variance of minimum of squared exponential random variable Given $Y_1 $ to $Y_n$  are  exponential r.v's with mean $\theta$ find $\operatorname{var}[\min(Y^2 )]$ with the help of gamma distribution.
attempt:
$\min(Y) $ is exponential with $(\text{mean} = \theta/n)$
i know that sum of $n$ exponentials is gamma with $\alpha = n$ and $\beta = \text{mean}$
The squared of $Y$ is throwing me off. Thanks for help
 A: Hints;


*

*For every nonnegative random variable $Z$, 
$$
E(Z)=\int_0^\infty P(Z\geqslant z)\,\mathrm dz,\qquad E(Z^2)=\int_0^\infty 2z\,P(Z\geqslant z)\,\mathrm dz.
$$

*For every independent random variables $(Z_k)_{1\leqslant k\leqslant n}$, the random variable $Z=\min\{Z_k\mid1\leqslant k\leqslant n\}$ is such that, for every $z$,
$$
[Z\geqslant z]=\bigcap_{k=1}^n[Z_k\geqslant z],
$$
hence, if furthermore the random variables $(Z_k)_{1\leqslant k\leqslant n}$ are identically distributed,
$$
P(Z\geqslant z)=P(Z_1\geqslant z)^n.
$$

*For every nonnegative random variable $Y$ and every nonnegative $z$,
$$
[Y^2\geqslant z]=[Y\geqslant\sqrt{z}].
$$

*If $Y$ is exponential with mean $\theta$, then, for every nonnegative $y$,
$$
P(Y\geqslant y)=\mathrm e^{-y/\theta}.
$$


Applying these facts to $Z=\min\{Y^2_k\mid1\leqslant k\leqslant n\}$ with $(Y_k)_{1\leqslant k\leqslant n}$ i.i.d. exponential with mean $\theta$, one gets for example
$$
E(Z^2)=\int_0^\infty 2z\,\mathrm e^{-n\sqrt{z}/\theta}\,\mathrm dz.
$$
The change of variable $z=\theta^2t^2/n^2$ yields
$$
E(Z^2)=\frac{4\theta^4}{n^4}\int_0^\infty t^3\,\mathrm e^{-t}\,\mathrm dt=\frac{24\theta^4}{n^4}.
$$
