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For underlying asset $$d S = r S dt + \sigma S d W + (J-1)Sd N$$ here $W$ is a Brownian motion, $N(t)$ is Poisson process with intensity $\lambda.$

Suppose $J$ is log-normal with standard deviation $\sigma_J,$ denote $$k = E[J-1]$$ then the value of vanilla call is $$\sum\limits_{n=0}^{\infty}\dfrac{(\lambda'\tau)^n}{n!}e^{-\lambda'\tau}V_{BS}(S,t;\sigma_n,r_n)$$ Here $$\lambda' = \lambda(1+k),\ \tau =T - t$$ $$\sigma^2_n = \sigma^2 + \dfrac{n\sigma_J^2}{\tau},\ r_n=r-\lambda k+\dfrac{n\log(1+k)}{\tau}$$ and $V_{BS}$ is Black-Scholes value of a call without jumps.

The result seems to be the weighted mean of vanilla calls, but how to deduce this conclusion? Or is there any reference?

Suppose $\log J \sim N(\mu,\sigma^2_J)$ and $Z$ is standard normal, then we have \begin{eqnarray*} \log S_T &=& \log S_0 + (r - \dfrac{1}{2}\sigma^2)T + \sigma\sqrt{T} Z + N(T)\log(J)\\ &\sim& \log S_0 + \dfrac{1}{2}N(T)\sigma_J^2 +N(T)\mu + (r - \dfrac{1}{2}\hat{\sigma}^2)T + \hat{\sigma}\sqrt{T}Z\\ &=& \log \hat{S}_0 + (r - \dfrac{1}{2}\hat{\sigma}^2)T + \hat{\sigma}\sqrt{T} Z \end{eqnarray*} we omit the representations of $\hat{S}_0,\ \hat{\sigma}.$ So I think the price at $0$ should be $$\sum\limits_{n=0}^{\infty}\dfrac{(\lambda T)^n}{n!}e^{-\lambda T}V_{BS}(\hat{S}_0,0;\hat{\sigma}_n,r)$$

Why the author change the risk free rate $r_n$ and intensity $\lambda',$ does he change the measure for both Brownian motion and Poisson process?

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  • 1
    $\begingroup$ See this question. $\endgroup$
    – Gordon
    May 6, 2017 at 16:28
  • $\begingroup$ Looks like the Merton model. $\endgroup$
    – nbbo2
    May 6, 2017 at 22:03
  • 1
    $\begingroup$ Possible duplicate of Formula for Merton jump diffusion call price $\endgroup$ May 6, 2017 at 22:39
  • $\begingroup$ please see the update @Gordon $\endgroup$
    – A.Oreo
    May 7, 2017 at 13:53
  • $\begingroup$ pls see the update @LocalVolatility $\endgroup$
    – A.Oreo
    May 7, 2017 at 13:53

1 Answer 1

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We assume that the process $\{J_t, \, t\ge 0\}$ is defined at the jump times of the Poisson process $\{N_t, \, t \ge 0\}$, and all the jump sizes are independent and identically distributed. That is, \begin{align*} Q_t \equiv \int_0^t (J_t-1) dN_t = \sum_{n=1}^{N_t} (J_i-1), \end{align*} where $J_i$, for $i=1, \ldots, \infty$, are independent and $\xi_i = \ln J_i \sim N\left(u, \sigma_J^2\right)$. Note that, the process \begin{align*} Q_t -\lambda k\, t \end{align*} is a martingale. Moreover, as the discount stock price process $\{e^{-rt}S_t, \ t\ge 0\}$ is a martingale, under the risk-neutral measure, we assume that the stock price process $\{S_t, \, t \ge 0\}$ satisfies an SDE of the form \begin{align*} dS_t = S_t \Big[(r-\lambda k)dt + \sigma dW_t + dQ_t\Big]. \end{align*} Then, \begin{align*} S_T &= e^{\left(r-\lambda k-\frac{1}{2}\sigma^2\right)T + \sigma W_T} \prod_{i=1}^{N_T} J_i = e^{\left(r-\lambda k-\frac{1}{2}\sigma^2\right)T + \sigma W_T + \sum_{i=1}^{N_T} \xi_i}. \end{align*} Consequently, $n\ge 0$, \begin{align*} S_T\, \mathbb{I}_{N_T=n} &=e^{\left(r-\lambda k + \frac{n u}{T}-\frac{1}{2}\sigma^2\right)T + \sigma W_T + \sum_{i=1}^n (\xi_i-u)}\\ &=e^{\left(r-\lambda k + \frac{n u}{T}-\frac{1}{2}\sigma^2\right)T + \sqrt{\sigma^2 + \frac{n \sigma_J^2}{T}}\sqrt{T} Z}\\ &=e^{\left(r_n -\frac{1}{2}\sigma_n^2\right)T + \sigma_n \sqrt{T} Z}, \end{align*} where $Z$ is a standard normal random variable, $\sigma_n= \sqrt{\sigma^2 + \frac{n \sigma_J^2}{T}}$, and \begin{align*} r_n &= r-\lambda k + \frac{n u}{T}-\frac{1}{2}\sigma^2 +\frac{1}{2}\sigma_n^2\\ &= r-\lambda k + \frac{n\ln(1+k)}{T}. \end{align*}

