How to find the expected maximum value of a exponentially distributed random variable?

Question

Let's say we have a random variable $X$ that is exponentially distributed with parameter $= 1$, what is the expected maximum value of $X$?

I have simulated this in R by finding the max value of $X$ with size $10000$ and parameter $1$, and repeated this process $10000$ times to get the average. The result seems to be somewhere between $9.77$ to $9.81$, but I have no clue on how to actually calculate this.

Answer 1

There is a classic result that states that the expected value of the maximum of $n$ independent and identically distributed exponential random variables $x \sim \lambda e^{-\lambda y}$ is given by:

$$ \mathbf{E}[x_n] = \sum^n_{i=1} {1\over i \lambda} = {H_n \over \lambda}$$

Where $H_n$ is the harmonic number. Here you state that you have $\lambda=1$. Wolfram gives the harmonic number for $n=$10000 as 9.7876 (5sf) which agrees with your result. see https://mikespivey.wordpress.com/2013/05/13/expectedmaxexponential/ for details of derivation

Answer 2

Suppose $X_1,\dots X_n$ are independent identically distributed RVs. Let $X^*=\max (X_1, \dots X_n)$.

Then: $$ F_{X^*}(x)=\mbox{Prob}(X_1<x,X_2<x, \dots X_n<x)=(F_X(x))^n $$

and so: $$ f_{X^*}(x)=\frac{d}{dx}(F_X(x))^n=n\times (F_X(x))^{n-1}\times f_X(x) $$ and: $$ \overline{X^*}=\int_Sx\times n\times(F_X(x))^{n-1}\times f_X(x)\; dx $$ Where $S$ is the support of $f$.