The jump diffusion model is defined as $$dS_t = \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\;\;\;\;\;\;\;(1)$$ , where ${V_i}$ is a sequence of iid non-negative random variables and it is independent of $W_t$. In the Merton's jump diffusion model , $log(V) \sim N(\mu_J, \sigma^2_J)$ and $ N_t$ is a poisson process with rate $\lambda$.

I was asked to apply Ito lemma to $(d \;logS_t)$ to obtain the following:

$$S_t = S_0 exp \left( \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t\right) \prod^{N_t}_{j=1}V_j \;\;\;\;\;\;\;(2)$$

I literally do not know how to solve this problem because of the term $d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)$. This is how far I got to:

$$dlogS_t = \frac{1}{S_t} \left( \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\right) - \frac{1}{2S_t^2} d[S,S]_t$$

What exactly is $d[S,S]_t$ ? I know that $$d[S,S]_t = \sigma^2 S^2_t dt + ...$$ But what is that "..." ?

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    $\begingroup$ Well, apply Itô's lemma to $d(\log S)$... What did you find for $d\langle S\rangle$, already? $\endgroup$ – Did Feb 11 '15 at 8:52
  • $\begingroup$ That's the part I have trouble with. I have no idea how to solve $d<S>$ ? Applying ito lemma to $d(log S)$ is straightforward. $\endgroup$ – mynameisJEFF Feb 11 '15 at 8:56
  • $\begingroup$ Which parts of $dS$ enter into $d\langle S\rangle$? Not many... (As an aside, note that if your trouble is that you cannot compute $d\langle S\rangle$, this is what your question should mention.) $\endgroup$ – Did Feb 11 '15 at 8:57
  • $\begingroup$ Sorry. I still cannot figure out the answer. Can you write an answer please ? $\endgroup$ – mynameisJEFF Feb 11 '15 at 9:12
  • $\begingroup$ @Did, please see update of my question $\endgroup$ – mynameisJEFF Feb 11 '15 at 9:19

I suspect that your SDE defining the $S_t$ process is problematic in the term "$S_t\text d\Big(\sum\limits_{i=1}^{N_t}(V_i-1)\Big)$". Intuitively, this term should capture instantaneous random jumps when they occur in any given realisation of the process. That is, if a jump occurs at time $t$, then the process jumps from $S_{t-}$ to $S_{t-}V_t$, where $S_{t-}$ is the left limit of the process at time $t$. So, it makes more sense for $S_t$ to be the jump-diffusion Levy process (so right-continuous, left-limit) defined by $$ \text dS_t = \mu S_{t-}\text dt + \sigma S_{t-}\text dW_t + \text d\Big(\sum\limits_{0\leqslant r\leqslant t}S_{r-}(V_r - 1){\bf 1}_{\{N_r-N_{r-}=1\}}\Big),\tag{1} $$

with jumps determined by a Compound Poisson process independent of $\{W_t\}_{t>0}$. That is, a Poisson process $\{N_t\}_{t>0}$ determines when jumps occur, and the size of such a jump at time $t$ is determined as $S_{t-}(V_t - 1)$ (where $\log V_t\sim\mathcal N(\mu_{J},\sigma^2_{J})\ $ for i.i.d. $V_t$).

Assuming $(1)$ for your SDE and $\mathbb E[\int_{0}^{T}S_{u-}^2\text du] < \infty$, Ito's lemma for jump-diffusion processes is applicable here. Note that, for a $\ C^{1,2}\ $ function $\ f:[0,T]\times\mathbb R\to\mathbb R$, the process $f(t,S_t)$ satisfies

$$ \begin{eqnarray*} f(t,S_t) - f(0,S_0) &=& \int_{0}^{t}(\partial_uf + \mu S_{u-}\partial_sf + \frac{1}{2}\sigma^2S_{u-}^2\partial_{ss}f)\text du \\ && + \sigma \int_{0}^{t}S_{u-}\partial_sf~\text dW_u + \sum_{0\leqslant r\leqslant t}(f(r,S_r)-f(r-,S_{r-})){\bf 1}_{\{N_{r}-N_{r-}=1\}}\,. \end{eqnarray*} $$

