$X, Y \sim Geom\left(p\right)$, 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!
It is not an "opposed", you use them together. It is simply a matter of:
$$\begin{align}\mathsf p_{\small X\mid X+Y}(i\mid n) =\dfrac{\mathsf p_{\small X, X+Y}(i,n)}{\mathsf p_{\small X+Y}(x)}\\[2ex]\mathsf p_{\small X+Y}(n)=\sum_{k=1}^{n-1} \mathsf p_{\small X,X+Y}(k,n)\end{align}$$
$X$ and $X+Y$ are not independent, but $X$ and $Y$ are, and surely the event $\{X=i, Y=n-X\}$ is identical to $\{X=i, Y=n-i\}$. $$\begin{align}\mathsf p_{\small X,X+Y}(i,n)&=\mathsf p_{\small X,Y}(i,n-i)\\[1ex]&=\mathsf p_{\small X}(i)\,\mathsf p_{\small Y}(n-i)\\[1ex]&= p^2(1-p)^{n-2}&&\big[1\leq i\lt n\big]\end{align}$$
Everything else should just fall into place.