probability of sequence of random variables We're going to look at a random process, which is a sequence of random variables that depends on time.
Let $X(t)$ = $A/t$ where A has the density $f_A(a)$ = $3/8$ $a^2$ for $ 0<=a <= 2$. Assume $t > 0$.
a)Find $F_X(x)$ and $f_X(x)$ (to get this, assume t is fixed-so you can assume t is a constant).
b) Find $E(X)$
So for part a, I know $F_X(x)$ = $P(X <= x)$. So this is rewritten as $P(A <= tx)$. So its the integral from $0$ to $tx$ of $3/8$ $a^2$, which is $ 1/8$ $a^3$ but I'm confused on if I'm supposed to plug values in for $tx$ and solve or just leave it as is. Then do I take the first derivative of $1/8$ $a^3$ to find $f_X(x)$? 
Also not sure how to find $E(X)$ since I have to include x in the integral, but my function is in terms of a.
Thanks!

a) You are on the right track.
$$ F_X(x) = P(X \leq x) = P(A \leq xt), $$
leading to
$$ P(A \leq xt) = \int_{-\infty}^{xt} \frac{3}{8} a^2 da = \int_{0}^{xt} \frac{3}{8} a^2 da = \frac{1}{8} a^3 |_0^{tx} = \frac{1}{8} t^3 x^3  $$ 
And then yes, the probability density function is the derivative of the cumulative distribution function (with respect to $x$!).
b) Your probability density function is in terms of $x$, so it should be straight forward to find $E(X)$ now.
$$ E(X) = \int_{-\infty}^{\infty}x f_X(x)dx = \int_0^{2/t}x f_X(x)dx $$
