theta}}$, $x>0$,
$X_1$ and $X_2$ are independent, and exponentially distributed.
Since $L$ is unbiased so I know $E[L] = \theta$, right?
Also, $\operatorname{Var}[x] = E[x^2] - E[x]^2$. But I'm struggling with finding $E[x^2]$. Any help will be appreciated. Thank you.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mathbb{E}\bracks{L} & \equiv \int_{0}^{\infty}{\expo{-x_{1}/\theta} \over \theta} \int_{0}^{\infty}{\expo{-x_{2}/\theta} \over \theta} \pars{{\pi \over 4}\root{x_{1}x_{2}}}\dd x_{1}\,\dd x_{2} \\[5mm] & = {\pi \over 4}\pars{\root{\theta} \int_{0}^{\infty}\expo{-x_{1}/\theta} \,\root{x_{1} \over \theta}\,{\dd x_{1} \over \theta}} \\[2mm] & \phantom{===\,} \pars{\root{\theta} \int_{0}^{\infty}\expo{-x_{2}/\theta}\,\root{x_{2} \over \theta} \,{\dd x_{2} \over \theta}} \\[5mm] & = {\pi \over 4}\ \underbrace{\pars{\int_{0}^{\infty}x^{1/2}\expo{-x}\dd x}^{2}} _{\ds{=\ \Gamma^{2}\pars{3/2}\ =\ {\pi/4}}}\ \theta\ =\ \bbx{{\pi^{2} \over 16}\,\theta} \\[1cm] \mathbb{E}\bracks{L^{2}} & \equiv \int_{0}^{\infty}{\expo{-x_{1}/\theta} \over \theta} \int_{0}^{\infty}{\expo{-x_{2}/\theta} \over \theta} \pars{{\pi \over 4}\root{x_{1}x_{2}}}^{2}\dd x_{1}\,\dd x_{2} \\[5mm] & = {\pi^{2} \over 16}\pars{\theta\int_{0}^{\infty} \expo{-x_{1}/\theta}\,{x_{1} \over \theta} \,{\dd x_{1} \over \theta}} \pars{\theta\int_{0}^{\infty} \expo{-x_{2}/\theta}\,{x_{1} \over \theta} \,{\dd x_{2} \over \theta}} \\[5mm] & = \bbx{{\pi^{2} \over 16}\,\theta^{2}} \\[1cm] \mbox{Var}\pars{L} & = \bbx{{\pi^{2} \over 16}\pars{1 - {\pi^{2} \over 16}}\theta^{2}} \end{align}