Ito's Formula Question: Understanding g and h Functions in Ito's Formula

[Gregg]

I have a question about the functions g and h in the Ito formula (below). The question is about finding

$F(Y(t))-F(Y(0)) = \sin(Y(t)) - \sin(Y(0))$

given that

$Y(t) - Y(0) = \int_0^t dW(t)$

Ito's formula:

$F(b,Y(b)) - F(b,Y(b)) = \int_a^b \frac{\partial F}{\partial t}(t,Y(t)) dt + \int_a^b (g(t) \frac{\partial F}{\partial x}(t,Y(t)) + \frac{1}{2}h^2(t) \frac{\partial^2 F }{\partial x^2} (t,Y(t)) dt + \int_a^b h(t) \frac{\partial f}{\partial x} (t,Y(t)) dW(t)$

For example, we have

$F(t,x) = \sin(x)$ so

$\frac{\partial F}{\partial t} = \frac{\partial Y}{\partial t} (t) \cos (Y(t))$ etc.

Plugging all the values in

$\sin(Y(t)) - \sin(Y(0))=\int_0^t \frac{\partial Y}{\partial t}(t) \cos(Y(t)) dt + \int_0^t g(t) \cos(Y(t)) - \frac{1}{2}h^2(t) \sin(Y(t)) dt + \int_0^t h(t) dW(t)$

does the $Y(t) - Y(0) = \int_0^t dW(t)$ set any extra condition on the $h(t)$ and $g(t)$ in the above expression? Or is it fine left as is, what are the two functions h,g? Are they mean and variance? This would mean that the mean of $Y_t$ is 0 and the variance is $1$? Does this mean that the RV a corresponding g and h? how do I work this out?

 

[Ray Vickson]

I have a question about the functions g and h in the Ito formula (below). The question is about finding

$F(Y(t))-F(Y(0)) = \sin(Y(t)) - \sin(Y(0))$

given that

$Y(t) - Y(0) = \int_0^t dW(t)$

Ito's formula:

$F(b,Y(b)) - F(b,Y(b)) = \int_a^b \frac{\partial F}{\partial t}(t,Y(t)) dt + \int_a^b (g(t) \frac{\partial F}{\partial x}(t,Y(t)) + \frac{1}{2}h^2(t) \frac{\partial^2 F }{\partial x^2} (t,Y(t)) dt + \int_a^b h(t) \frac{\partial f}{\partial x} (t,Y(t)) dW(t)$

For example, we have

$F(t,x) = \sin(x)$ so

$\frac{\partial F}{\partial t} = \frac{\partial Y}{\partial t} (t) \cos (Y(t))$ etc.

Plugging all the values in

$\sin(Y(t)) - \sin(Y(0))=\int_0^t \frac{\partial Y}{\partial t}(t) \cos(Y(t)) dt + \int_0^t g(t) \cos(Y(t)) - \frac{1}{2}h^2(t) \sin(Y(t)) dt + \int_0^t h(t) dW(t)$

does the $Y(t) - Y(0) = \int_0^t dW(t)$ set any extra condition on the $h(t)$ and $g(t)$ in the above expression? Or is it fine left as is, what are the two functions h,g? Are they mean and variance? This would mean that the mean of $Y_t$ is 0 and the variance is $1$? Does this mean that the RV a corresponding g and h? how do I work this out?

What, exactly, do you mean by "finding" F(Y(t))? Do you want the expected value? The probability distribution? Something else?

 

[Gregg]

$F(Y(t))-F(Y(0)) = \sin(Y(t)) - \sin(Y(0))$ this is the relationship between the RHS and the Ito formula which is given in terms of $F(b,Y(b) - F(a, T(a))$ it should read


$F(t, Y(t))-F(0, Y(0)) = \sin(Y(t)) - \sin(Y(0))$

I am to use the Ito formula to get an expression for the RHS in terms of Riemann and Ito integrals. My question is, once we have that form, what is the significance of the functions g,h, can their values be computed from the given information?