Distribution function of a Bernoulli variable

Question

Let $X_1, X_2,\dots, X_n$ be i.i.d. real-valued random variables. Let $F_X$ be the (common) distribution function of these random variables.

1. Find the distribution function of the random variable $Y=\max\{X_1,X_2,\dots,X_n\}$ in terms of $F_X$.

We have $F_Y(x)=\mathbb{P}(Y\le x)=\mathbb{P}(X_1\le x,\dots,X_n\le x)\overset{X_i\, i.i.d.}=\mathbb{P}(X_1\le x)\cdot\cdot\cdot\mathbb{P}(X_n\le x)=F_X(x)^n$

2. In case when the $X_i$’s are Bernoulli random variables with parameter $p$, that is $\mathbb{P}(X_i = 0) = p$, $\mathbb{P}(X_i = 1) = 1 − p$, find the expectation of $Y$ .

My approach: Is it right to say, since $\mathbb{P}(X_i=0)=p$ and $\mathbb{P}(X_i=1)=1-p$, that $F_X(0)=\mathbb{P}(X\le0)=p$, $F_X(1)=\mathbb{P}(X\le1)=p(1-p)?$ I don't know, it starts getting confusing. Maybe this is not relevant for the exercise?

$\mathbb{E}(Y)=\sum^\infty_{k=0}k\mathbb{P}(Y=k)$? Since Bernoulli distribution is discrete. Any help?

Answer 1

A full form of a Bernoulli cdf is $$F_X(1)=1,\,F_X(0)=1-p\implies F_X(x)=(1-p)^{(1-\lfloor x\rfloor)^+}\mathbf{1}_{[0,\infty)}(x)$$ Therefore $$F_Y(x)=(F_X(x))^n=(1-p)^{n(1-\lfloor x\rfloor)^+}\mathbf{1}_{[0,\infty)}(x)$$ This means that $$P_Y(0)=(1-p)^n,\,P_Y(1)=1-(1-p)^n$$ and therefore $E[Y]=1-(1-p)^n$.