Is $F_{\max(X_1, ..., X_n)} = F_{U[0,1]}^n$? Is $F^{-1}=F^{1/n}$?

Question

Suppose that I have that $X_1, ..., X_n$ are iid random variables with distribution function $F$ and $U$ is distributed uniformly on $[0,1]$. If I have that $F(\max(X_1, ..., X_n)) = F_U^n$, that is, the CDF of the maximum is equal to the nth power of the distribution of the uniform random variable (over (0,1)), then how can I find the inverse cdf? I know that it should be $F^{1/n}$, but am unable to see the mathematics behind it because I am not sure how to find the inverse of the CDF. Would anyone have any ideas or tips? Thanks!

Answer 1

Because of the monotonicity of $F$, the independence of the $X_i$'s and because of the fact the they are identically distributed, we have

$$P(F_{X_1}(\max(X_1,X_2,...,x_n))<x)=P(\max(X_1,X_2,...,X_n)<F_{X_1}^{-1}(x))=$$ $$P(X_1<F_{X_1}^{-1}(x)\cap X_2<F_{X_1}^{-1}(x)\cap \cdots \cap X_n <F_{X_1}^{-1}(x))=$$ $$=P(X_1<F_{X_1}^{-1}(x))^n.$$

And $$P(X_1<F^{-1}_{X_1}(x))=F_{X_1}(F_{X_1}^{-1}(x))=x.$$

(If $0\le x\le 1$.) This is called uniform distribution over the unit interval.

So,

$$F_{\max(X_1,X_2,...,X_n)}(x)=P(F_{X_1}(\max(X_1,X_2,...,X_n))<x)=x^n$$

for $0\le x\le 1$.

This does not mean that $\max(X_1,X_2,...,X_n)$ is the product of $n$ uniformly distributed random variables. This means simply that the distribution of $\max(X_1,X_2,...,X_n)$ equals the $n^{th}$ power of the uniform distribution over the unit interval.

As far as the inverse of $F_{\max(X_1,X_2,...,X_n)}(x)=x^n$. Taking the $n^{th}$ root of both sides we get that

$$F_{\max(X_1,X_2,...,X_n)}^{\frac1n}(x)=x.$$

The definition of the inverse of $G(x)$ is that $G^{-1}(G(x))=x$. In this sense

$$F_{\max(X_1,X_2,...,X_n)}^{\frac1n}=F_{\max(X_1,X_2,...,X_n)}^{-1}.$$