Why is the sum of two independent geometrically distributed RVs not geometrically distributed again? Let X = Geo(1/4) and Y = Geo(1/4) (both independent) and let Z = X + Y. 
Why is X + Y not Geo(0.5) for example and how could I, as an example, calculate P(Z = 3)? 
 A: The easiest way to get at the distribution of $Z$ is to condition on the value of one of $X$ or $Y$, first. Let's say we condition on $X=x$. We adopt the convention that $x\in\{0,1,2,\ldots\}$. Now,
\begin{align}
P(Z=z) &= \sum_{x=0}^{z}P(Z=z|X=x)P(X=x)\\
&= \sum_{x=0}^{z}P(Y=z-x)P(X=x) \quad \text{[Independence of X and Y]}\\
&= \sum_{x=0}^{z}(0.75)^{z-x}(0.25)(0.75)^{x}(0.25)\\
&= (z+1)(0.75)^{z}(0.25)^{2}
\end{align} 
It is now a simple task to substitute $z=3$ in the expression above, to evaluate $P(Z=3)$. You can note that the resultant distribution above, is called the negative binomial distribution.
A: 
Why is $X+Y$ not $\operatorname{Geo}(0.5)$ ? 

Notice that 
$$P(X+Y=z \mid X = k) = P(X = k \cap Y = z-k)=P(X=k)P(Y=z-k) \tag{1}$$
Using the probability of a geometric distribution, that is
$$P(X=k) = p(1-p)^{k-1}\tag{2}$$
Also
$$P(Y=z-k) = p(1-p)^{z-k-1}\tag{3}$$
Replace $(2),(3)$ in $(1)$, you get:
$$P(X+Y=z) =p(1-p)^{k-1}p(1-p)^{z-k-1}=p^2(1-p)^{k-1+z-k-1} $$
which is 
$$P(X+Y=z \mid X = k) = p^2(1-p)^{z-2}$$
Now marginalize over $k$ as 
$$P(X+Y=z) = \sum_k P(X+Y=z \mid X = k)  = \sum_{k=1}^{z-1} p^2(1-p)^{z-2}$$
Notice that:
\begin{align} P(X+Y = z) &=\sum_{k=1}^{z-1} p^2(1-p)^{z-2} \\
& = p^2(1-p)^{z-2}\sum_{k=1}^{z-1} 1 \\
& =  p^2(1-p)^{z-2}(z-1) \\
& \neq (2p)(1-2p)^z
\end{align}
where $p = 0.25$ in your case.
