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Let f be a deterministic function in $L^2[a,b]$ and $X_t=X_a+\int^t_af(s)dB(s)$.

Show that $\int^b_af(t)X_tdB(t)=\frac{1}{2}(X_b^2-X_a^2-\int^b_af(t)^2dt)$.

My attempt:

$\frac{1}{2}(X_b^2-X_a^2)=X_a\int^b_af(t)dB(t)+\frac{1}{2}(\int^b_af(t)dB(t))^2$.

On the other hand:

$\int^b_af(t)X_tdB(t)=\int^b_af(t)X_adB(t)+\int^b_af(t)(\int^t_af(s)dB(s))dB(t)$.

So it remains to show that:

$\int^b_af(t)(\int^t_af(s)dB(s))dB(t)=\frac{1}{2}(\int^b_af(t)dB(t))^2-\frac{1}{2}\int^b_af(t)^2dt$

I learn only about the basic definition of stochastic integral. Is there any suggestions or hints? Also, I am wondering why f should be deterministic. What will go wrong if f is adapted and being in $L^2$? Thank you.

1 Answers 1

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I am not sure whether you know Ito's formula, which is a useful tool to prove such identity.

Let $dX_t=f(t)dB_t$ and apply Ito's formula to $X_t^2$ gives you

$$dX_t^2=2X_tdX_t+(dX_t)^2=2X_tf(t)dB_t+f^2(t)dt,$$ then integrate $dX_t^2$ from $a$ gives you the desired result.


For an elementary proof, you may refer to my comment for this post. As you can see from the above argument, if $f$ is adapted and satisfies suitable integrablity assumptions, a similar identity also holds. But for an elementary proof, if $f$ is a deterministic function, we can start first with $f$ be a indicator function of $[s,t]$, for which the identity can be established with a direct calculation. But if $f$ is $L^2(\Omega\times [a,b])$, there is no such simple function for us to start with.

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    I know that it can be solved by Ito's formula, but I still would like to know whether there are other solutions that is more elementary.2017-02-14
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    @Simo Here is my **thought** about an elementary approach. We assume $a=X_a=0$. Then we can show by a direct calculation that if $f=1_{[s,t]}$ for $a$f=\sum_n f_n$ where $f_n$ are indicator functions for disjoint intervals, the identity holds. Then we approxiamte $f$ by such $f_n$ in $L^2$ sense and pass to limit. – 2017-02-14