Therefore, \begin{align*} e^{-rT}E\left((S_T-K)^+ \right) &=\sum_{n=0}^{\infty}e^{-rT}E\left((S_T-K)^+ \mathbb{I}_{N_T=n}\right)P(N_T=n)\\ &=\sum_{n=0}^{\infty}e^{(-r+r_n)T}e^{-r_nT}E\left((S_T-K)^+ \mathbb{I}_{N_T=n}\right)P(N_T=n)\\ &=\sum_{n=0}^{\infty}\frac{(\lambda T)^n}{n!}e^{-\lambda T}e^{(-r+r_n)T}C(S_0, K, r_n, \sigma_n, T)\\ &=\sum_{n=0}^{\infty}\frac{(\lambda T)^n}{n!}e^{-\lambda T}e^{\left(-\lambda k + \frac{n\ln(1+k)}{T}\right)T}C(S_0, K, r_n, \sigma_n, T)\\ &=\sum_{n=0}^{\infty}\frac{(\lambda T)^n}{n!}e^{-\lambda(1+k) T}e^{\left(-\lambda k + \frac{n\ln(1+k)}{T}\right)T}C(S_0, K, r_n, \sigma_n, T)\\ &=\sum_{n=0}^{\infty}\frac{(\lambda' T)^n}{n!}e^{-\lambda' T}C(S_0, K, r_n, \sigma_n, T), \end{align*} where $\lambda'=(1+k)\lambda$ and $C(S_0, K, r_n, \sigma_n, T)$ is the Black-Scholes option price.

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  • $\begingroup$ Thanks a lot, why here is $d S = S_t\left((r- \lambda k)d t +\sigma d W + (J-1)d N_t\right)$? Under the original measure, I think it should be $d S = S_t\left((r+ \lambda k)d t +\sigma d W + ((J-1)d N_t - \lambda kd t\right) $ $\endgroup$
    – A.Oreo
    May 8, 2017 at 2:18
  • $\begingroup$ If you already have it in the end, you don't need it at the drift to have the martinaglity. $\endgroup$
    – Gordon
    May 8, 2017 at 11:04
  • $\begingroup$ but the given condition of $dS$ on the top have no $\lambda kdt$ in both $d N$ and $r,$ then the result should not be the given one. And I think $dS$ should write as $dS =S(rdt + \sigma(dW + \dfrac{\lambda k}{\sigma}dt) + ((J-1)dN - \lambda kdt),$ then change the measure. But we can no longer obtain the representation of $r_n$ bove. $\endgroup$
    – A.Oreo
    May 8, 2017 at 12:52
  • $\begingroup$ Sorry, I think my deduction is right, and after changing the measure, we have your representation of $dS.$ The last question is the method in my update not right? Since the discounted value of $S$ is not martingale, and we can't simple replace by $\hat{S}$ and $\hat{\sigma}.$ $\endgroup$
    – A.Oreo
    May 8, 2017 at 13:15
  • $\begingroup$ You can still work with the original process you provided and calculate the option price in a mathematical sense. However, note that, as $e^{-rt}S_t$ is not a martingale, the term $-\lambda k$ has to be added to the drifted to make it appropriate. $\endgroup$
    – Gordon
    May 8, 2017 at 13:16

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