I'm new to the concept of joint conditionals, and I want to make sure that a move I made is valid and logic.
$$X\sim Geom(0.21) \rightarrow P(X=k)=(0.79)^{k-1}\times0.21$$ $$Y|X\sim Poisson(x+1)\rightarrow P(Y=y|X=x)=\frac{e^{-(x+1)}(x+1)^y}{y!}$$ $$P(Y=0) = ? $$ then we know that $P(Y=y) = \sum_{x=1}^{\infty}P(Y=y|X=x)P(X=x)$ from the Law of total probability and since X distributes geometrically its support $x \in {[1,2,3...]}$ so we sum from $x=1$.
My question is whether I can decide that $y=0$ before I sum, and instead of getting a general function $P(Y=y)$ I'll immediately get the value of $P(y=0)$.
$$P(Y=0) = \sum_{x=1}^{\infty}{P(Y=0|X=x)P(X=x)} = \sum_{x=1}^{\infty}\frac{e^{-(x+1)}(x+1)^0}{0!}\times {(0.79)^{x-1}\times0.21}=\sum_{x=1}^{\infty}\frac{(0.79)^{x-1}\times0.21}{e^{(x+1)}}=\frac{0.21}{0.79\times e}\sum_{x=1}^{\infty}(\frac{0.79}{e})^x =0.04006$$
Was it valid to already set the value $Y=0$ before finding the general probability function in order to avoid the sum of a very unpleasant form? Do moves like that might lead to mistakes or to totally wrong values in general?