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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

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He Davide! You're of course right, this isn't rigorous at all! I was just hoping this would maybe give some people a hint into the right direction... –  Mad Si Nov 24 '11 at 14:00
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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 $$\lim_{t\to\infty}\frac 1{\sqrt{\lambda t}}E(N_t-\lambda t)^+=\frac 1{\sqrt{2\pi}}.$$

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Nice! Thanks a lot for your proof! Regards, Si –  Mad Si Nov 24 '11 at 13:58
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