Consider the Poisson Process $\{N(t); t \geq 0\}$ with intensity $\lambda$. For $0 < s < t$, determine
a) $\operatorname E[2^{N(t)}e^{-\lambda t} \mid N(s) = k]$ (Hint: $\operatorname E[e^{xN(t)}] = e^{-\lambda t(1-e^x)})$
b) $\operatorname{Var}(N(t) \mid N(s) = k).$
This is what I've tried:
Edit: the last implication below was wrong. $\operatorname{E}[2^{N(t)}e^{-\lambda t}\mid N(s)=k]$ doesn't have to be equal to 1.
a) First I calculated $\operatorname E[2^{N(t)}e^{-\lambda t}]{:}$
\begin{align} & = e^{-\lambda t}E[2^{N(t)}] \\[10pt] & = e^{-\lambda t}E[e^{\ln(2)N(t)}] \\[10pt] & = e^{-\lambda t}e^{\lambda t} \text{ using the hint} \\[10pt] & = 1. \end{align}
Then, using $\operatorname{E}[X] = \sum_{y} \operatorname{E}[X\mid Y=y]f_Y(y)$:
$\operatorname{E}[2^{N(t)}e^{-\lambda t}] = \sum_{k=0}^{\infty} \operatorname{E}[2^{N(t)}e^{-\lambda t}\mid N(s)=k]e^{-\lambda s}\frac{(\lambda s)^k}{k!}$
$\implies 1 = e^{-\lambda s} \sum_{k=0}^{\infty} \operatorname{E}[2^{N(t)}e^{-\lambda t}\mid N(s)=k]\frac{(\lambda s)^k}{k!}$
$\implies e^{\lambda s} = \sum_{k=0}^{\infty} \operatorname{E}[2^{N(t)}e^{-\lambda t}\mid N(s)=k]\frac{(\lambda s)^k}{k!}$
$\implies \sum_{k=0}^{\infty} \operatorname{E}[2^{N(t)}e^{-\lambda t}\mid N(s)=k]$ should equal 1 because $e^{\lambda s} = \sum_{k=0}^{\infty}\frac{(\lambda s)^k}{k!}$.
On the second try I found $\operatorname{E}[2^{N(t)}e^{-\lambda t}\mid N(s)=k] = 2^k e^{-\lambda s}$ (thanks to Michael Hardy's answer):
\begin{align} \operatorname{E}[2^{N(t)}e^{-\lambda t}\mid N(s)=k] & = \operatorname{E}[2^{N(t) - N(s) + N(s)}e^{-\lambda t}\mid N(s)=k] \\[10pt] & =e^{-\lambda t}\operatorname{E}[2^{N(t) - N(s)}\mid N(s)=k]\cdot\operatorname{E}[2^{N(s)}\mid N(s)=k] \\[10pt] & =e^{-\lambda t}\operatorname{E}[2^{N(t-s)}]\cdot 2^k \\[10pt] & =2^k e^{-\lambda t}e^{\lambda (t-s)}\\[10pt] & =2^k e^{-\lambda s}. \end{align}
Is this correct?
b) $\operatorname{Var}(N(t) \mid N(s) = k)$ \begin{align} & = \operatorname{Var}(N(t) - N(s) + N(s) \mid N(s) = k) \\ & = \operatorname{Var}(N(t) - N(s) \mid N(s) = k) + \operatorname{Var}(N(s) \mid N(s) = k)\\ & = \operatorname{Var}(N(t) - N(s)) + 0\\ & = \lambda (t-s). \end{align}
Could this be correct?
Any help or answers would be greatly appreciated.
$\newcommand{\v}{\operatorname{var}}$The number of arrivals in the time interval $[0,s]$ and the number of arrivals in the time interval $[s,t]$ are independent.
Therefore \begin{align} \v(N(t) \mid N(s)=k) & = \v(N(t)-k\mid N(s)=k) = \v(N(t)-N(s)\mid N(s)=k) \\[10pt] & = \v(N(t) - N(s)) \text{ by the aforementioned independence} \\[10pt] & = t-s \text{ since } N(t)-N(s) \sim \operatorname{Poisson}(t-s). \end{align}
Similarly \begin{align} & \operatorname E(2^{N(t)} e^{-\lambda t}\mid N(s)=k) = \operatorname E(2^{(N(t) - N(s)) + k} e^{-\lambda t} \mid N(s)=k) = 2^k e^{-\lambda t} \operatorname E(2^{N(t) - N(s)}), \end{align} and then use the hint.