# Expectation of Minimum of nn i.i.d. uniform random variables.

$X_1, X_2, \ldots, X_n$ are $n$ i.i.d. uniform random variables. Let $Y = \min(X_1, X_2,\ldots, X_n)$. Then, what’s the expectation of $Y$(i.e., $E(Y)$)?

I have conducted some simulations by Matlab, and the results show that $E(Y)$ may equal to $\frac{1}{n+1}$. Can anyone give a rigorous proof or some hints? Thanks!

To calculate the expected value, we’re going to need the density function for $Y$. To get that, we’re going to need the distribution function for $Y$. Let’s start there.

By definition, $F(y) = P(Y \leq y) = 1 - P(Y > y) = 1 - P(\text{min}(X_1, \ldots, X_n) > y)$. Of course, $\text{min}(X_1, \ldots X_n) > y$ exactly when $X_i > y$ for all $i$. Since these variables are i.i.d., we have $F(y) = 1 - P(X_1 > y)P(X_2>y)\ldots P(X_n>y) = 1 - P(X_1 > y)^n$. Assuming the $X_i$ are uniformly distributed on $(a, b)$, this yields

We take the derivative to get the density function.

Now $E(Y) = \int_{-\infty}^{\infty} y f(y) dy$. The integral is straightforward; I’ll leave the details to you. I calculate $E(Y) = \frac{b+na}{n+1}$.