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.

enter image description here

  • $\begingroup$ 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 $\endgroup$ Apr 7 '16 at 13:15

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^-} $)

  • $\begingroup$ 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 $S_t\neq S_{t-}$. $\endgroup$ Apr 7 '16 at 13:44
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.