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EdisonCruise
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Joined: September 15th, 2012, 4:22 am

How to make ito's forumula for jump-diffusion a martingale

November 20th, 2019, 7:40 am

\(dY_t=Z_{N_t} dN_t \) is a compound Poisson process with intensity \(\lambda\) and \( Z_{N_t} \) is a random variable for the jump size. \(dW_t \) is Brownian motion.
The jump diffusion process \(X_t\) is defined as
$$ dX_t=\nu_t dt+u_t dW_t + \eta_t dY_t$$
So the Ito's lemma for this jump diffusion process is 
$$ df(X_t)=\nu_t f'(X_t)dt+u_t f'(X_t) dW_t+ \frac{1}{2}u^2_t f''(X_t)+(f(X_t)-f(X_t-))dN_t$$
The compensated Poisson process of \( dN_t \) is \( d\hat{N_t}=dN_t-\lambda dt \), which is a martingale. Then how to use \(d\hat{N_t} \) to make \( df(X_t) \) a martingale? 
A simple subsitution as below seems somehow not correct:
$$ df(X_t)=\nu_t f'(X_t)dt+u_t f'(X_t) dW_t+  \frac{1}{2}u^2_t f''(X_t)+\lambda(f(X_t)-f(X_t-))dt+(f(X_t)-f(X_t-))d\hat{N_t}$$

According to page 26 of http://people.ucalgary.ca/~aswish/JumpProcesses.pdf, maybe it should be something like this:
$$ df(X_t)=\nu_t f'(X_t)dt+u_t f'(X_t) dW_t+  \frac{1}{2}u^2_t f''(X_t)+\lambda (E[(f(X_t)-f(X_t-)]-E[Z_{N_t}]\eta_tf'(X_t))dt+(f(X_t)-f(X_t-))d\hat{N_t}$$
but I can't understand why.
 
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Alan
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Re: How to make ito's forumula for jump-diffusion a martingale

November 20th, 2019, 9:53 pm

I think you're making this harder than it needs to be. 

If you want [$]V_t  \equiv f(t,X_t)[$] to be a martingale, first add a [$]f_t dt[$] term to the 2nd line. (I added a t-dependence because in applications, typically V will be the value of a derivative security with a known value formula at [$]t=T[$], some expiration/maturity). Then, take the [$]E_{x,t}[ \cdots][$] expectation of the amended second line and set the resulting coefficient of dt to be zero. Here [$]E_{x,t}[ \cdots][$] is the time-t expectation, conditional on [$]X_{t-}=x[$]. Doing so yields

[$]f_t + \frac{1}{2} u^2 f_{xx} + \nu f_x + \lambda \int (f(t,x+\xi) - f(t,x)) \, q(\xi) \, d \xi = 0[$],

where [$]q(\xi)[$] is the jump-size density and subscripts (with the exception of the one on V) denote partial derivatives. Also, to avoid confusion about the meaning of subscripts, here [$]u = u(t,x)[$] and [$]\nu = \nu(t,x)[$], two given functions. If the jump-size density also depends upon [$]t[$], you can write [$]q(t,\xi)[$] instead.

If [$]f(t,x)[$] satisfies this PIDE, then  [$]V_t  \equiv f(t,X_t)[$]  is a martingale.

As a check, take [$]dX_t = \nu \, dt + c \, dN_t[$], where [$]\nu[$] and [$]c[$] are constants. Suppose [$]f(t,x)=x[$]. Since [$]q(\xi) = \delta(\xi-c)[$], using the Dirac delta, the PIDE implies that [$]\nu + \lambda \, c = 0[$], which means that [$]dX_t = c \, (dN_t - \lambda \, dt)[$], as expected.  
 
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EdisonCruise
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Joined: September 15th, 2012, 4:22 am

Re: How to make ito's forumula for jump-diffusion a martingale

November 21st, 2019, 12:28 am

Thank you so much Alan. That's my also my understanding of this problem before reading page 26 of http://people.ucalgary.ca/~aswish/JumpProcesses.pdf, based on which after vanishing the \(dt\) term, there should be an additional \( -E [Z_{N_t}] \eta_t f'(X_t) \)  term in the \(f(t,x) \) PIDE.
 
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FaridMoussaoui
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Location: Genève, Genf, Ginevra, Geneva

Re: How to make ito's forumula for jump-diffusion a martingale

November 21st, 2019, 12:39 am

Go to the basics. There is a chapter on Jump Processes in Shreve book "Stochastic Calculus for Finance II". Very well explained.
For advanced material, check Rama & Tankov book.
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