Let $X ∼ Geometric(p)$, and let $M > 0$ be a positive integer. Determine the expected value of $Z = \min(X, M)$.
I know that this question has already been answered here. While I am ok with the solution (which I eventually came up with), my initial solution was incorrect. Yet, I was unable to find which step was incorrect.
It would be very helpful if someone could point out the mistake in my thought process, because it looks like I may have some hole in my understanding.
So my solution goes as follows:
First we want to compute the PMF of $Z = \min(X, M)$. We have:
$$ P(Z = z) = P(\min(X, M) = z) = \begin{cases} 0, & \text{if } z > M \\ P(X = z), & \text{if } z \leq M \end{cases} $$
This is not correct though, because the probabilities do not add up to 1. To make sure that the probabilities add up to 1, we must divide by the their sum (normalization):
$$ P(\min(X, M) = z) = \begin{cases} 0, & \text{if } z > M \\ \frac{P(X = z)}{\sum\limits_{i=0}^{M}P(X = i)}, & \text{if } z \leq M \end{cases} = \begin{cases} 0, & \text{if } z > M \\ \frac{p(1-p)^z}{\sum\limits_{i=0}^{M}p(1-p)^i}, & \text{if } z \leq M \end{cases} $$
but $\sum\limits_{i=0}^{M}p(1-p)^i$ is equal to $1 - (1 - p)^{M + 1}$, which gets us the closed form:
$$ P(\min(X, M) = z) = \begin{cases} 0, & \text{if } z > M \\ \frac{p(1-p)^z}{1 - (1 - p)^{M + 1}}, & \text{if } z \leq M \end{cases} $$
Now we want to compute $P(Z \geq z)$:
$$ P(Z \geq z) = \sum_{i=z}^{\infty} P(Z = i) = \sum_{i=z}^{\infty} P(\min(X, M) = i) = \begin{cases} 0, & \text{if } z > M \\ \sum\limits_{i=z}^{M} P(\min(X, M) = i), & \text{if } z \leq M \end{cases} $$
which evaluates to
$$ P(Z \geq z) = \begin{cases} 0, & \text{if } z > M \\ \sum\limits_{i=z}^{M} \frac{p(1-p)^z}{1 - (1 - p)^{M + 1}}, & \text{if } z \leq M \end{cases} = \begin{cases} 0, & \text{if } z > M \\ \frac{(1 - p)^z - (1 - p)^{M + 1}}{1 - (1 - p)^{M + 1}}, & \text{if } z \leq M \end{cases} $$
Now we are ready to compute $\mathop{\mathbb{E}}[Z]$ using the Tail sum formula for expectation:
$$ \mathop{\mathbb{E}}[Z] = \sum_{z=1}^{\infty} P(Z \geq z) = \sum_{z=1}^{M} P(Z \geq z) = \sum_{z=1}^{M} \frac{(1 - p)^z - (1 - p)^{M + 1}}{1 - (1 - p)^{M + 1}} = \frac{\frac{1 - p - (1- p)^{M + 1}}{p} - M(1 - p)^{M + 1}}{1 - (1 - p)^{M + 1}} $$
which is obviously not the right answer. Thanks in advance.
