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In the textbook by Shreve in sec. 11.7.2 a jump-diffusion process is introduced. More precisely

$$ dS_t = \alpha\,S_t\,dt+\sigma\,S_t\,dW_t+S_{t-}\,d\left(Q_t-\beta\,\lambda\,t\right)\quad (1) $$

where $Q_t = \sum_{i=1}^{N_t}Y_i$ and $N_t$ is Poisson with intensity $\lambda$. The process is re-written as

$$ dS_t = (\alpha-\beta\,\lambda)\,S_t\,dt+\sigma\,S_t\,dW_t+S_{t-}\,dQ_t\quad(2). $$

The problem is that, a part from time instants in which there is no jump and hence $S_t=S_{t-}$, I cannot go from (1) to (2), because if there is a jump of size $Y_i$ at time $t$ it holds that

$$ \frac{S_t-S_{t-}}{S_{t-}} = Y_i\rightarrow S_t = S_{t-}\,(Y_i+1). $$

and so I get

$$ dS_t = \alpha\,S_t\,dt+\sigma\,S_t\,dW_t+S_{t-}\,dQ_t-S_{t-}\,\beta\,\lambda\,dt\neq (\alpha-\beta\,\lambda)\,S_t\,dt+\sigma\,S_t\,dW_t+S_{t-}\,dQ_t. $$

Here there is a snapshot of the textbook.

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  • $\begingroup$ Can you be more specific how do you have the inequality? It seems to me that is simply a rearrangement of terms. $\endgroup$ – Gordon Apr 7 '16 at 12:57
  • $\begingroup$ I report here what is written in the textbook $\endgroup$ – AlmostSureUser Apr 7 '16 at 13:15
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Could it be that your problem is only due to the $t^-$ notation convention?

Think of it that way, it is only worth distinguishing $S_{t^-}$ from $S_t$ at a jump time. Elsewhere, knowing that Brownian motion paths are continuous, you'll always have $S_t = S_{t^-}$.

Thus you could also write the SDE:

$$\frac {dS_t}{S_{t^-}} = \alpha dt+\sigma dW_t+ d\left(Q_t-\beta\,\lambda\,t\right)$$

Or simply drop the $t^-$ notation altogether if it bothers you. Just remember that we are dealing with processes with càdlàg paths (right continuous $S_t $, left limit $S_{t^-} $)

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  • $\begingroup$ Something is not clear in your answer. The equation $$ S_{t^-} d(Q_t - \beta\lambda t) = S_{t^-} dQ_t - S_t \beta\lambda dt $$ seems wrong to me. ps = I agree that going from (1) to (2) at time instants in which there are no jumps is ok, the problem is when there is a jump and so $S_t\neq S_{t-}$. $\endgroup$ – AlmostSureUser Apr 7 '16 at 13:44
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    $\begingroup$ Sometimes ago, you posted a question concerning the solution to Merton's SDE, I suggested you to read ntu.edu.sg/home/nprivault/MA5182/…. See the exercise 1) a), it is exactly what I just described. $\endgroup$ – Quantuple Apr 7 '16 at 13:59
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    $\begingroup$ You should really think in terms of adding two components: one has discontinuous paths (the compounded Poisson part) and the other one is continuous. As far as the continuous part is concerned, you can interchangeably use $S_t$ or $S_{t^-}$ because of continuity. Please re-read the answer and try to pin point what is not clear to you so that someone can further help. $\endgroup$ – Quantuple Apr 7 '16 at 14:00
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    $\begingroup$ PS: the equation you write is of course right since for the continuous part $S_t = S_{t^-}$, but the Shreve notation is specifically made to distinguish what happens at a jump. $\endgroup$ – Quantuple Apr 7 '16 at 14:05
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    $\begingroup$ As you observed, the SDE tells us you that, at a jump time: $S_t = S_{t^-} (Y_i + 1)$, thus you cannot have $Y_i < -1$ if you want the asset price to remain positive. This is simply a consequence of the manner in which you define the SDE. $\endgroup$ – Quantuple Apr 7 '16 at 15:22

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