Find Conditional PDF and joint PDF from given independent random variables.

Question

Given X: is a uniformly distributed R.V on [a,b], Y is a exponential R.V with parameter $\lambda$.

X, Y are independent and Z = X+Y, Is the given information enough to determine?

1. $f_{Z|Y}(z|y)$

2. $f_{Z,Y}(z,y)$

If it is enough, please help me to suggest how to derive it.

Answer 1

Yes, saying they are independent and giving their distributions fully specifies the joint behavior of $X$ and $Y,$ so in principle, you have enough information to compute anything pertaining to them.

We can compute the conditional CDF as: $$P(Z\le z\mid Y=y) = P(X+Y\le z\mid Y=y) \\= P(X+y\le z\mid Y=y) \\= P(X+y\le z) \\= P(X\le z-y)$$ where I used the condition to change $Y$ to $y$ and then used independence to drop the conditional. And you know the distribution of $X$...

You can get the first answer from this, and the second follows from the first by definition.

Answer 2

Well $f_{Z\mid Y}(z\mid y)=\left\lVert\dfrac{\partial [z-y,y]}{\partial [z,y]}\right\rVert\cdot f_{X\mid Y}(z-y\mid y)$

Now, since $X\perp Y$ therefore $f_{Z\mid Y}(z\mid y)=f_{X}(z-y)$

So $f_{Z,Y}(z,y)$ is clearly $f_X(z-y)f_Y(y)$ and hence you may determine $f_Z(z)$.