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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.


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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

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

Sorry for my english.

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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.

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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 $dN_s$.

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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

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