3
$\begingroup$

We know that the elasticity of substitution is defined as $$ e=\frac{d \ln(x_2/x_1)}{d \ln(MRS_{12})}=\frac{\frac{d(x_2/x_1)}{x_2/x_1}}{\frac{d(MU_1/MU_2)}{MU_1/MU_2}} $$

When we compute ES for CES functions, MRS is function of $x_2/x_1$, so we can let $z=x_2/x_1$ and do the math easily.

But for other utility/production functions, such as quasilinear $u(x_1,x_2)=2x_1^{0.5}+x_2$, what am I supposed to do?

I believe ES will be depending on $(x_1,x_2)$. So maybe the question can be asked as "What is ES of $u(x_1,x_2)=2x_1^{0.5}+x_2$ at $(x_1,x_2)=(1,1)$?"

$\endgroup$
1
1
$\begingroup$

The formula you already have there is a general formula for elasticity of substitution, but I can see that it might be difficult to apply to your problem here given that $MU_{x_2}=1$. There is also another way how formula for elasticity of substitution can be expressed. You can use 'partial derivative formula' (e.g. see Sydsaeter et al. EMEA pp 430) which is the inelegant 'monstrosity' given below. For elasticity of substitution between $x_2$ and $x_1$ it will be given by:

$$\sigma_{x_2x_1} = \frac{−F_1^′F_2^′(x_1 F_1^′ + x_2 F_2^′)}{x_1x_2[(F_2^′)^2F_{11}^{''} − 2F_1^′F_2^′F_{12}^{''}+ (F_1^′)^2F_{22}^{''}]}, \text{ for } F (x_1, x_2) = c$$

So in your case:

$$\sigma_{x_2x_1} = \frac{−x_1 (x_1^2 + x_2)}{x_1x_2[1^2\cdot 1 ]}= \frac{-x_1^2-x_2}{x_2}$$

More specifically for $(x_1, x_2) = (1,1)$: $\sigma_{x_2x_1} = -2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.