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# Conditional probability mass function of the sum of independent geometric random variables.

$$X, Y \sim Geom\left(p\right)$$, and $$X, Y$$ are independent. Find $$p_{X | X+Y}\left(i | n\right)$$.

My approach:

pmf of geometric distribution: $$p_X(i) = {p(1-p)^{i-1}}$$

Then, by the law of total probability: $$p_{X | X+Y}\left(i | n\right) = \frac{p_{X, X+Y}\left(i,n\right)}{p_{X+Y}\left(n\right)}$$

[Is using the law of total probability here correct as opposed to using Bayes theorem?]

Let $$Z = X + Y$$. Then, the CDF of Z is
$$F_Z\left(n\right) = Pr\left(Z \leq n\right) = Pr\left(X+Y \leq n\right)$$

Then, by conditioning on X, we can get:
$$Pr\left(X+Y \leq n\right) = \sum_{i=1}{Pr\left(X+Y \leq n\right | X = i)Pr\left(X=i\right)}$$

Then, $$Pr\left(X=i\right)$$ is simply the pmf of $$X$$ for a geometric random variable, which we know. Then since X takes the value $$i$$, we can rewrite this as

$$Pr\left(X+Y \leq n\right) = \sum_{i=1}{Pr\left(Y \leq n-i | X=i\right)Pr\left(X=i\right)}$$

Since $$X, Y$$ are independent, then $$Pr\left(Y \leq n-i | X=i\right) = Pr\left(Y \leq n-i\right)$$, which is the CDF for $$Y$$, $$F_Y(n-i)$$. Thus, we have

$$\sum_{i=1}{F_Y(n-i) p_{X}(i)}$$

Then by differentiating with respect to n to get the pmf from the CDF [Is the interchange of the differential and the summation allowed here since the upper bound of the summation is technically infinity?]:
$$p_{Z}(n) = p_{X+Y}(n) = \frac{d}{dn} \sum_{i=1}{F_Y(n-i) p_{X}(i)} = \sum_{i=1}{\frac{d}{dn} F_Y(n-i) p_{X}(i)} = \sum_{i=1}{p_Y(n-i) p_{X}(i)}$$

Thus, $$p_{X+Y}(n) = p^2 (1-p)^{n-i-1} (1-p)^{i-1}) = p^2 (1-p)^{n-2}$$. Then, all that is needed is the joint pmf $$p_{X, X+Y}\left(i, n\right)$$

I am stuck with how to find the joint pmf because $$X$$ and $$X+Y$$ are not independent. Any help is appreciated!