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ralfbuesser
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simple integral

May 11th, 2012, 10:06 pm

Can someone help with the following integral:Specifically, what is the derivative of G w.r.t X?Thanks in advanceRalf
 
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Alan
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simple integral

May 12th, 2012, 2:26 pm

Last edited by Alan on May 11th, 2012, 10:00 pm, edited 1 time in total.
 
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Alan
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simple integral

May 12th, 2012, 2:47 pm

Let's first see if we agree on an example: take S(t) = W(t), a BM and q(.) = 1, and make the upper integration limit c.ThenG(t,X) = e^(-r t) int(X,c) W(u) dW(u) = -e^(-r t)[ (1/2) W(X)^2 - (1/2) X + "constant"]Do you agree that this is the result of the integration? [The "constant" is actually a random variable with zero expectation, but more importantly, independent of X]Now take S(t) = W(t) and q(S) = 1/S, so G(t,X) = e^(-r t) int(X,c) dW(u) = -e^(-r t) [ W(X) + "constant"].Agree with that one?So, really, your X is a temporal variable, probably better to write X = u. When we try to develop the derivative wrt u, the constants drop out. The problem is dW(u)/du doesn't really exist, although sometimes it is formally manipulated. So, I will guess your derivative doesn't exist in the case where S(t) is a diffusion process. If S(t) is not a diffusion process, but simply an ordinary function of time, so that the integral is not a stochastic integral, then the ordinary calculus rule works.
Last edited by Alan on May 11th, 2012, 10:00 pm, edited 1 time in total.
 
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croot
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simple integral

May 12th, 2012, 11:24 pm

I don't see where this comes from, but there is one integral sign and one dSt sign, so the integral is a stochastic integral innit, well those are seldom differentiable, unless say it's defined as a Stieltjes ..
 
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ralfbuesser
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simple integral

May 13th, 2012, 6:50 pm

Really its the first part of the Breeden-Litzenberger result, i.e. call=exp(-rt)*int_K^infty (S-K)q(S)dSand I was looking only at the first term with S. So essentially I wonder whether the spot rate S in the integral is simply ignored when taking the derivative of the call price w.r.t. to the strike price, or whether I have to account for this too since the strike price is the lower bound in my integral.Thanks for your help.
 
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croot
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simple integral

May 13th, 2012, 7:00 pm

ooh haha I guess I should have figured as $e^{-rt}$ wasn't inside the integral. -1 cookies to me.
 
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acastaldo
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simple integral

May 13th, 2012, 8:03 pm

OK, so it is not a stochastic integral but a real one then dG/dX = -exp(-r*t) X q(X) using D.U.I.S.
Last edited by acastaldo on May 12th, 2012, 10:00 pm, edited 1 time in total.
 
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Alan
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simple integral

May 13th, 2012, 11:15 pm

I will admit to using dS_t as a dumy integration variable before -- after this thread, never again!
 
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ralfbuesser
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simple integral

May 14th, 2012, 4:21 pm

Ok thx much. The D.U.I.S is what I was looking for. And I suppose this is not a stochastic integral because we are talking about spot at expiry?
 
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acastaldo
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simple integral

May 14th, 2012, 5:28 pm

Quotethis is not a stochastic integral because we are talking about spot at expiry? Yes. The notation dS_t gave the impression that time is advancing, the random process S_t is varying over time and the integration is carried out over these random changes. Alan solved that but the derivative did not appear to exist. But actually t is fixed and it is a simple matter of integrating over all possible stock prices at expiration.
 
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Alan
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simple integral

May 14th, 2012, 5:32 pm

In a stochastic integral the integrand and/or the differential are random variables (indeed, stochastic processesas acastaldo notes), so the result is a random variable:example: int W_t dW_t, with W_t a Brownian motion.In an ordinary integral (your case), the differential is simply a differential of a dummy variable and the integrand is a deterministic function of that (and perhaps other parameters). The result may either be a number or a determinstic function of the integration limits. example: your integral.
Last edited by Alan on May 13th, 2012, 10:00 pm, edited 1 time in total.
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