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I want to calculate the following: $$ \int_0^t (T-x)b(x)dW(x) $$ at $t=T$ where, $W(x)$ is a Brownian Motion.

I know there is no explicit way to calculate this (as we would for a deterministic integration), but is there a way to show what possible value it might be close to? Because I would like to then further calculate $$e^{\int_0^t (T-x)b(x)dW(x)}$$ at $t=T$, and I would then check if that is 1 (.ie. $e^0=1$)

coffee-raid
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  • That exponential is one if and only if your stochastic integral is zero. That is equivalent to the variance of it being zero. Can you proceed? – Kurt G. Aug 02 '24 at 17:33

1 Answers1

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If the function $b(x)$ is deterministic, then

$$I_{t}:=\int_{0}^{t}(T-x)b(x)dW(x)$$

is a Gaussian random variable with zero mean (Itō Integral has expectation zero)

$$E[I_{t}]=0$$

and variance given by Itô isometry

$$E[I^2_{t}]=\int_{0}^{t}((T-x)b(x))^{2}dx.$$

So since it is Gaussian we get that

$$E[e^{I_t}]=exp(\frac{1}{2}E[I^2_{t}])=exp(\frac{1}{2}\int_{0}^{t}((T-x)b(x))^{2}dx).$$

So for this to be zero you request that the nonnegative integrand is actually zero almost everywhere $((T-x)b(x))^{2}=0$ a.e. x.

Thomas Kojar
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