If $ X = \sqrt{Y_{1} Y_{2}} $, then find a multiple of $ X $ that is an unbiased estimator for $ \theta $. 
Problem: Suppose that $ (Y_{1},Y_{2},Y_{3},Y_{4}) $ denotes a random sample of size $ 4 $ from a population with an exponential distribution whose probability density function $ f $ is given by
  $$
f(y) =
\begin{cases}
\dfrac{1}{\theta} e^{− \frac{y}{\theta}} & \text{if $ y > 0 $}; \\\\
0                                        & \text{elsewhere}.
\end{cases}
$$
  Let $ X = \sqrt{Y_{1} Y_{2}} $. Then find a multiple of $ X $ that is an unbiased estimator for $ \theta $.
Hint: Use your knowledge of the gamma distribution and the fact that $ \Gamma \! \left( \dfrac{1}{2} \right) = \sqrt{\pi} $ to find $ \mathsf{E} \! \left( \sqrt{Y_{1}} \right) $. Recall that the $ Y_{i} $’s are independent random variables.

I am told that $ \mathsf{E} \! \left( \sqrt{Y} \right) = \dfrac{1}{2} \sqrt{\pi \theta} $. Could anyone help me with the intermediate steps?
Thanks!
 A: We want to calculate $E(\sqrt{Y_1Y_2})$. So it is enough to calculate $E(\sqrt{Y})$, where $Y$ is exponential with mean $\theta$. Thus we want 
$$\int_0^\infty \frac{\sqrt{y}}{\theta}e^{-y/\theta}\,dy.\tag{A}$$
Let $\frac{y}{\theta}=z^2/2$. Then $dy=\theta z\,dz$, and $\sqrt{y}=z\sqrt{\theta}/\sqrt{2}$. Thus we want 
$$\int_0^\infty \frac{1}{\sqrt{2}}\sqrt{\theta}z^2e^{-z^2/2}\,dz.\tag{1}$$
Rewrite this as 
$$\frac{1}{2}\sqrt{\pi}\sqrt{\theta}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} z^2e^{-z^2/2}\,dz.$$
We recognize the integral as $E(Z^2)$, where $Z$ is standard normal. Thus Integral (1) is equal to $\frac{1}{2}\sqrt{\pi\theta}$.  
Now computing the appropriate constant should be straightforward. 
Added: If we want to use knowledge of the Gamma distribution with parameter $k=1/2$ and $\theta$, we can proceed as follows.
If $W$ is a random variable with this distribution, then $W$ has density function
$$\frac{1}{\Gamma(1/2)\theta^{1/2}}x^{-1/2}e^{-x/\theta}$$
on $(0,\infty)$. 
We know that $E(W)=k\theta=\frac{\theta}{2}$. This tells us that
$$\frac{\theta}{2}=\int_0^\infty\frac{1}{\Gamma(1/2)\theta^{1/2}}x^{1/2}e^{-x/\theta}\,dx.$$ 
Using this equation, we can readily evaluate $\int_0^\infty \frac{\sqrt{y}}{\theta}e^{-y/\theta}\,dy$. 
Another way to evaluate the integral by using facts about Gamma distributions is to use the fact that the Gamma distribution with parameters $k=3/2$, $\theta$ has density function
$$\frac{1}{\Gamma(3/2)\theta^{3/2}}x^{1/2}e^{-x/\theta}$$
on $(0,\infty)$. Thus the integral from $0$ to $\infty$ of the above function is $1$. 
Some detail: We already saw, in the main answer, that the integral (A) is equal to $\frac{1}{2}\sqrt{\pi\theta}$. It follows that $E(X)=E(\sqrt{Y_1Y_2})=\frac{\pi}{4}\theta$. Thus the constant $c$ such that $cX$ is an unbiased estimator of $\theta$ is given by $c=\frac{4}{\pi}$.

Now we go back to the already solved problem of evaluating the integral (A). The change of variable $y=u\theta $ transforms the integral to 
$$\sqrt{\theta}\int_0^\infty u^{1/2}e^{-u}\,du.$$
We recognize this as $\sqrt{\theta}\Gamma(3/2)$.  Note that in general $\Gamma(z+1)=z\Gamma(z)$. (This is a one step integration by parts.) So $\Gamma(3/2)=\frac{1}{2}\Gamma(1/2)=\frac{1}{2}\sqrt{\pi}$.

We can hide the integration by parts, as mentioned earlier, by quoting a result about the Gamma distribution, parameters $k=\frac{1}{2}$ and $\theta$.    
From the material on the Gamma with parameters $\frac{1}{2}$, $\theta$ obtained earlier, and quoting a result about the mean of this Gamma, we had
$$\frac{\theta}{2}=\int_0^\infty\frac{1}{\Gamma(1/2)\theta^{1/2}}x^{1/2}e^{-x/\theta}\,dx.$$ 
Multiply both sides by $\theta^{3/2}\Gamma(1/2)$, that is, by $\theta^{3/2}\sqrt{\pi}$. On the right, we obtain the integral (A) we have been evaluating in various ways. On the left we obtain $\frac{1}{2}\sqrt{\pi}\theta^{1/2}$.
Remark: All these integration results are closely related. For example, the usual way we find $\Gamma(1/2)$ is to make the change of variable $x=t^2$, and then use the polar coordinates trick to evaluate $\int_{-\infty}^\infty e^{-t^2}\,dt$. 
