# Distribution of sample maximum from exponential distribution

$$x_1,x_2,...,x_n \sim \exp(\mu=1)$$ where $$x_i$$ are independently identically distributed. What is the distribution of $$z_n = max(x_1,x_2,...,x_n)-\ln(n)$$? Below is my work , I am uncertain whether it is correct or not.

$$F(x) = P(X\leq x) = 1-\exp(-x)$$

\begin{align}P(z_n\leq\delta) &= P(max(x_1,x_2,...,x_n)-\ln(n)\leq\delta)\\ &=\prod_{i=1}^{n} P(x-\ln(n)\leq\delta) = \prod_{i=1}^{n}P(x\leq\delta+\ln(n))\\ &= F(\delta+\ln(n))^n = [1-\exp(-\delta-\ln(n))]^n = [1-\exp(-\delta)/n]^n\end{align}

• If you want to determine the CDF of the sample maximum why are subtracting ln(n) from the sample maximum & trying to determine its distribution? – Michael R. Chernick Dec 15 '19 at 22:13

## 1 Answer

Ignoring the typos, your solution is correct (i.e. what you've found is the CDF of $$Z_n$$). Note that, by the way, $$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n=e^x$$ So, the result will approach to $$e^{-e^{-\delta}}$$ as $$n$$ goes to $$\infty$$.

• Kudos , p.s you forgot the negative sign. $exp(-e(-\delta))$ due to factorization. – Amirul Dec 15 '19 at 22:10