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I am trying to find first and second order conditions for the following problem:

$$P(t) = V(t)e^{-rt}- \int^t_0 me^{-r\tau}d \tau $$

I managed to find the first order condition:

$$P'(t) = V'(t)e^{-rt} - rV(t)e^{-rt} - me^{-rt} =0 \\ V'(t) = rV(t)+m $$

According to our textbook this is correct, but I have problem finding SOC. We learned before that SOC are found by taking second order derivative and examining if it is > or < than 0.

When I try to find $P''(t)$ I get:

$$P''(t) = V''(t)e^{-rt} - rV'(t)e^{-rt}- rV'(t)e^{-rt} + r^2V(t)e^{-rt} + rme^{-rt}$$

But the textbook says that the second order condition should be:

$$D=V''(t) -rV'(t)$$

this looks like the derivative of $V'(t) = rV(t)+m \implies V'(t) - rV(t) - m = 0 $

but I don't understand why that is correct. Previously when deriving SOC we always just took the 2nd derivative of the function so how come that does not work here?

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    $\begingroup$ Your formula for $P''(t)$ looks correct. To find whether it is > or < 0 at the point where $P'(t)=0$, you can substitute $V'(t)=rV(t)+m$ and also note that $e^{-rt}>0$. $\endgroup$ – Adam Bailey Feb 15 at 19:38
  • $\begingroup$ hi, thanks. But why then the textbook says the SOC is given by: $V′′(t)−rV′(t)>0$? I guess I was probably not clear in my question, I know how to check $P''(t)$, but the credited answer is different and I can't believe this is typo since it is too different $\endgroup$ – WilliamT Feb 15 at 22:23
  • $\begingroup$ Can I ask your textbook out of curiosity? $\endgroup$ – EB3112 Feb 16 at 8:51
  • $\begingroup$ @EB3112 it is Essential Mathematics for Economic Analysis by Hammond and other authors $\endgroup$ – WilliamT Feb 16 at 9:33
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Note that, from your SOC: \begin{align} P''(t) &= V''(t)e^{-rt} - rV'(t)e^{-rt}- rV'(t)e^{-rt} + r^2V(t)e^{-rt} + rme^{-rt}\\ &=e^{-rt}\Bigl[\underbrace{\color{red}{V''(t)-rV'(t)}}_{=D}-r\underbrace{\bigl[V'(t)-rV(t)-m\bigr]}_{=0\text{ from FOC}}\Bigr] \end{align}

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  • $\begingroup$ Thanks, it makes sense I guess. But I don't understand why then we don't include $e^{-rt}$ in D, I know it has to be positive but is it proper to just say that SOC are given by D just because $e^{-rt}>0$? $\endgroup$ – WilliamT Feb 16 at 9:36
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    $\begingroup$ @WilliamT: You're right, and I think the author of the answer key was just being sloppy. $\endgroup$ – Herr K. Feb 16 at 15:28

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