Compute the density fZ(z) for Z

[sneaky666]

1.
Let X~Geometric(1/4), and let Y have probability function:
pY(y)=
1/6 if y=2
1/12 if y=5
3/4 if y=9
0 otherwise

Let W=X+Y. Suppose X and Y are independent. Compute pW(w) for all w in R.

For this i am not sure i think its
summation from K=0 to infinity (PY=w-K)(1/4)(3/4)^K

where
(PY=w-K) = 1/6 if w-K=2
(PY=w-K) = 1/12 if w-K=5
(PY=w-K) = 3/4 if w-K=9
(PY=w-K) = 0 otherwise

Is this right?

2.
Suppose X has density fX(x)=(x^3)/4 for 0<x<2, otherwise fX(x)=0, and Y has density fY(y)=(5y^4)/32 for 0<y<2, otherwise fY(y)=0. Assume X and Y are independent, and let Z=X+Y.
Compute the density fZ(z) for Z.

For this is it
the integral from z-2 to 2 (x^3)/4 * (5(z-x)^4)/32 dx

Is this right?

 

[nate808]

For the first problem, your approach is correct. However, you should also consider the possibility that w-K is negative, in which case pY(w-K) would be 0. So the summation should be from K=0 to w instead of infinity.

For the second problem, your approach is also correct, but you need to consider the limits of integration. Since X and Y can take values from 0 to 2, the limits of integration should be from max(0, z-2) to min(2, z). This is because if z is less than 2, then the range of values for X and Y that would result in z would be smaller. So the correct integral would be:

fZ(z) = integral from max(0, z-2) to min(2, z) (x^3)/4 * (5(z-x)^4)/32 dx