If $X_1,\ldots,X_n\sim Uniform(0,1)$ and $Y=\max[X_1,\ldots,X_n]$, find $E(Y)$.
I don't really have any good starting point. I guess Y is a function of the X's and so I might be able to apply the theorem about the expectation of a function, but I don't know how to deal with this since max is a discrete function.
Can anybody help me please?
First you need to obtain the distribution of the variable $Y$:
$$F(y)=P(\max{(X_1,X_2,...,X_n)}\leq y)=P(X_i\leq y,i=1,...,n)$$
Where $F$ is the cumulative distribution. Assuming that your variables are independent:
$$P(X_i\leq y,i=1,...,n)=\left(P(X_1\leq y)\right)^n=y^n$$
Then the density is:
$$f(y)=F'(y)=ny^{n-1}$$
Therefore the expectation is:
$$E(Y)=\int_0^1 y ny^{n-1}dy=\int_0^1 ny^{n}dy=\frac{n}{n+1}$$
If your $X_i$ are independent, you could use $$P(\max X_i < t) = P(\forall X_i < t) = \prod P(X_i < t)$$ to get the cumulative distribution function of $Y$.
