# Finding distribution of a maximum of random variables

Take $Y_1,Y_2,...,Y_n \sim_{i.i.d.} Gaussian(0,1)$ and Take $X = \max\{ Y_1,Y_2,...,Y_n\}.$ I would like to find the distribution of $X$ as well its expected value and variance.

I can see that $$P(X = x) = P(Y_1 \leq x \wedge Y_2 \leq x \wedge ... \wedge Y_n \leq x),$$ Which by i.i.d. $$=\prod_{i=1}^n P(Y_i \leq x),$$ and thus this is a product of $n$ CDFs of standardized normal distributions evaluated at $x$. However, this seems very complicated for evaluating $E[X]$ and $V[X]$ since we do not currently have a parameterized distribution of $X$. Anyone familar with what family of distributions that $X$ is under? Along with that, I could use some help evaluating $E[X]$ and $V[X]$.

• I believe this is covered theoretically in Kendall, Stuart & Ord, but not in a way I have found computationally convenient. For a particular $n,$ it is easy to simulate. – BruceET Sep 19 '16 at 23:44
• Order Statistics (David & Nagaraja,2003) Chapters 2 and 3 will answer many of your questions. – Ganesh Sep 19 '16 at 23:47
• For a more practical viewpoint, Embrechts, Klüppelberg & Mikosch (1997) is also worth taking a look at. – Furrer Sep 20 '16 at 0:00

Here is a simulation in R using a million samples of size $n = 10$ from a standard normal population.

m = 10^6; n = 10;  x = rnorm(m*n)
DTA = matrix(x, nrow=m) # each row a sample of size n
mx = apply(DTA, 1, max) # m-vector of maximums
mean(mx);  sd(mx)
## 1.537666             # aprx E(max)
## 0.5869767            # aprx SD(max)


The histogram below shows the simulated distribution of $X$ for $n = 10.$ Skewness becomes greater as $n$ increases.

For $n = 5$ we have $E(X) \approx 1.164$ and $SD(X) \approx 0.670.$ For $n = 15$ we have $E(X) \approx 1.73$ and $SD(X) \approx 0.55$ (based on 100,000 samples).