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First note, I had asked a similar question here, but the thread seems to have died, so I'll revive it here with more details. As a simplification of my real problem, I want to compute $\mathbb{E}[\int_o^T N_{t-}dS_t]$, where $S_t$ is a $(\mu,\sigma)$-geom. Brownian motion, and $N_t$ an independent Poisson process with const. intensity $\lambda$. Hence $\mathbb{E}[\int_o^T N_{t-}dS_t] = \mathbb{E}[\int_o^T N_{t-}S_t(\mu dt + \sigma dW_t)]$. Ito formula shows that the latter part of the integral should be square integrable, with integrable quadratic variation, so it is zero in expectation. So we should get

$\mathbb{E}[\int_o^T N_{t-}dS_t] = \mu\int_0^T\underbrace{\mathbb{E}[N_{t-}]}_{=\mathbb{E}[N_t-\mathbb{1}_{\Delta N_t \neq 0}]=\lambda t}\mathbb{E}[S_t]dt = \int_0^T \mu\lambda t e^{\mu t}dt$

$N_{t-}$ is a simple, left-continuous process, so I tried to confirm that numerically by calculating (MC)

$\mathbb{E}\left[ \sum_{i=0}^{N_t} i(S_{\tau_{i+1}}-S_{\tau_i})\right]$,

with $\tau_0=0$, $\tau_{N_t+1}=T$ and $\tau_i$ the exp-$(\lambda)$-distributed jump times. But values don't seem to converge to the analytic solution above. Is anything wrong?

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    $\begingroup$ Your final formula doesn't involve T. But is this really of interest to research level mathematicians? It just seems like youre asking someone to check your calculations. $\endgroup$
    – George Lowther
    Commented Apr 5, 2012 at 18:42
  • $\begingroup$ It probably isn't. But thank you anyway. I thought this requires only a glimpse. And it did. Indeed, there was only a $=$ missing. So again, thank you. $\endgroup$
    – Pierre
    Commented Apr 5, 2012 at 19:09

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I don't see why your approximation of the stochastic integral works. Shouldn't you take more points in the interval $[0,T]$ and not only the jump times to get a better approximation?

[Edit: I've updated the response with a better approximation result which can be derived from Th.21 p. 64 Protter's Book]

We have have the following approximation theorem for the stochastic integral: Let $X$ a semimartingale and $H$ a càdlag or càglàd adapted process. Let $t_0^n = 0 \leq t_1^n \leq \ldots \leq t_n^{p_n} = T$ a sequence of subdivisions $[0,T]$ such that $\lim_{n \to \infty} \sup_{k=0}^{p_n} t_{k+1}^n - t_{k}^n = 0$ then:

$$ \lim_{n \to \infty} \sum_{k=0}^{p_n} H_{t_{k}^n} (X_{t_{k+1}^n} - X_{t_{k}^n}) = \int_0^T H_{s^-} dX_s \quad \text{in probability}$$

So i think a finer subdivision of the interval should allow you to approximate better the integral.

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  • $\begingroup$ Well in this special case, as $H$ is constant between two jumps $\tau_i,\tau_{i+1})$, most of the terms would cancel out. However, can you give me a reference of the above theorem? Shouldn't this limit exist only for finite variation process $X$? - cf Protter p. 44 "Stochastic Integration and Differential Equations". $\endgroup$
    – Pierre
    Commented Apr 6, 2012 at 10:10
  • $\begingroup$ I have updated the response with a consequence of a Th II.21 from Protter's book p. 64. $\endgroup$
    – Felipe Olmos
    Commented Apr 6, 2012 at 17:04

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