# Memoryless property implies $E(\max\{X_1,X_2,\ldots,X_n\})- E(\min\{X_1,X_2,\ldots,X_n\})=E(\max\{X_1,X_2,\ldots,X_{n-1}\})$

$X_1,X_2,\ldots,X_n \sim \mathrm{Exp}(\lambda), \quad \text{i.i.d.}$

How to show the following equation by using memoryless property:

$$E(\max\{X_1,X_2,\ldots,X_n\}) - E(\min\{X_1 , X_2, \ldots, X_n\}) =E(\max\{X_1, X_2, \ldots ,X_{n-1}\}).$$

Actually I can prove this by writing both sides to a form of integration, just do transforming and calculation stuff, which is a normal way to prove it. Is there any easy explanation to this equation that do not need much calculation? By using memoryless property.

• Let $M_n=\max\{X_1,X_2,...,X_n\}$, then $M_n\ne M_{n-1}$ exactly when $X_n=M_n$, which by symmetry happens with probability $\frac1n$. When this happens, by the lack of memory, $X_n-M_{n-1}$ is exponential with mean $\lambda$. Thus, $E(M_n-M_{n-1})=\frac1n\cdot\frac1{\lambda}$. On the other hand, $\min\{X_1,X_2,...,X_n\}$ is exponential with parameter $n\lambda$, QED. – Did Oct 26 '15 at 21:29