# Poisson distribution and probability distributions

Suppose $X$ has the $\mathrm{Poisson}(5)$ distribution considered earlier.

Then $P(X \in A) = \sum_{j\in A} \frac{e^{-5}5^j}{j!}$, which implies that $L(X) = \sum^\infty_{j=0} \left(\frac{e^{-5}5^j}{j!}\right)\delta_j$, a convex combination of point masses. The following propostion shows that we have $E(f(X)) = \sum_{j=0}^\infty\frac{f(j)e^{-5}5^j}{j!}$ for any function $f : \mathbb{R} \rightarrow \mathbb{R}$.

Prop. Suppose $\mu = \sum_i \beta \mu_I$ where $\{\mu_i\}$ are probability distributions, and $\{\beta_i\}$ are non-negative constants (summing to 1, if we want $\mu$ to also be a probability distribution). Then for Borel-measurable functions $f : \mathbb{R} \rightarrow \mathbb{R}$,

$$\int fd\mu = \sum_i \beta_i \int f \, d\mu_i,$$

provided either side is well-defined.

Using this proposition:

Let $X \sim \mathrm{Poisson}(5)$.

(a) compute *E*$(X)$ and *Var*$(X)$.

(b) compute *E*$(3^X)$.

• $\displaystyle\mathbb E(f(X) = \sum_{j=0}^\infty \frac{f(j)e^{-5}5^j}{j!}$. It's not correct without "$\displaystyle\sum_{j=0}^\infty$". – Michael Hardy Dec 3 '12 at 0:33
• good catch, thanks – dmcqu314 Dec 3 '12 at 0:36
• Why did you create this exact duplicate of the question here? math.stackexchange.com/questions/250413/… – Learner Dec 4 '12 at 3:33

The three answers are $E[X]=\lambda$, $E[X^2]=\lambda+\lambda^2$ (from the previous two, you could compute the variance) and $E[\exp(3X)]=\exp(\lambda(e^3-1))$. As a hint, the major tool here is to use Taylor series expansion for the exponential function.
I noticed there was an edit with a new question (or I misread the previous one). The answer for $E[3^X]$ is $\exp(2 \lambda)$.
I will give as an example how to compute $E[3^X]$. Writing the formula for the expectation \begin{eqnarray*} E \left[ 3^X \right] & = & \sum_{k = 0}^{\infty} 3^k P \left[ X = k \right]\\ & = & \sum_{k = 0}^{\infty} 3^k \mathrm{e}^{- \lambda} \frac{\lambda^k}{k!}\\ & = & \mathrm{e}^{- \lambda} \sum_{k = 0}^{\infty} \frac{\left( 3 \lambda \right)^k}{k!}\\ & = & \mathrm{e}^{- \lambda} \underbrace{\sum_{k = 0}^{\infty} \frac{\left( 3 \lambda \right)^k}{k!}}_{= \mathrm{e}^{3 \lambda}}\\ & = & \mathrm{e}^{2 \lambda} \end{eqnarray*}
You see that the main tool here was the expansion of $\exp(3 \lambda)$.