Marginal Density Functions and Probabilities The time (in hours) Y1 and Y2 it takes for a computer to do two different job has a density function
given by
$f(y1, y2) = 3y1$ if $0 < y2 < y1 < 1$  
a. (8 pts) Find the marginal densities for the time it takes to do each of the two jobs.
b. (8 pts) If you know the second job took 1/2 hour, what is the conditional density for the time it
takes to do the first job and what is its conditional expectation?
c. (8 pts) What is the probability that it will take more than 1 hour for the computer to do both of
these jobs (sum total of their times)?
 A: Note: Since the questions may be homework, we leave out many of the details. There should be enough information below to find the answers. 
We will use $X,Y$ and $x,y$ instead of subscripts.
The joint density lives in the triangle with corners $(0,0)$, $(1,0)$, $(1,1)$. Draw the triangle. 
a) To find the marginal density of $X$, we "integrate out" $y$. Note that $y$ travels from $0$ to $x$. Thus for $0\lt x\lt 1$,
$$f_X(x)=\int_0^x 3x\,dy.$$
The integration is trivial, since $3x$ does not even involve $y$. We get $3x^2$. For completeness, note that $f_X(x)$ is $0$ if $x\lt 0$ or $x\gt 1$. The densities at $0$ and $1$ can be defined in any way you like. 
The marginal density of $Y$ is calculated in a similar way.
b) We note that the conditional density of $X$ given $Y=b$ is given by
$$\frac{f_{X,Y}(x,b)}{f_Y(b)}.$$
You will have computed the only missing ingredient in part a).
For the conditional expectation of $X$ given $Y=1/2$, let $g(x)$ be the conditional density of $X$ given $Y=1/2$, and find the usual integral. We have to be a little careful: Since $x\gt y$, the conditional density is $0$ except when $x$ is between $1/2$ and $1$. In essence you will be integrating from $1/2$ to $1$.
c) We want to find the probability that $X+Y\gt 1/2$. Draw the line $x+y=1/2$. We need to integrate the joint density over the part of the triangle that is above the line $x+y=\frac{1}{2}$.
It will be a little easier to find the probability that $X+Y\le \frac{1}{2}$, and subtract the result from $1$.  
To find $\Pr(X+Y\le \frac{1}{2})$, integrate the joint density over the part of the triangle below the line $x+y=\frac{1}{2}$. You can integrate first with respect to $y$, $y=0$ to $\frac{1}{2}-x$, then with respect to $x$ from $0$ to $\frac{1}{2}$.
Or else you can integrate first with respect to $x$. 
