Geometric distribution with a capped number of trials - finding expectation and prior predictive distribution So I am modeling a random variable which follows a geometric distribution with probability $\theta$ except that the total number of trials is capped at some value $n$. I.e., the probability mass function is given by:
$ f(x)=\begin{cases} 
      (1-\theta)^{x-1}\theta & 1 \leq x < n
       \\ 
      (1-\theta)^{n-1} & x = n \\
      0 & x > n
   \end{cases}
$
This could represent an experiment in which one flips a coin until the first heads or until $n$ flips whichever comes first.
Firstly, I am interested in knowing the expectation of x when $\theta$ is known: $\mathbb{E}(x | \theta)$
Further, suppose $\theta$ is unknown and I have a prior distribution: $\theta$ ~ $Beta(a,b)$.
I would like to find the marginal expectation of $\mathbb{E}(x)$ over all values of $\theta$ which is given by,
$\mathbb{E}(x) = \int_0^1 \mathbb{E}(x | \theta) p(\theta) d\theta$
As well as the prior predictive distribution for $x$:
$p(x) = \int_0^1 p(x | \theta) p(\theta) d\theta$
It seems like this would have been studied somewhere before, but searches for "truncated geometric distribution" or "capped geometric distribution" aren't finding anything. 
 A: Here's a few things that might help. I left the derivation of the prior predictive distribution left for you because it follows some of the same tricks I used towards the end.
First, the expectation is a sum, not an integral.
\begin{align*}
\mathbb{E}(x | \theta) &= \sum_{x=1}^n x f(x) \\
&= \sum_{x=1}^{n-1} x (1-\theta)^{x-1}\theta + n(1-\theta)^{n-1} \\
&= \frac{\theta + \theta n (1 - \theta)^n - \theta (1 - \theta)^n + (1 - \theta)^n - 1 + n(1-\theta)^{n-1}(\theta-1)\theta}{(\theta - 1) \theta}  \\
&= \frac{\theta + \theta n (1 - \theta)^n - \theta (1 - \theta)^n + (1 - \theta)^n - 1 - n(1-\theta)^{n}\theta}{(\theta - 1) \theta}  \\
&= \frac{\theta  - \theta (1 - \theta)^n + (1 - \theta)^n - 1 }{(\theta - 1) \theta}  \\
&= \frac{\theta  + (1 - \theta)^n(1 - \theta)  - 1 }{(\theta - 1) \theta}  \\
&= \frac{(1 - \theta)^n(1 - \theta)  - (1-\theta) }{(\theta - 1) \theta}  \\
&= \frac{(1 - \theta)\{(1 - \theta)^n- 1 \} }{-(1- \theta) \theta}  \\
&= \frac{ 1-(1 - \theta)^n  }{\theta}  ,
\end{align*}
Second, if $\theta \sim \text{Beta}(a,b)$ then
$$
\mathbb{E}(x ) = \mathbb{E}\left[\mathbb{E}(x | \theta)\right]
$$
where the outer expectation is an integral, not a sum.
So, if $\alpha > 1$, we could use
$$
\mathbb{E}(\theta^{-1}) = \frac{1}{\text{B}(a,b)}\int_0^1\theta^{a-2}(1-\theta)^{b-1}d\theta = \frac{\text{B}(a-1,b)}{\text{B}(a,b)}
$$
and
$$
\mathbb{E}(\theta^{-1}(1-\theta)^n) = \frac{1}{\text{B}(a,b)}\int_0^1\theta^{a-2}(1-\theta)^{b+n-1} d\theta= \frac{\text{B}(a-1,b+n)}{\text{B}(a,b)}.
$$
We are using the fact that a beta density integrates to $1$.
