5
$\begingroup$

Lets have the next jump difussion Stochastic Process: $$S_t = S_0 e^{\sigma W_t + (v-\frac{\sigma ^2}{2})t}\prod_{i=1}^{N_t}(1+J_i)$$

where $W_t$ is the Brownian Motion, hence $G_t \equiv e^{\sigma W_t + (v-\frac{\sigma ^2}{2})t}$ is the Geometric Brownian Motion, $N_t$ is the Poisson Process and $R_t \equiv \prod_{i=1}^{N_t}(1+J_i)$ is the Multiplicative Poisson Compound Process.

Suppose there exists a Martingale Probability $\mathbb{Q}$ and that under $\mathbb{Q}$ the Girsanov Theorem hipótesis valid. Moreover suppose that under $\mathbb{Q}$ $N_t$ has a Poisson Rate $\widehat{\lambda}$.

In this context I have to price the European Put Option of S_t, that is

$$P=e^{-r(T-t)}\mathbb{E}_\mathbb{Q}((k-S_T)_+|S_t)$$

I have thought this way, but I don´t know if it is correct. \begin{eqnarray} \mathbb{E}_\mathbb{Q}((k-S_T)|S_t)_+& = & \mathbb{E}_\mathbb{Q}((k-S_0 e^{\sigma W_T + (v-\frac{\sigma ^2}{2})T}\prod_{i=1}^{N_T}(1+J_i))_+|S_t)\\ & = & \mathbb{E}_\mathbb{Q}((k-S_t e^{\sigma (W_T-W_t) + (v-\frac{\sigma ^2}{2})(T-t)}\prod_{i={N_{t}+1}}^{N_T}(1+J_i))_+)\\ & = & \mathbb{E}_\mathbb{Q}(\mathbb{E}_\mathbb{Q}((k-S_t e^{\sigma (W_T-W_t) + (v-\frac{\sigma ^2}{2})(T-t)}\prod_{i={N_{t}+1}}^{N_T}(1+J_i))_+|N_T-N_t=n))\\ & = & \mathbb{P}_\mathbb{Q}(N_T-N_t=n))(\mathbb{E}_\mathbb{Q}((k-S_t e^{\sigma (W_T-W_t) + (v-\frac{\sigma ^2}{2})(T-t)}\prod_{i={1}}^{n}(1+J_i)_+) \end{eqnarray}

So finally, $$\mathbb{P}_\mathbb{Q}(N_T-N_t=n)=e^{-\widehat{\lambda}(t-t)}\frac{(\widehat{\lambda} (T-t))^n}{n!}$$

And $\mathbb{E}_\mathbb{Q}((k-S_t e^{\sigma (W_T-W_t) + (v-\frac{\sigma ^2}{2})(T-t)}\prod_{i={1}}^{n}(1+J_i)_+)$ can be calculated using Black-Sholes usual formula.

Is this okay or is it another way? Thanks! :)

$\endgroup$
  • $\begingroup$ One issue is the lack of specification of the distribution of the jumps $J_i$. This is rather important, seeing as how, for a sizeable enough negative jump, your price process becomes negative. $\endgroup$ – ocstl Jun 18 '15 at 12:37
  • $\begingroup$ @Edin_91, your solution is perfect.Congratulation. $\endgroup$ – user16651 Jun 29 '15 at 19:27
1
$\begingroup$

There are three main issues. As per my comment, one is the lack of specification for the distribution of the jumps (I'll assume that there is a $J_0 = 0$ at time 0 (otherwise, the process doesn't account for no jumps). Unless $P (J \leq -1) = 0$, your price process is problematic, and the Girsanov theorem is not applicable. To see why:

$S_t = S_0 e^{\sigma W_t + (\nu - \frac{\sigma^2}{2}) t} \prod_{i=1}^N (1 + J_i) \\ = S_0 e^{\sigma W_t + (\nu - \frac{\sigma^2}{2}) t + \sum_{i=1}^N \log (1 + J_i)}$

Obviously, the Radon-Nikodym derivative cannot be derived unless $(1 + J_i) \gt 0$.

Assuming a proper distribution for the jumps, the second issue concerns your risk-neutral measure. Assuming that you're using $W_t$ rather than another brownian motion for the sake of simplicity, your new measure doesn't seem to take into account the jumps, and it is by no means certain that you're dealing with a martingale.

Finally, there is an issue when going from the expectation on the number of jumps to the probability of $n$ jumps.

$P = E_{\mathbb{Q}} \left[ E_{\mathbb{Q}} \left[ (k - X_T \prod_{i = 0}^n (1 + J_i))^+ \mid N_T - N_t = n \right] \right] \\ = \sum_{n = 0}^\infty P_{\mathbb{Q}} (N_T - N_t = n) E_{\mathbb{Q}} \left[ (k - X_T \prod_{i = 0}^n (1 + J_i))^+ \right]$

where $X_T$ is obviously the price process without the jumps. While the expectation can be calculated using BS (with the proper risk-neutral measure), the unboundedness on the number of jumps can be problematic, though a truncation can give a reasonable approximation.

$\endgroup$
  • 1
    $\begingroup$ Good points +1. But your last equation makes no sense. There should be no conditioning in the first term and sum should be infinite. Also I think you are assuming you are in the Merton model when you say the.expectation can be computed using BS. $\endgroup$ – AFK Jun 30 '15 at 22:08
  • $\begingroup$ Thanks, and you're right, it makes no sense as is. As for the applicability of BS, I don't think it requires the Merton model, but it would require conditioning on the total effet of the jumps. Bit messy. $\endgroup$ – ocstl Jul 1 '15 at 11:44
0
$\begingroup$

You have to make further assumptions on the distribution of $J_i$s. For example, if $J_i$s are iid normal, your option pricing problem becomes that of Merton (1976) and the solution to it is an infinite sum. If $J_i$s are assumed to be double exponential, you end up with Kou (2004) model and it has an analytical solution. Furthermore, there are three different sources of randomness in the price process. Do you assume that $W_t, N_t$, and $J_i$ are independent from each other?

$\endgroup$

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.