Let X~Bernoulli(θ) and Y~Geometric(θ), with X and Y independent. Let Z=X+Y. What is the probability function of Z?
I am getting
PX(1) = θ
PX(0) = 1-θ
PX(x) = 0 otherwise
pY(y) = θ(1-θ)^y for y >= 0
pY(y) = 0 otherwise
What would PZ(z) be?
Let X~Bernoulli(θ) and Y~Geometric(θ), with X and Y independent. Let Z=X+Y. What is the probability function of Z?
I am getting
PX(1) = θ
PX(0) = 1-θ
PX(x) = 0 otherwise
pY(y) = θ(1-θ)^y for y >= 0
pY(y) = 0 otherwise
What would PZ(z) be?
i did that research and the only thing i could come up with is
PX(X=1) = θPX(X=0) = 1-θPX(X=x) = 0 otherwise
PY(Y=y>=0) = θ(1-θ)^y
PY(Y=y) = 0 otherwiseso (X=k) and (Y=z-k) since Z = X+Y
so
PZ(Z=z)=
summation from -inf to inf
θ^2 * (1-θ)^(z-1)
if x=1,y=1
summation from -inf to inf
θ * (1-θ)^(z+1)
if x=0,y=0
0
otherwise
Is this right?