A question about poisson processes Reading through the book "Brownian Motion & Stochastic Calculus" by Karatzas and Shreve, 
I found the following exercise (problem 3.9, page 15): 
Let $ \ N \ $ be a poisson process with intensity $ \lambda > 0 $ (this means, in particular, that $ N_t $ is poisson-$\lambda t$-distributed, i.e. $ P(N_t = k ) = \exp(-\lambda t) \frac{(\lambda t)^k}{k!}, \ \forall \ k \geq 0$)
Use Stirling's approximation to show that  $\  \lim_{t \to \infty} (1/\sqrt{\lambda t} ) \ E( N_t - \lambda t )^+ = \frac{1}{2 \pi}$.
Trying to prove the claim, I started out like this:
$$ \frac{1}{\sqrt{\lambda t}} E( N_t - \lambda t )^+ = \frac{1}{\sqrt{\lambda t}} \exp(-\lambda t) \ 
\sum_{k \geq \lambda t} \ (k - \lambda t) \frac{(\lambda t)^k}{k!}  $$
$$ \approx \frac{1}{\sqrt{\lambda t}} \exp(-\lambda t) \ 
\sum_{k \geq \lambda t} \ (k - \lambda t) \frac{(\lambda t)^k}{\sqrt{2 \pi k} \left( \frac{k}{e} \right) ^k}$$
Does anybody know how to finish the proof?
Thanks a lot for your help! Regards, Si
 A: We have 
\begin{align*}
\frac 1{\sqrt{\lambda t}}E(N_t-\lambda t)^+&=\frac{e^{-\lambda t}}{\sqrt{\lambda t}}\sum_{k=\lfloor \lambda t\rfloor+1}^{+\infty}\frac{(\lambda t)^k}{k!}(k-\lambda t)\\
&=\frac{e^{-\lambda t}}{\sqrt{\lambda t}}\left(\sum_{k=\lfloor \lambda t\rfloor+1}^{+\infty}\frac{(\lambda t)^k}{k!}k-\sum_{k=\lfloor \lambda t\rfloor+1}^{+\infty}\frac{(\lambda t)^k}{k!}(\lambda t)\right)\\
&=\frac{e^{-\lambda t}}{\sqrt{\lambda t}}(\lambda t)\left(\sum_{k=\lfloor \lambda t\rfloor+1}^{+\infty}\frac{(\lambda t)^{k-1}}{(k-1)!}-\sum_{k=\lfloor \lambda t\rfloor+1}^{+\infty}\frac{(\lambda t)^k}{k!}\right)\\
&=\frac{e^{-\lambda t}}{\sqrt{\lambda t}}(\lambda t)\left(\sum_{j=\lfloor \lambda t\rfloor}^{+\infty}\frac{(\lambda t)^j}{j!}-\sum_{j=\lfloor \lambda t\rfloor+1}^{+\infty}\frac{(\lambda t)^j}{j!}\right)\\
&=\sqrt{\lambda t}e^{-\lambda t}\frac{(\lambda t)^{\lfloor \lambda t\rfloor}}{\lfloor \lambda t\rfloor !},
\end{align*}
and using Stirling's approximation we get
\begin{align*}
\frac 1{\sqrt{\lambda t}}E(N_t-\lambda t)^+&\overset{t\to\infty}{\sim}\sqrt{\lambda t}e^{-\lambda t}(\lambda t)^{\lfloor \lambda t\rfloor}\left(\frac e{\lfloor \lambda t\rfloor}\right)^{\lfloor \lambda t\rfloor}\frac 1{\sqrt{2\pi \lfloor \lambda t\rfloor}}\\
&=\sqrt{\frac{\lambda t}{\lfloor \lambda t\rfloor}}\exp\left(\lfloor \lambda t\rfloor-\lambda t+\lfloor \lambda t\rfloor\ln \frac{\lambda t}{\lfloor \lambda t\rfloor}\right)\frac 1{\sqrt{2\pi}}.
\end{align*}
Let $f(t)$ the fractional part of $\lambda t$, and $F(t)=\lfloor \lambda t\rfloor$ 
\begin{align*}
\frac 1{\sqrt{\lambda t}}E(N_t-\lambda t)^+&\overset{t\to\infty}{\sim}\frac 1{\sqrt{2\pi}}
\sqrt{1+\frac{f(t)}{F(t)}}\exp\left(-f(t)+F(t)\ln\left(1+\frac{f(t)}{F(t)}\right)\right),
\end{align*}
and since 
$$-f(t)+F(t)\ln\left(1+\frac{f(t)}{F(t)}\right)=-f(t)+f(t)+f(t)o(1)=f(t)o(1),$$
and $\lim_{t\to+\infty}\frac{f(t)}{F(t)}=0$, we finally get 
$$\lim_{t\to\infty}\frac 1{\sqrt{\lambda t}}E(N_t-\lambda t)^+=\frac 1{\sqrt{2\pi}}.$$
