question about using ito's lemma to a function

[wuti]

This question is an example question in the book. Basically it applies Ito's lemma to a function Y.
The thing i don't understand is why when we use dy/dt, we do not consider d(z(t))/dt, since y=e^-(mu*t)*Z(t), when we want to find dy/dt, do we also need to consider Z'(t) as well? But why the book doesn't consider Z'(t)?







Below is the formula for Ito's lemma, in case you don't know.

 

[mario99]

This question is an example question in the book. Basically it applies Ito's lemma to a function Y.
The thing i don't understand is why when we use dy/dt, we do not consider d(z(t))/dt, since y=e^-(mu*t)*Z(t), when we want to find dy/dt, do we also need to consider Z'(t) as well? But why the book doesn't consider Z'(t)?

View attachment 38326





Below is the formula for Ito's lemma, in case you don't know.
View attachment 38327

[imath]\displaystyle \frac{\partial Y}{\partial t} \neq Z'(t)[/imath]
Why do you want to do that?

If you want to use [imath]\displaystyle Z'(t)[/imath], the lemma should give us [imath]\displaystyle X'(t)[/imath]. Since the lemma didn't say anything about [imath]\displaystyle X'(t)[/imath], you cannot use [imath]\displaystyle Z'(t)[/imath].

The example should define [imath]\displaystyle Y = F(Z,t) = e^{-\mu t}Z[/imath].

Following the lemma will give us these results:

[imath]\displaystyle F_X = F_Z = \frac{\partial Y}{\partial Z} = e^{-\mu t}[/imath]

[imath]\displaystyle F_{XX} = F_{ZZ} = \frac{\partial^2 Y}{\partial Z^2} = 0[/imath]

[imath]\displaystyle F_t = \frac{\partial Y}{\partial t} = -\mu e^{-\mu t} Z = -\mu e^{-\mu t} Z(t)[/imath]

[imath]\displaystyle a(X,t) = a(Z,t) = \mu Z = \mu Z(t)[/imath]

[imath]\displaystyle b(X,t) = b(Z,t) = \sigma [/imath]

 

[wuti]

[imath]\displaystyle \frac{\partial Y}{\partial t} \neq Z'(t)[/imath]
Why do you want to do that?

If you want to use [imath]\displaystyle Z'(t)[/imath], the lemma should give us [imath]\displaystyle X'(t)[/imath]. Since the lemma didn't say anything about [imath]\displaystyle X'(t)[/imath], you cannot use [imath]\displaystyle Z'(t)[/imath].

The example should define [imath]\displaystyle Y = F(Z,t) = e^{-\mu t}Z[/imath].

Following the lemma will give us these results:

[imath]\displaystyle F_X = F_Z = \frac{\partial Y}{\partial Z} = e^{-\mu t}[/imath]

[imath]\displaystyle F_{XX} = F_{ZZ} = \frac{\partial^2 Y}{\partial Z^2} = 0[/imath]

[imath]\displaystyle F_t = \frac{\partial Y}{\partial t} = -\mu e^{-\mu t} Z = -\mu e^{-\mu t} Z(t)[/imath]

[imath]\displaystyle a(X,t) = a(Z,t) = \mu Z = \mu Z(t)[/imath]

[imath]\displaystyle b(X,t) = b(Z,t) = \sigma [/imath]

Thanks so much Mario99! Really appreciate your help!!
I mentioned Z'(t) because Y(t)=e^(-mu*t)*Z(t), and if we want to get Y'(t), i.e. derivative of Y against t, do we need to consider Z(t) as well (i.e. Z'(t))? because Y(t) consists of: (1) e^(-mu*t), (2)Z(t). So isn't it both (1) and (2) has a derivative with respect to t if we want to find Y'(t)?
So what i mean is let's say f(x)=g(x)*h(x), in this case f'(x)=g'(x)*h(x)+g(x)*h'(x). Does this same rule apply to example 5.8 when calculating Y'(t) is what i meant...
excuse me I am not familiar with latex formula.

