# Variance of minimum of N random variables

Let $X_1,X_2,\dots,X_N$ be i.i.d. random variables with support $[0,M]$, and with density and distribution functions $f_X(x)$ and $F_X(x)$ respectively. Given $Y= \min_{i\in \{1,\dots,N\}} X_i$ , how does one obtain the variance of $Y$?

My approach so far:

$F_Y(y) = 1 - [1-F_X(y)]^N$, hence $f_Y(y) = Nf_X(y)[1-F_X(y)]^{N-1}$.

So we have

\begin{align} E[Y^2] &= \int_0^M y^2 Nf_X(y)[1-F_X(y)]^{N-1} dy\\ &= \Big[-y^2 [1-F_X(y)]^{N}\Big]_0^M + \int_0^M 2y [1-F_X(y)]^{N} dy\\ &= 2\int_0^M y [1-F_X(y)]^{N} dy \end{align}

and similarly

\begin{align}E[Y] &= \int_0^M y Nf_X(y)[1-F_X(y)]^{N-1} dy\\ & = \Big[-y [1-F_X(y)]^{N}\Big]_0^M + \int_0^M [1-F_X(y)]^{N} dy\\ &= \int_0^M [1-F_X(y)]^{N} dy. \end{align} Hence,

$$\text{var}(Y) = E[Y^2] - (E[Y])^2 = 2\int_0^M y [1-F_X(y)]^{N} dy - \Big[\int_0^M [1-F_X(y)]^{N} dy\Big]^2.$$

I feel something is wrong about the final expression I have obtained for $\text{var}(Y)$. The first term depends $2\int_0^M y [1-F_X(y)]^{N} dy$ on the magnitude of the values of $y$, while the second term $\int_0^M [1-F_X(y)]^{N} dy$ doesnt depend on the magnitude of the values of $y$ (it just integrates probabilities over the entire support of $y$). Hence, I feel that by scaling $Y$, the first term can be made arbitrarily small, while the second term remains constant. This could result in a negative variance, which is impossible! Could anyone tell me if I am going wrong, and if so, where I'm going wrong?

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Note that scaling $Y$ does scale the variance of $Y$. Indeed, $Var(aY) = |a|^2 Var(Y)$. – nullUser Jul 13 '13 at 20:03
Your expression behaves well under scaling. You kind of forgot about the $dy$, maybe because it is so small. But if we let $y=kz$, then in the expectation of the square we get a $(kz)(k\,dz)$ and in the square of the expectation we get a $(k\,dz)^2$, so each gets scaled by $k^2$. – André Nicolas Jul 13 '13 at 20:43
Could anyone tell me if I am going wrong, and if so, where I'm going wrong? To sum up, your computations are perfectly valid and yield the correct result but your interpretation of them is slightly erroneous. – Did Jul 14 '13 at 9:11