Find the joint and conditional distributions of $Z=X+Y$? 
Suppose that $X$ and $Y$ are independent and identically distributed:
  $$P (X = k) = P (Y = k) = ρ (1 − ρ)^k$$
  for $k = 0, 1, \dots$ and let $Z := X + Y$. Find the joint distribution of $(X, Z)$ and find the conditional distribution of $X$, given $Z = n$.

I need a little help just setting this up.
If I understand the question, I'm looking for $P (X = k, Z = n)$ and $P (X = k \mid Z = n)$. I'm a little confused, though, because I've never done this when I have a variable defined as a linear function of the other random variables. I also don't really understand what $n$ is.
Any hints or help getting started is appreciated.
 A: More than setup: If $Z=X+Y=n$, then necessarily $0\le X,Y\le n$. So, for $k>n$ you have that $$P(X=k,\ Z=n)=0$$ For $0\le k\le n$ (for example, say $n=5$, i.e. $X+Y=5$ so, $X$ might be equal to $0$ and $Y$ equal to $5$ or $X$ equal to $1$ and $Y$ equal to $4$ etc) you have 
\begin{align}P(X=k,\ Z=n)&=P(X=k,\ X+Y=n)=P(X=k, Y=n-k)\\[0.2cm]&=P(X=k)P(Y=n-k)=ρ(1-ρ)^kρ(1-ρ)^{n-k}=ρ^2(1-ρ)^n\end{align} and therefore by the formula of conditional probability $$P(X=k \mid Z=n)=\frac{P(X=k, Z=n)}{P(Z=n)}=\frac{ρ^2(1-ρ)^n}{P(Z=n)}$$ In order to proceed you can calculate the denominator by the law of total probability as $$P(Z=n)=\sum_{k=0}^{n}P(X+Y=n \mid X=k)P(X=k)=\sum_{k=0}^{n}P(Y=n-k)P(X=k)$$ or you can observe that the denominator is just a constant with respect to $k$ and hence it does not affect (up to a constant) the distribution of $X\mid Z=n$ which is solely determined by the numerator. However, the numerator is independent of $k$ as well which implies that the distribution of the random variabel $X\mid Z=n$ is discrete uniform on $0\le k \le n$. Hence $$P(X=k \mid Z=n)=\frac{1}{n+1}$$ for $0\le k\le n$ and $0$ else. 
A: So $n$ would be a fixed value of $Z$ which gives us information regarding the value of $X$, since whichever values $X=x$ and $Y=y$ satisfy it has to happen that $x+y=z$.
To find the distribution of Z, we ask:
$$
P(Z=z)=P(X+Y=z)=\sum_{y=0}^{\infty} P(Y=y)P(X=z-y)
$$
That is because the probability of Z is equal to the sum of the probabilities of each of its inverse images through the transformation $g(x,y)=x+y$.
To find the joint distribution, we ask:
$$
P(Z=z, X=x)=P(X+Y=z, X=x)=P(Y=z-y, X=x)=P(Y=z-y)P(X=x)
$$
And finally, for the conditional:
$$
P(X=x|Z=z)=\frac{P(X=x,Z=z)}{P(Z=x)}
$$
Since by the product rule $P(X=x,Z=z)=P(Z=z)P(X=x|Z=z)$.
