Variance and expected value of $2^X$ I have 2 similar problems:

*

*given $X \sim Po(3)$, find the variance of $2^X$

*given $X \sim B\left(200, \frac{1}{9}\right)$, find the expected value of $2^{X+3}$
I'm not really certain what I'm expected to calculate should it be $Var(2^X)$ and $E[2^{X+3}]$ respectively or should it be $2^{Var(X)}$ and $2^{E[X+3]}$. If it is the second one then: for (1) $$X \sim Po(3)\Longrightarrow\lambda=3\Longrightarrow Var(X)=\lambda = 3 \Longrightarrow 2^{Var(X)}=2^3$$  and for (2) $$X \sim B\left(200, \frac{1}{9}\right)\Longrightarrow E[X+3]=\frac{200}{9} + 3\Longrightarrow 2^{E[X+3]}=2^{\frac{200}{9} + 3}$$
But if it's the first one I don't know how I can find the variance and expected value with the given information

after reading the answers I did for (1): $$ E[2^X]=\sum^\infty_{i=0}2^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(2\lambda)^i}{i!}=e^{-\lambda}\cdot e^{2\lambda} = e^\lambda=e^3$$
and same way
$$ E[(2^X)^2]=E[(4^X)]=\sum^\infty_{i=0}4^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(4\lambda)^i}{i!}=e^{-\lambda}\cdot e^{4\lambda} = e^{3\lambda}=e^9 $$
hence
$$ Var(X) = E[(2^X)^2] - (E[2^X])^2 = e^9 - (e^3)^2 = e^9 - e^6$$
as for (2): $$ E[2^X]=\sum^{200}_{i=0}2^i\binom{200}{i}(\frac{1}{9})^i(1-\frac{1}{9})^{200-i} = \sum^{200} _{i=0}\binom{200}{i}(\frac{2}{9})^i (\frac{8}{9})^{200-i} =(\frac{2}{9} + \frac{8}{9})^{200} = (\frac{10}{9})^{200} $$
so $$ E[2^{X+3}]=E[8\cdot 2^X]=8\cdot E[2^X]=8(\frac{10}{9})^{200} $$
 A: The probability-generating function (PGF) of a random variable $X$ is given by $$\mathcal P_X(t) = \operatorname{E}[t^X]$$ whenever this expectation is well-defined.  For $X \sim \operatorname{Poisson}(\lambda)$, we have
$$\begin{align}
\mathcal P_X(t) &= \sum_{x=0}^\infty t^x \Pr[X = x] \\
&= \sum_{x=0}^\infty t^x e^{-\lambda} \frac{\lambda^x}{x!} \\
&= e^{-\lambda} \sum_{x=0}^\infty \frac{(\lambda t)^x}{x!} \\
&= e^{-\lambda} e^{\lambda t} \sum_{x=0}^\infty e^{-\lambda t} \frac{(\lambda t)^x}{x!} \\
&= e^{\lambda (t-1)},
\end{align}$$
where in the last step, we recognize that the summand is the probability mass function for a Poisson random variable with parameter $\lambda t$, thus the sum equals $1$.
Similarly, I leave it as an exercise to show that if $X \sim \operatorname{Binomial}(n,p)$, then
$$\mathcal P_X(t) = ((1-p) + pt)^n = (1 + p(t-1))^n.$$
Hint:  use the binomial theorem $$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k.$$

With these in hand, we can now easily evaluate the desired moments.  We have
$$\operatorname{Var}[2^X] = \operatorname{E}[(2^X)^2] - \operatorname{E}[2^X]^2 = \operatorname{E}[4^X] - \operatorname{E}[2^X]^2 = \mathcal P_X(4) - (\mathcal P_X(2))^2,$$ and $$\operatorname{E}[2^{X+3}] = \operatorname{E}[8 \cdot 2^X] = 8 \operatorname{E}[2^X] = 8 \mathcal P_X(2).$$  I leave it to you to evaluate these for their respective distributions.
A: You need to calculate $E[2^{X+3}]$ and $\mathrm{Var}[2^X]$. That is what is asked.
To calculate $E[2^X]$, you'll need to work out $\int_x 2^x p(x) dx$, where $p(x)$ is the pdf of $B\left(200, \frac19\right)$. Similar for the variance.
A: Find variance of $2^X$ for X ~ Poi(3).
First, what is $E(2^X)$? Intuitively, this is just $2^3$. Let’s check:
$$\sum_{k=0}^\infty2^k\frac{e^{-3}3^k}{k!} = \frac{e^{-3}}{e^{-6}}\sum_{k=0}^\infty\frac{e^{-6}6^k}{k!} = e^3.$$
Well, that was surprising. In fact, using the same calculations, $E(b^X)$ for X ~ Poi(a) is $\exp(ab-a)=\exp(a(b-1))$.
I invite you to use similar calculations to find $E((2^X)^2)=E(2^{2X})$ and finish the calculation. Feel free to post your solution as I’d be interested in the result.
