How do you prove that this is a Martingale

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SUMMARY

The discussion centers on proving that the process defined by Xt=log(1+W(t)²)-∫₀ᵗ(1-W(s)²)/(1+W(s)²)²ds is a martingale, where W(t) represents Brownian Motion. The key approach involves applying Itô's formula, specifically the expression dX=∂X/∂W dW + (1/2)∂²X/∂W² dW². The condition for dX to be a martingale is established by demonstrating that the drift term, represented by (1/2)∂²X/∂W² dt, equals zero, leading to the conclusion that ∂²X/∂W² must also equal zero.

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So the following process involves W(t) which is Brownian Motion, and I need to prove that it is a martingale.

Xt=log(1+W(t)2)-∫0t(1-W(s)2)/(1+W(s)2)2ds

The problem I am having is the integral. My professor did a lot of integrals w.r.t. W(t), but he didn't do very many integrals where W(t) was in the integrand and we were differentiating w.r.t. t.

I feel like I am going to use Ito's formula/rule, but I'm not sure how. I'm still a bit unclear on what is "allowed" with derivatives and integrals of Brownian motion, and w.r.t. Brownian motion. Any recommendations or suggestions as to a direction would be great.
 
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Well Ito says that
dX=\frac{\partial X}{\partial W}dW+\frac{1}{2}\frac{\partial^2 X}{\partial W^2}dW^2
for your function of W. Since
dW^2=dt
that means the process for dX is a martingale if the dt term is zero. That is, dX is a martingale if it has "zero drift." So now we must show that
\frac{1}{2}\frac{\partial^2 X}{\partial W^2}dt=0
or
\frac{\partial^2 X}{\partial W^2}=0
Hope it helps
 

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