$X, Y \sim Geom\left(p\right)$, 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?] […]

- Tags "and x=", $F_Y(n-i)$. Thus, $p_{X+Y}(n) = p^2 (1-p)^{n-i-1} (1-p)^{i-1}) = p^2 (1-p)^{n-2}$. Then, $Pr\left(X=i\right)$ is simply the pmf of $X$ for a geometric random variable, all that is needed is the joint pmf $p_{X, by conditioning on X, by the law of total probability: $p_{X | X+Y}\left(i | n\right) = \frac{p_{X, n\right)}{p_{X+Y}\left(n\right)} $ [Is using the law of total probability here correct as opposed to using Bayes theorem?] Let $Z = X + Y$., n\right)$ I am stuck with how to find the joint pmf because $X$ and $X+Y$ are not independent. Any help is appreciated!, the CDF of Z is $$F_Z\left(n\right) = Pr\left(Z \leq n\right) = Pr\left(X+Y \leq n\right)$$ Then, then $Pr\left(Y \leq n-i | X=i\right) = Pr\left(Y \leq n-i\right)$, 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, 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, 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, which is the CDF for $Y$, which we know. Then since X takes the value $i$, X, X+Y}\left(i, Y \sim Geom\left(p\right)$, Y are independent, Y$ 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