$E[\min(X,2)]$ for$ X\sim Po(3)$

Question

Let $X\sim Po(3), W=\min(X,2)$

Find $E[W]$.

My attempt:

$W=\begin{cases} X & X \leq 2 \\ 2 & X \gt 2 \end{cases}$

So $$P(W=k)=P(\min(X,2)=k)=P(\{\min(X,2)=k\} \cap \{X\le2\})+P(\{\min(X,2)=k\} \cap \{X\gt2\}) $$ (because the sets are disjoint)

$$=\begin{cases} P(X=k) & k=0,1,2 \\ 0 & \text{else} \end{cases} + \begin{cases} 1 & k=2 \\ 0 & \text{else} \end{cases} $$

The problem with this is that, among others, $P(W=2) > 1$ so clearly I have done some things wrong.

My questions are:

• Where did I go wrong?

• How can I solve this problem?

Thanks for your help!

Answer 1

Second question

To efficiently find the expectation $E[W]$, I suggest you to divide it into two cases $\{X \le 2\}$ and $\{ X > 2 \}$.

\begin{align} E[W] &= E[W1_{\{X \le 2\}}] + E[W1_{\{ X > 2 \}}] \\ &= E[X1_{\{X \le 2\}}] + E[21_{\{ X > 2 \}}] \\ &= \sum_{k = 0}^2 k\frac{3^k}{k!} e^{-3} + 2E[1_{\{ X > 2 \}}] \\ &= \left( 3 + 9 \right) e^{-3} + 2P(X > 2) \\ &= 12 e^{-3} + 2 \left[ 1 - \left( 1 + 3 + \frac92 \right) e^{-3} \right] \\ &= 2 - 5e^{-3} \end{align}

First question

First term in the sum is correct

The transition from $P(\{\min(X,2)=k\} \cap \{X\le2\})$ to $\begin{cases} P(X=k) & k=0,1,2 \\ 0 & \text{else} \end{cases}$ is OK, since in the event $\{X\le2\}$, $\{\min(X,2)=k\} = \{ X = k\}$. When $k \in \{0,1,2\}$, $\{ X = k\}$ is a subset of $\{X\le2\}$, so $\{X\le2\}$ can be omitted in the calculation of the probability $P(\{\min(X,2)=k\} \cap \{X\le2\})$. Otherwise, these two events in the probability calculation has no intersection, one gets zero.

Second term in the sum is incorrect

The transition from $P(\{\min(X,2)=k\} \cap \{X>2\})$ to $\begin{cases} \color{red}{1} & \color{red}{k=2} \\ 0 & \text{else} \end{cases}$ is incorrect. In fact, $$\{\min(X,2) = k\} = \{ X \le k\} \cap \{ k \ge 2\}.$$ When $k=2$, $\{X>2\} \cap \{ X \le k\} = \varnothing$, so $$P(\{\min(X,2)=k\} \cap \{X>2\}) = P(\varnothing) = 0.$$