There is a task:
Let $ X1, X2 $ – independent random variables with the same density $\quad {2x \over{a^2} } \quad 0 < x < a $.
Let additionally $ \quad Y = max(X1, X2),\quad Z = min(X1, X2) $.
A) What is the value of expected value and variance of r.v. $X1$.
B) What is the form of density and expected value $Y$?
C) Find density and expected value af random variable $Z$.
Soo... I found the value of expected value and variance of r.v $X1$ and they are:
$E(X1) = {2a \over 3}$
$V(X1) = {a^2 \over 18} $
Then I found expected value and density of Y as follow:
$ F_y (Y) = P(Y<y) = P(max(X1,X2)<y) = P(X1<y, X2<y) = P(X1<y)P(X2<y) = F_x(y)^2 $
$f(y) = (F_y(y))' = 2(F_x'(y)) = 2f_x(y) = 2 * {2y \over a^2}.$
So then the expected value of Y is $ \quad{4a \over 3 }$
And to find value of expected value and density of $Z$ I need to find cummulative distribution function for $Z$ and it is:
$F_z(z) = P(Z<z) = P(min(X1, X2)<z) = 1-P(min(X1,X2)>z) = 1-P(X1>z)P(X2>z) = 1 - [(1-P(X1<z))(1-P(X2<z))] = 1-(1-F_x(z))(1-F_x(z))] = 1-(1-F_x(z))^2 $
And then almost in the same way as for $Y$, I can find what I need.
Is it right? Should I find value of $a$ to solve the problem? If I should find value of $a$, how can I do it?
What you found for expectation and variance of $X_1$ is correct.
There is a mistake in finding $f_Y(y)$.
You achieved correctly that $F_{Y}(y)=F_X(y)^2$ but differentiating leads to $f_Y(y)=2F_X(y)f_X(y)$.
Application of chain-rule.
It is not needed to go for finding values of parameter $a$.