# Tag Info

## New answers tagged stochastic-calculus

4

You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable $W_t$ is stochastic and hence $S_t$ is as well. So, to derive $S_t$ from $dS_t$, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it. If you want to do it the other way ...

0

A martingale must have constant expectation, such that adding a deterministic finite variation process $(b-r)dt$ would break the martingale property (except for when its a constant, which it is not by multiplication with $dt$). Hence the finite variation process must be eliminated under $Q$ for LRS to be an (equivalent) martingale measure, and as shown the ...

1

I saw a quote from Brigo & Mercurio "IR models" (page 26, 2.1 No-Arbitrage in Continuous Time) . May be it will help you to find answer: Harrison and Pliska (1983) proved the following fundamental result. A financial market is (arbitrage free and) complete if and only if there exists a unique equivalent martingale measure.

2

I am rather a fan of mathematical/statistical software for doing numerical finance (R/Matlab). But returning to your question: The commercial software UNRISK is based on mathematica, a computer algebra system. Usually you can use the Unrisk functions right in mathematica and price financial derivatives there. There also exists Jave interfaces if you want ...

1

Note that $\{W_t \mid t \geq 0\}$ is a martingale. Then, for $0<p<q<r$, \begin{align*} E(W_pW_qW_r) &= E\Big( E(W_pW_qW_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q E(W_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q^2\Big)\\ &=E\Big(W_p(W_q-W_p+W_p)^2\Big)\\ &=E\Big(W_p(W_q-W_p)^2+W_p^3+2W_p^2(W_q-W_p) \Big)\\ ...

2

\begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 ...

5

The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows: $$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$ where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + ... 6 We know that$(\tilde{W}_t) := (-W_t)$is also a Wiener process so $$E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r]$$ and that implies that$E[W_pW_qW_r] = 0$. 2 I think you are on the right track here. You made a sign error in the first line, unfortunately: $$E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 - W_qW_p^2]=\\ E[(W_r-W_q)W_p^2]+E[W_pW_q^2]= E[W_pW_q^2]$$ The first term is$0$by independence (as$p<\text{min}(r,q)$and the square does not affect independence). To take care of the second term we do the ... 1 You can use that$f(t,W_t)\in C^2$is Martingale iff:$$\partial_t f+\frac{1}{2}\partial_{WW}f= 0$$ We get:$$\partial_t f=-3W_t$$$$\partial_{WW}f=6W_t$$ Finally: $$-3W_t+3W_t= 0$$ q.e.d. The proof of theorem follows by writing out$f(t,W_t)\$ via Ito formula. Proof of theorem:

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