Too stupid to understand random variable questions?

Question

I have two excercises:

1.) Let $X_1,X_2,X_3$ be independent uniformly distributed random variables on $[0,1]$. What is the density function of $X_1+X_2+X_3$?

2.) Let $X_1,...,X_4$ be independet uniformly distributed random variables on $[0,1]$. What is the expectation value of $Y:=max\{X_1,...,X_4\}$?

To 1.) Well uniformly distributed r.v. means $P_{X_i}(A)=\int _A 1 dt$. So I would say that the constant density function stays the same (which is equal to 1) for the sum. This is perfectly reconcilable with the convolution. But this seems to be somehow to easy for an excercise, so am I missing anything?

To 2.) Well, my idea was to derive $P(Y \le t)$, which is the same as the product of those things for $X_1,...,X_4$(since they are independent). Therefore we have(by taking the derivative) $f_Y(t)=P(Y \le t)'=4t^3$ for $t \in [0,1]$ and $f_Y(t)=0$ elsewhere.

In principle I would say: $E(Y)= \int_0^1 t f_Y(t) dt$

But this cannot be correct here, since we have $E(Y):=\int Y(t) f(t) dt$ and we would require $Y$ to be differentiable in order to derive the equation two lines above. So, how can I proceed?

Answer 1

When it comes to the second question: I asked a similar question once: Density function of $\max(X_1,\dots,X_n)$.

It is more abstract but I guess it could be interesting for you.

Answer 2

As to the first question, lets look at the CDF of $X$: $$F(x) = P(X>x)= P(X_3>x-X_1-X_2)$$ This is just the volume of the intersection of a cube and a plane: $$F(x) = 1-x^3/6\quad \to \quad p(x) = -x^2/2$$