Jump to content

HJM model


Gost91

Recommended Posts

Hi there!

I state that I'm an electronic engineer (undergraduate), then the my knowledges about the world of economics are almost null.
A colleague asked to me an help about one point of the proof of the theorem 1 in this paper.

The critical point is how to obtain [latex]\text{ d}Y_t [/latex] that at first glance it seemed to me enough trivial.

Unfortunately I get a wrong result, so I ask if is possible, without using advanced tools like stochastic calculus, get the solution, and in that case, how to get it.

I post my (absolutely not rigorous) calculations.

 

[latex]

\begin{aligned}

\text{d}Y_t &= \lim_{\tau \to 0} \,\, [ Y_{t+\tau}-Y_{t} ] \\ &= \lim_{\tau \to 0} \left[- \int_{t+\tau}^s f(t+\tau,u) \text{ d}u -\left( - \int_t^s f(t,u) \text{ d}u\right) \right] \\ &=\lim_{\tau \to 0} \left[- \left( \int_t^s f(t+\tau,u) \text{ d}u - \int_t^{t+\tau} f(t+\tau,u) \text{ d}u \right) -\left( - \int_t^s f(t,u) \text{ d}u\right) \right] \\

&=\lim_{\tau \to 0} \left[ \int_t^{t+\tau} f(t+\tau,u) \text{ d}u -\left( \int_t^s f(t+\tau,u)-f(t,u) \text{ d}u\right) \right] \\

&=\lim_{\tau \to 0} \int_t^{t+\tau} f(t+\tau,u) \text{ d}u - \int_t^s \lim_{\tau \to 0} \, [f(t+\tau,u)-f(t,u)] \text{ d}u \\ &=f(t,s)\text{ dt}- \int_t^s \frac{\partial f}{\partial t} (t,u) \text{ d}t \text{ du} \end{aligned}

[/latex]

 

The question is: why it appear the total differential of [latex] f(t,u) [/latex] under the sign of integral?

Thank you in advance and sorry for my English.

Link to comment
Share on other sites

Thanks for the reply mathematic, but the problem insist.

For my calculations, the result is

[latex] \text{d}Y_t=f(t,t) \text{ d}t- \int_t^s \frac{\partial f}{\partial t} (t,u) \text{ d}t \text{ d}u \tag{1} [/latex]

 

according to the Leibniz Integral rule. In fact, for that formula

 

[latex]

\begin{aligned}

\frac{ \text{d}Y_t}{\text{d}t} &= -\left ( f(t,s)\frac{\text{d} s}{\text{d} t}-f(t,t)\frac{\text{d} t}{\text{d} t}+\int_t^s \frac{\partial f}{\partial t} (t,u) \text{ d}u \right) \\

&=f(t,t)- \int_t^s \frac{\partial f}{\partial t} (t,u) \text{ d}u

\end{aligned} \tag{2}

[/latex]

 

where I have observed that [latex] s[/latex] is costant with respect to [latex] t[/latex].

By multiplying [latex] \text{d}t[/latex] each terms of [latex] (2) [/latex] we obtain [latex] (1) [/latex].

 

But in the paper the author have writed

[latex] \text{d}Y_t=f(t,t) \text{ d}t- \int_t^s \text{ d}f (t,u)\text{ d}u [/latex]

 

The previus expression is different respect [latex] (1) [/latex] because

[latex] \text{ d}f (t,u) =\frac{\partial f}{\partial t} (t,u) \text{ d}t +\frac{\partial f}{\partial u} (t,u) \text{ d}u [/latex]

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.