 

[mario99]

If you will take the partial derivative [imath]\displaystyle \frac{\partial Y}{\partial t}[/imath], we don't care about the parameter [imath]t[/imath] of [imath]Z(t)[/imath], we can write [imath]Z(t) = Z[/imath]. [imath]\displaystyle \left(\frac{\partial Y}{\partial t} = \frac{\partial \ [ \ e^{-\mu t}Z \ ]}{\partial t}\right).[/imath]

But if you will take the complete derivative [imath]\displaystyle \frac{dY}{dt}[/imath], you have to write [imath]\displaystyle Z(t)[/imath]. [imath]\displaystyle \left(\frac{ dY}{dt} = \frac{d \ [ \ e^{-\mu t}Z(t) \ ]}{dt}\right)[/imath] and here you have to use the product rule:

[imath]\displaystyle \frac{d \ [ \ e^{-\mu t}Z(t) \ ]}{dt} = -\mu e^{-\mu t}Z(t) + e^{-\mu t}Z'(t)[/imath]

 

[wuti]

thanks again. But i still don't understand why we don't care about paramater t of Z(t) when we take partial derivative of [imath]\frac{\partial Y}{\partial t}[/imath]? So basically Z(t) is a function with t as its variable too why we don't consider? Sorry to ask again but i am stuck.

 

[mario99]

thanks again. But i still don't understand why we don't care about paramater t of Z(t) when we take partial derivative of [imath]\frac{\partial Y}{\partial t}[/imath]? So basically Z(t) is a function with t as its variable too why we don't consider? Sorry to ask again but i am stuck.

If we have a variable like [imath]Z(t)[/imath], [imath]Z[/imath] is called the dependent variable while [imath]t[/imath] is called the independent variable. Partial derivative cares only about the dependent variables. It differentiates only with respect to the dependent variables and treat all other variables like constants. (No chain rule can be applied to Partial Derivatives.)

In other words,

[imath]\displaystyle \frac{\partial [ \ e^{-\mu t}Z(t) \ ]}{\partial t}[/imath] here every thing is treated like constants including [imath]Z(t)[/imath] except [imath]e^{-\mu t}[/imath].

[imath]\displaystyle \frac{\partial [ \ e^{-\mu t}Z(t) \ ]}{\partial Z}[/imath] here every thing is treated like constants including [imath]e^{-\mu t}[/imath] except [imath]Z(t)[/imath].

 

[wuti]

If we have a variable like [imath]Z(t)[/imath], [imath]Z[/imath] is called the dependent variable while [imath]t[/imath] is called the independent variable. Partial derivative cares only about the dependent variables. It differentiates only with respect to the dependent variables and treat all other variables like constants. (No chain rule can be applied to Partial Derivatives.)

In other words,

[imath]\displaystyle \frac{\partial [ \ e^{-\mu t}Z(t) \ ]}{\partial t}[/imath] here every thing is treated like constants including [imath]Z(t)[/imath] except [imath]e^{-\mu t}[/imath].

[imath]\displaystyle \frac{\partial [ \ e^{-\mu t}Z(t) \ ]}{\partial Z}[/imath] here every thing is treated like constants including [imath]e^{-\mu t}[/imath] except [imath]Z(t)[/imath].

amazing! got it. Thank you so much Mario99!!

 

[mario99]

amazing! got it. Thank you so much Mario99!!

Correction: (No chain rule can be applied to the independent variables of a variable which is treated like a constant in Partial Derivatives.)

For example:

[imath]\displaystyle \frac{\partial \ [ \cos(t^2) Z(t^2) \ ]}{\partial t}[/imath]

You will apply the chain rule to [imath]\displaystyle \cos(t^2)[/imath]

But

You will not apply the chain rule to [imath]\displaystyle Z(t^2)[/imath] because it is treated like a constant.

Another example for the sake of completeness:

[imath]\displaystyle \frac{d \ [ \cos(t^2) Z(t^2) \ ]}{d t}[/imath]

Here you will apply the chain rule for both of them.