Let a random variable X has a following distribution:
$$f_X(x)=\frac{x}{\theta}\exp(\frac{-x^2}{2\theta}),x>0.$$
It can be shown that $\mathbb E(X)=(\theta\pi/2)^{(1/2)},$ $\mathbb E(X^2)=2 \theta$ and $\mathbb E(X^4)=8 \theta^2$ .
I have to find method of moments(MoM) estimator for $\theta$ and the bias of the MoM estimator.
I found MoM estimator for $\theta$ to be:
$$\bar X=(\tilde\theta\pi/2)^{(1/2)}$$ $$\Rightarrow \bar X^2=\tilde\theta\pi/2$$ $$\Rightarrow \tilde\theta=2\bar X^2/\pi.$$
But I am having difficulty to find the bias. I tried as follows:
Bias of MoM $=\mathbb E(\tilde\theta)-\theta=\mathbb E(2\bar X^2/\pi)-\theta=\frac{2}{\pi}\mathbb E(\bar X^2)=\frac{2}{\pi}\mathbb E[(1/n\sum_{i=1}^{n}X_i)^2]=\frac{2}{\pi n^2} \mathbb E[(\sum_{i=1}^{n}X_i)^2]$
Could you please tell me how can I simplify the $E[(\sum_{i=1}^{n}X_i)^2]$ term?