How do I compute the expectation value of the following random variable?

Question

I have the following problem:

We have given $X_1,...,X_n$ i.i.d. random variables. And we define $F_X$ to be the distribution function of these random variable. Let $$Y=\operatorname{max}\{X_1,...,X_n\}$$ In the lecture we have just shown that $$F_Y(x)=F_X(x)^n$$. Now I need to assume that the $X_i$'s are Bernoulli random variables, so $$\Bbb{P}(X_i=0)=p,~~~\Bbb{P}(X_i=1)=1-p$$I need to compute $\Bbb{E}(Y)$.

I thought about the following:

$$\Bbb{E}(Y)=0\cdot \Bbb{P}(Y=0)+1\cdot \Bbb{P}(Y=1)=\Bbb{P}(Y=1)=\Bbb{P}(\operatorname{max}\{X_1,...X_n\}=1)\leq\Bbb{P}(X_1=1,...X_n=1)\stackrel{i.i.d}{=}\Bbb{P}(X_1=1)...\Bbb{P}(X_n=1)=(1-p)^n$$

But somehow I don't think this works because of this inequality. Is there maybe someone who can give me a hint how to try it?

Thanks for your help

Answer 1

If $X_i$ are iid Bernoulli with $\Pr[X_i = 1] = 1-p$, then $$Y = \max(X_1, \ldots, X_n) = \begin{cases} 0, & X_1 = X_2 = \ldots = X_n = 0 \\ 1, & \text{otherwise}. \end{cases}$$ That is to say, $Y = 0$ if and only if all of the $X_i$ are zero. So $\Pr[Y = 0] = p^n$ and $\Pr[Y = 1] = 1 - p^n$. Thus $$\operatorname{E}[Y] = 0 \Pr[Y = 0] + 1 \Pr[Y = 1] = 1 - p^n.$$

Answer 2

$Y$ takes only the values $0$ and $1$. It is $0$ iff $X_i=0$ for all $i$. So $EY=(1)P(Y=1)+(0)P(Y=0)=1-P(X_i=0 \, \,\forall i)=1-p^{n}$.