In my STAT 210A class, we frequently have to deal with the minimum of a sequence of independent, identically distributed (IID) random variables. This happens because the minimum of IID variables tends to play a large role in sufficient statistics.

In this post, I’ll briefly go over how to find the minimum of IID uniform random variables. Specifically, suppose that $X_1,\ldots,X_n \sim {\rm Unif}(0,1)$ and let $T = \min_i X_i$. How do we find $\mathbb{E}[T]$? To compute this, observe that

\begin{align} P\left(\min_i X_i \le t\right) &= 1 - P\left(\min_i X_i \ge t\right) \\ &= 1 - P(X_1 \ge t, \ldots, X_n \ge t) \\ &= 1 - \prod_{i=1}^n P(X_i \ge t) \\ &= 1 - \prod_{i=1}^n [1 - P(X_i \le t)] \\ &= 1 - [1 - P(X_i \le t)]^n, \end{align}

so the next step is to determine $P(X_i \le t)$. Due to the IID assumption, it doesn’t matter which $X_i$ we use. Note also that to avoid adding cumbersome indicator functions, assume that $t \in [0,1]$.

The value $P(X_i \le t)$ is easier to compute because it directly relates to the cumulative distribution function (CDF) of a uniform random variable. For ${\rm Unif}(0,1)$, the CDF is simply $F_X(x) = x$ if $x \in (0,1)$. This means

$P\left(\min_i X_i \le t\right) = 1 - [1 - P(X_i \le t)]^n = 1 - [1 - t]^n,$

so that by differentiating with respect to $t$, the density function is $f_T(t) = n(1-t)^{n-1}$ (the notation here with the $T$ as the subscript is to denote the variable whose density is being defined).

The reason why we need the density is because of the definition of the expectation:

$\mathbb{E}[T] = \int_0^1 t f_T(t) dt = n \int_0^1 t(1-t)^{n-1}dt.$

To compute this, we integrate by parts. Set $u=t, du=dt$ and $dv = (1-t)^{n-1}dt, v = -(1-t)^{n}/n$. We get

\begin{align} \int_0^1 t(1-t)^{n-1}dt &= \left(-\frac{t(1-t)^n}{n} \Big|_{t=0}^{t=1}\right) - \int_0^1 -\frac{(1-t)^n}{n}dt \\ &= \left[-\frac{(1)(1-(1))^n}{n} - -\frac{(0)(1-(0))^n}{n} \right] + \frac{1}{n} \int_0^1 (1-t)^ndt \\ &= \frac{1}{n} \left[ -\frac{(1-t)^{n+1}}{n+1}\Big|_{t=0}^{t=1}\right] \\ &= \frac{1}{n} \left[ -\frac{(1-(1))^{n+1}}{n+1}- -\frac{(1-(0))^{n+1}}{n+1}\right] \\ &= \frac{1}{n} \frac{1}{n+1}. \end{align}

Combining this with the extra $n$ factor we left out (you didn’t forget that, did you?) we get $\mathbb{E}[T] = \frac{1}{n+1}$. Notice that as $n \to \infty$ the expected value of the minimum of these uniform random variables goes to zero. In addition, this expectation is always in $(0,1/2]$ for $n \ge 1$. Thus, the answer passes the smell test and seems reasonable.