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.

It is the partial derivative wth respect to t multiplied by dt, which is the limit of the expression in the previous line.

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]

It looks like a typo on the author's part, since he is looking at the t derivative, so it should have been a partial under the integral sign.