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I'm solving an exercise from a last year exam. Suppose we have an Poisson process $(N_t)$ with parameter $\lambda=\frac{1}{3}$ given with respect to a filtration $(\mathcal{F}_t)$.

The first question is:

1) for $t\ge 0$ show $N_t^3=\int_0^t (3N^2_{s-}+3N_{s-}+1)dN_s$ with the hint $(a+b)^3=a^3+3a^2b+3ab^2+b^3$.

and the second question

2) define the process $M_t:=N_t^3-\frac{t}{3}-\int_0^t(N^2_{s-}+N_{s-})ds$ and show that it is a local martingale. There is also hint, which says: we should use $1)$ and that the compensated Poisson process $X_t:=N_t-\lambda t$ is a martingale.

For the first one I tried quite everything. I used $Itô$ to solve it, but I never used the hint. So maybe my solution is wrong. How could you solve this with the hint? Here is what I've done

let $f(x)=x^3$, hence applying Itô yield,

$$N^3_t=\int_0^t3N^2_{s-}dN_s+\frac{1}{2}\int_0^t 6 N_{s-}d[N]_s+\sum_{0<s\le t}(f(N_s)-f(N_{s-}))$$ using $[N]=N$, we get $$N^3_t=\int_0^t3N^2_{s-}+3N_{s-}dN_s+\sum_{0<s\le t}(f(N_s)-f(N_{s-}))$$ hence the question reduces to show that $\sum_{0<s\le t}(f(N_s)-f(N_{s-}))=N_t$. But at this point I struggle. Why is this true? Clearly $f(N_s)-f(N_{s-})$ is either $0$, or $(n+1)^3-n^3=3(n^2+n+1)$, why should this, summing up, be $N_t$?

For the second question, I simply put the equation from $1)$ for $N^3_t$ in, but this does not help a lot. I also recognized that $\frac{t}{3}=\lambda t$, so I guess there is the point, where we have to use the result for the compensated poisson process.

However some help would be appreciated. Thanks in advance.


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up vote 1 down vote accepted

2.) For the second process you have $$ M_t = N_t^3 -\lambda t - \lambda \int\limits_0^t (3N^2_{s-}+3N_{s-})\mathrm ds = N_t^3 - \int\limits_0^t (3 N^2_{s-}+3N_{s-} +1)\mathrm d(\lambda \cdot ds) $$ $$ = \int\limits_0^t (3 N^2_{s-}+3N_{s-} +1)\mathrm d(N_s - \lambda \cdot ds) $$ which is an integral w.r.t. martingale.

1.) I think, it was a right way just to apply Ito to find $\mathrm dN^3_t$. I am not sure, whether you have the right Ito formula for the Poisson process. Forget about the stochastic framework and think about $N_t$ path-wise, as fortunately one can do this for Poisson processes. Then $$ N^3_t = N_0^3+\sum_{s\leq t}(N^3_s - N^3_{s-}) $$ where you can apply the hint to find that the latter sum is $$ \sum_{t\geq s:\;\mathrm dN_s = 1}(3N_{s-}^2+3N_{s-}+1) = \int_0^t (3N_{s-}^2+3N_{s-}+1)\mathrm dN_s $$ where you only used the fact that $N_t$ is a pure jump process, rather than the stochastic nature of it.

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thanks for your help. I think I would not get $2.)$. I think my solution for 1 is not correct, see my edit in the question. maybe you can help me with this,too? – user20869 Jan 11 '13 at 9:16
@hulik: updated. One think to remember when doing Ito calculus for Poisson process is that there you can do everything path-wise, stochastic aspect does not affect at all. – Ilya Jan 11 '13 at 9:30
thans for the update: some questions: why is $N_t^3=\sum_{s\le t}(N_s^3-N_{s-}^3)$ (we can forget about $N_0^3$ since this is zero). How did you apply the hint? And why is this sum over $s:dN_s=1$ equal an integral? – user20869 Jan 11 '13 at 9:37
@hulik 1. The formula $X_t = X_0 + \sum_{s\leq t}\Delta X_s$ holds for any piecewise-constant function $X:[0,t]\to\Bbb R$. In particular, for $N_t$ it also follows from the Ito formula. By the way, can you provide the formula that you have in mind and the source for it? $\phantom{x}\\$ 2. The hint I applied in the same way as you did: $$ \Delta N^3_s = N^3_s - N^3_{s-} = (N_{s-}+1)^3 - N_{s-}^3. $$ – Ilya Jan 11 '13 at 9:50
3. Yet again, for any piece-wise constant function $X:[0,t]\to \Bbb R$ it holds that $$ \int_0^t f(X_{s-})\mathrm dX_s = \sum_{s\leq t} f(X_{s-})\Delta X_s = \sum_{s:\;\mathrm dX_s >0}f(X_{s-}) \Delta X_s $$ where the first equality holds just by the definition of the integral (Stiltjes, if I'm correct) and the second one holds just because we eliminated all zero terms from the summation. – Ilya Jan 11 '13 at 9:52

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