X1,X2,…,Xn are n i.i.d. uniform random variables. Let Y=min. 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!

**Answer**

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

F(y) = \left\{

\begin{array}{ll}

1 – \left(\frac{b-y}{b-a}\right)^n & : y \in (a, b)\\

0 & : y < a\\

1 & : y > b

\end{array}

\right.

We take the derivative to get the density function.

f(y) = \left\{

\begin{array}{ll}

\frac{n}{b-a} \left(\frac{b-y}{b-a}\right)^{n-1} & : y \in (a, b)\\

0 & : \text{otherwise}

\end{array}

\right.

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}.

**Attribution***Source : Link , Question Author : jet , Answer Author : Alex G.*