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

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

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:

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

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.