I am new here I need help of the enlightened to solve a couple of probability problems. I've been trying to tackle them but so far been unsuccessful.
1) Three independent random variables $x, y, z$ ~ Uniform [1, 3]. Find the probability function for the maximum of X, Y, Z.
As far as I understood i need to find $f(c) = \mathbb{P}(\max(x,y,z) \leq c)$ for any $c$ ∈ ℝ. How to solve it?
2) Two independent random variables $x, y$~ Uniform [0, 1]. Find the probability function of the random variable $z = x-y$ Compute the expectation of $z$.
Same as above, do I need to find $f(c) = \mathbb{P} ( z \leq c)= \mathbb{P} (x-y \leq c)$ for any $c$ ∈ ℝ? I have no idea how to compute the expectation, though. How to solve them?
Would much appreciate your help, many thanks in advance.
(1) The complicated answer would be to tell you to look up "order statistics". However, this is a bit simpler. The maximum is $\le c$ if and only if each of the random variables is $\le c$. Since they are independent, that's just the probability that $X\le c$ times the probability that $Y\le c$ times the probability that $Z\le c$. But each of those is $0$ if $c\le 1$, $(c-1)/2$ if $1\le c \le 3$, and $1$ if $c\ge 3$. Thus, the probability distribution function $F(c)$ for $max(X,Y,Z)$ is $0$ if $c\le 1$, $(c-1)^3/8$ if $1\le c\le 3$ and $1$ if $c\ge 3$.
If in fact what you needed was the density $f$ of $max(X,Y,Z)$ that's easy now since it's just the derivative of $F$, i.e., the density is $\frac{3(c-1)^2}{8}$ if $1\le c \le 3$ and $0$ otherwise.
(2) You don't actually need to compute the distribution of $Z$ here. Just use that expectation is linear. $E(Z)=E(X-Y)=E(X)-E(Y)=0.5-0.5=0$.