Gost91 Posted September 2, 2015 Share Posted September 2, 2015 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 More sharing options...
mathematic Posted September 3, 2015 Share Posted September 3, 2015 It is the partial derivative wth respect to t multiplied by dt, which is the limit of the expression in the previous line. Link to comment Share on other sites More sharing options...
Gost91 Posted September 3, 2015 Author Share Posted September 3, 2015 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 More sharing options...
mathematic Posted September 3, 2015 Share Posted September 3, 2015 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. Link to comment Share on other sites More sharing options...
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