Question. Let $ X_1,\cdots,X_n $ be i.i.d. random variables with exponential distributions, $ X_i\sim Exp(\lambda) $. Find $E\left[ \max(X_1,\cdots,X_n) \right]$.
My approach. Let's find a recursive relation in terms of $ n $. Define \begin{align*} &Y_1 = X_1,\\ &Y_2 = \max(X_1,X_2),\\ &\vdots\\ &Y_n = \max(X_1,\cdots,X_n). \end{align*} We have the relation \begin{align*} &Y_n = \begin{cases} X_n\ &{\rm if}\ X_n>Y_{n-1},\\ Y_{n-1}\ &{\rm otherwise}, \end{cases}\implies Y_n = Y_{n-1} + (X_n-Y_{n-1})1_{\{X_n>Y_{n-1}\}}\\ &\implies E[Y_n] = E[Y_{n-1}] + E[(X_n-Y_{n-1})1_{\{X_n>Y_{n-1}\}}]. \end{align*} Now, using the independence of $ X_n $ and $ Y_{n-1} $, we have \begin{align*} E[(X_n-Y_{n-1})1_{\{X_n>Y_{n-1}\}}]& = \int_{0}^{\infty}\int_{0}^{\infty}(x-y)1_{\{x>y\}}f_{X_{n}}(x)f_{Y_{n-1}}(y)\ dx dy\\ & = \int_{0}^{\infty}\int_{y}^{\infty}(x-y)f_{X_{n}}(x) dx\ f_{Y_{n-1}}(y) dy\\ & = \int_{0}^{\infty}\dfrac{\exp(-\lambda y)}{\lambda} f_{Y_{n-1}}(y) dy \end{align*} So, \begin{align*} E[Y_n] = E[Y_{n-1}] + \int_{0}^{\infty}\frac{\exp(-\lambda y)}{\lambda} f_{Y_{n-1}}(y) dy. \end{align*} But I do not know what to do with the inetgral.
