Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to evaluate the following stochastic integral?

$$\int_0^t M_{s^-}^2 dM_s$$

where $M_t = N_t - \lambda t$ is a compensated Poisson process.

I tried to apply Ito's formula to $M_t^2$ but still cannot solve it. Any help appreciated.

References

share|improve this question
    
Hi what do you mean exactly when you say "solve"? –  TheBridge May 9 '11 at 20:56
    
is this question still actual for you? –  Ilya Jun 10 '11 at 14:57
add comment

3 Answers

You can see this book Introduction to Stochastic Integration p.109 and following; example 7.6.3 answers your question.

Sorry for my english.

share|improve this answer
add comment

By Shreve (Stochastic Calculus in Finance II) we use formula $$ X_t Y_t = X_0Y_0 +\int\limits_0^t X_{s-}dY_s + \int\limits_0^t Y_{s-}dX_s + [X,Y](t) $$ so by taking $X = M$ and $Y = M^2$ we obtain $$ \int\limits_0^t M^2_{s-}dM_s = M^3_s - \int\limits_0^t M_{s-}d(M^2_s) - [M,M^2](t) $$ where the latter term is given by $ [M,M^2](t) = \sum\limits_{s\leq t}\Delta M_s\Delta M^2_s $ and $\Delta X_s = X_s - X_{s-}$.

Simply, $\Delta M_s = \Delta N_s$ since $\Delta (-\lambda s) = 0$ due to the continuity of this term. There is a little bit of work with $\Delta M^2_t$. First of all, $$ \Delta M^2_t = \Delta (N^2_t - 2\lambda t N_t + \lambda^2 t^2) = \Delta N^2_t - 2\lambda t \Delta N_t. $$

Further, $$ \Delta N^2_t = N^2_{t} - N^2_{t-} = (N_{t} - N_{t-})(N_{t} + N_{t-}) = \Delta N_t (2N_{t-}+\Delta N_t) $$

Recall, that $(\Delta N_t)^2 = \Delta N_t$, so $ \Delta N^2_t = (2N_{t-}+1)\Delta N_t. $

Now $ \Delta M^2_t = (2N_{t-}-2\lambda t+1)\Delta N_t, $ so we obtain $$\sum\limits_{s\leq t}\Delta M_s\Delta M^2_s = \int\limits_0^t(2N_{s-}-2\lambda s+1)dN_s.$$

Now we deal with $ \int\limits_0^t M_{s-}d(M^2_s)$. Using the very first formula for $X = Y = M$ we obtain $$ M^2_t = 2 \int\limits_0^t M_{s-}dM_s+\sum\limits_{s\leq t}(\Delta M_s)^2 =2 \int\limits_0^t M_{s-}dM_s+N_t, $$ so $ \int\limits_0^t M_{s-}d(M^2_s) = 2\int\limits_0^t M^2_{s-}dM_s + \int\limits_0^t M_{s-}dN_s$.

We can put it on the right-hand side: $$ 3\int\limits_0^t M^2_{s-}dM_s = M^3_t - \int\limits_0^t (M_{s-}+2N_{s-}-2\lambda s+1)dN_s. $$ Equivalently: $$ 3\int\limits_0^t M^2_{s-}dM_s = M^3_t - \int\limits_0^t (3N_{s-}-3\lambda s)dN_s -N_t. $$

To calculate $\int\limits_0^t N_{s-}dN_s$ we apply the very first formula to $X = Y = N$ and obtain
$$ \int\limits_0^t N_{s-}dN_s = \frac{1}{2}(N^2_t - N_t). $$

Finally, we have $$ \int\limits_0^t M^2_{s-}dM_s = \frac{1}{3}M^3_t - \frac{1}{2}N^2_t - \frac{1}{6}N_t-\lambda\int\limits_0^ts\,dN_s $$ and I don't think that the latter integral can be simplified.

share|improve this answer
add comment

I am not an expert on the calculus of jump processes, but you may make some progress by applying Ito's formula to $M_s^3$ to give

$$d(M_s^3) = 3M_s^2 dM_s + 3M_s dM_s^2$$

and hence

$$\int_0^t M_s^2 dM_s = \int_0^t d(\tfrac{1}{3}M_s^3) - M_s dM_s^2 = \tfrac{1}{3} M_t^3 - \int_0^t M_s dM_s^2$$

The exact interpretation of the integral term depends on the interpretation of $dM_s=dN_s-\lambda ds$, and hence on the interpretation of $dN_s$.

share|improve this answer
    
I got this far but I am having trouble with substituting the poisson process $N_s$ into it. –  user957 May 9 '11 at 18:18
    
you forgot about quadratic variation and presented the wrong formula for stochastic differential. –  Ilya Jun 10 '11 at 11:32
add comment

Your Answer

 
discard

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.