This is simply Ito's lemma for a diffusion process but, in addition, the final term adds the change in the process due to the jumps determined by the compound Poisson process. So, choosing $f(S_t):=\log S_t$ gives us

$$ \begin{eqnarray*} \log\Big(\frac{S_t}{S_0}\Big) &=& (\mu-\frac{1}{2}\sigma^2)t + \sigma W_t + \sum_{0\leqslant r\leqslant t}{\bf 1}_{\{N_{r}-N_{r-}=1\}}\log\Big(\frac{S_{r-}V_r}{S_{r-}}\Big) \\ &=&(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t + \sum_{0\leqslant r\leqslant t}{\bf 1}_{\{N_{r}-N_{r-}=1\}}\log V_r\,. \end{eqnarray*} $$ Therefore, in your notation,

$$ S_t = S_0 e^{(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t}\prod_{i=1}^{N_t}V_i\,. $$

  • $\begingroup$ I just want to ask in the part where you have $log(\frac{S_{r-} V_r}{S_{r-}})$, shouldn't it be $log(\frac{S_{r-} (V_r- 1)}{S_{r-}})$ ? $\endgroup$ – mynameisJEFF Mar 30 '15 at 10:54
  • $\begingroup$ Using $S_r = S_{r-}V_r$ when a jump occurs at time $r$, we have $$f(r,S_r)-f(r-,S_{r-}) = \log(S_r)-\log(S_{r-}) = \log\Big(\frac{S_r}{S_{r-}}\Big) = \log\Big(\frac{S_{r-}V_r}{S_{r-}}\Big)$$ $\endgroup$ – ki3i Mar 30 '15 at 11:53
  • $\begingroup$ Yes , I understand that part, as it is based on the ito doeblein formula for jump process. However, I don't understand why is the minus 1 missing ? $\endgroup$ – mynameisJEFF Mar 30 '15 at 11:55
  • $\begingroup$ @mynameisJEFF, If you understand that part, then why do you think there should be a "-1"? $\endgroup$ – ki3i Mar 30 '15 at 11:57
  • $\begingroup$ O. -1 is a constant so it doesn't constitute a jump. Am I right ? $\endgroup$ – mynameisJEFF Mar 30 '15 at 11:58

In your case

the quadratic variation can be obtained by formally squaring the SDE $ d[S,S]_t = dS_t \cdot dS_t =\left( \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right) \right)^2 $ using the fact that $ (\mu S_t)^2 dt^2 = 0 $ and $ \mu S_t dt \cdot \sigma S_t dW_t =0 $ we get

$ d[S,S]_t = \sigma^2 S_t^2 dt + 2 [ d\left(\sum^{N_t}_{i=1}(V_i - 1)\right), \mu S_t dt ] + 2 [ d\left(\sum^{N_t}_{i=1}(V_i - 1)\right), dW_t ] + [ d\left(\sum^{N_t}_{i=1}(V_i - 1)\right), d\left(\sum^{N_t}_{i=1}(V_i - 1)\right) ]_t $.

Since $V_i $ are i.i.d. $ [dV_i, dt ]=0$ and $ [dV_i, dW_t ]=0$ and $ [dV_i, dV_j ]=\delta_{i\,j}$.
For the last term we also have $[dV_i , dV_i ] = dV_i$. Then finally

$ dlogS_t = \frac{1}{S_t} \left( \mu S_t dt + \sigma S_t dW_t + S_t d \left(\sum^{N_t}_{i=1}(V_i - 1)\right)\right) - \frac{1}{2S_t^2} \sigma^2 S^2_t dt = $

$ dlogS_t = \left( \mu - \frac{1}{2 } \sigma^2 \right) dt + \sigma \; dW_t + d\left(\sum^{N_t}_{i=1}(V_i - 1)\right) $.

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    $\begingroup$ $[dV_i-1, dV_i-1]=0$ ... why is that? $\endgroup$ – saz Mar 23 '15 at 7:01
  • $\begingroup$ A Poisson process has quadratic variation equal to itself. This is because it is a pure jump process with jump sizes equal to 1. If $V_t$ is the Poisson process then the only contribution to the quadratic variation $[V,V]$ comes from the jumps. $\endgroup$ – user48672 Mar 23 '15 at 22:28
  • $\begingroup$ my mistake. I overlooked. $\endgroup$ – user48672 Mar 23 '15 at 22:30
  • $\begingroup$ Is this answer wrong? $\endgroup$ – user357269 Dec 11 '16 at 12:27

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