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Gost91

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About Gost91

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  1. 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]
  2. 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.
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