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*$(3^X)$.

  • 1
    $\begingroup$ $\displaystyle\mathbb E(f(X) = \sum_{j=0}^\infty \frac{f(j)e^{-5}5^j}{j!}$. It's not correct without "$\displaystyle\sum_{j=0}^\infty$". $\endgroup$ – Michael Hardy Dec 3 '12 at 0:33
  • $\begingroup$ good catch, thanks $\endgroup$ – dmcqu314 Dec 3 '12 at 0:36
  • $\begingroup$ Why did you create this exact duplicate of the question here? math.stackexchange.com/questions/250413/… $\endgroup$ – 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 $\exp(3 \lambda)$.

  • $\begingroup$ Let me know if the hint is clear or sufficient. $\endgroup$ – Learner Dec 3 '12 at 2:34
  • $\begingroup$ I'm still somewhat confused as to how to initially approach the problem. The book we have isn't the best in describing how to approach the exercises. $\endgroup$ – dmcqu314 Dec 4 '12 at 2:04
  • $\begingroup$ Ok, I solved when of the examples in full. Please see the edit above. $\endgroup$ – Learner Dec 4 '12 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.