Mean of the minimum of a geometric random variable and an integer

Question

Let $M$ be a positive integer and $X\sim Geo(p)$

Compute the mean of $Y:=min(X,M)$

My attempt was if $Y=min(X,M)$ then for $1 \leq n<m$ I got $P(Y<M)=\sum_{n=1}^{m-1}P(Y=n)=1-(1-p)^{m-1}$ so $P(Y=m)=1-P(Y<m)=(1-p)^{m-1}$ and

$P(Y=n)=\begin{cases} (1-p)^{n-1}p, n=1...m-1 \\ (1-p)^{m-1}, n=m \end{cases}$

Got stuck here. A hint is to do it by cases, $X=M,X<M,X>M$ but I don't know how to do it this way. Please, be detailed in your explanations and thanks a lot.

Answer 1

Let $M$ be a positive integer and $X \sim Geo(p)$. In order to calculate the expectation of $Y = min(X,M)$ we proceed by observing that taking the minimum of a random variable and a number actually "flattens" the possible realization of the variable to $M$ for $x>M$. Hence we can compute: \begin{equation} \mathbb{E}(Y) = \sum_{k=1}^{M-1}P(X=k)k + \sum_{j=M}^{\infty}P(X=j)M \end{equation} Where $P(X=z)=(1-p)^{z-1}p$. Now the above can be expressed as follows: \begin{align} \sum_{k=1}^{M-1}P(X=k)k &= \frac{Mp(1-p)^{M}-p(1-p)^{M}+(1-p)^{M}+p-1}{(p-1)p} \\ \sum_{j=M}^{\infty}P(X=j)M &= (1-p)^{M-1}M \end{align} Summing the two you get: \begin{equation} \mathbb{E}(Y) = \frac{1-(1-p)^{M}}{p} \end{equation}