two question about poisson processes

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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.

hulik

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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.