Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 8 down vote accepted

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 $$\lim_{t\to\infty}\frac 1{\sqrt{\lambda t}}E(N_t-\lambda t)^+=\frac 1{\sqrt{2\pi}}.$$

share|cite|improve this answer
Nice! Thanks a lot for your proof! Regards, Si – Mad Si Nov 24 '11 at 13:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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