variance of minimum of squared exponential random variable

Question

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

Answer 1

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 $$ E(Z^2)=\frac{4\theta^4}{n^4}\int_0^\infty t^3\,\mathrm e^{-t}\,\mathrm dt=\frac{24\theta^4}{n^4}. $$