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The inter temporal elasticity of substitution between two periods of time is $$\sigma=-\frac{d[\ln(c(s)/c(t))]}{d[\ln(U'(c(s))/U'(c(t)))]}$$ I need to prove that $$\lim_{s \to t}\sigma = -\frac{U'(c(t))}{U''(c(t))c(t)}$$ So, I have the following by the definition of total differential:

$$\sigma=-\frac{ \frac{dc(s)}{c(s)}-\frac{dc(t)}{c(t)}}{ \frac{U''(c(s))}{U'(c(s)}dc(s)- \frac{U''(c(t))}{U'(c(t)}dc(t) }$$ Then if $s$ tends to $t$ we have an indetermination $0/0$ so, we use L'Hôpital Rule. That is my problem, I don't know how to derivate with respect to $c(s)$.

wythagoras
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Marcos Fabian C
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1 Answers1

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To close this one: As I wrote in my initial comment, one should consider the derivative with respect to $s$, not $c(s)$. This first means that anything containing $t$ has a zero derivative. So we have

for the numerator

$$\frac {d\ln(c(s)/c(t))}{ds} = \frac {c'(s)}{c(s)}$$

and for the denominator

$$\frac{d[\ln(U'(c(s))/U'(c(t)))]}{ds} = \frac {U''(c(s))\cdot c'(s)}{U'(c(s))}$$

Forming the quotient of the two we get after simplifying and adding the minus sign the desired

$$\lim_{s \to t}\sigma = -\frac{U'(c(t))}{U''(c(t))c(t)}$$

Alecos Papadopoulos
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