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I have the following quasi-linear utility function given: $u_0 = f(x_1) + x_2$ (with $f'>0$,$f''<0$).

I know that the indifference curves are vertically parallel, which means that the slope is independent of the consumption of $x_2$. I suppose that there is no income effect, but how can i show this?

Cheers

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You can show this concerning the optimization problem with the objective function $U_0 = f(x_1) + x_2$ and the budget restriction $M - p_1 x_1 - p_2 x_2 = 0$. Using the Lagrangian, this leads you to $$ f'(x_1) = \frac{p_1}{p_2} \quad \text{or} \quad f'^{-1}(\frac{p_1}{p_2}) = x_1^{*} = D_1(p) $$ You can see that in this special case the optimum quantity of $x_1^{*}$ (Marshallian demand function) does not depend on the income $M$ $$ \frac{\partial D_1}{\partial M} = 0, $$ The income effect is therefore zero, and you will not consume a different amount of $x_1^{*}$ if the income $M$ varies.

Some further considerations: Based on the Marshallian $D_i(p, M) = x_i^{*}$ and Hicksian $H_i(p, u) = x_i^{*}$ demand function, you can show some interesting properties of this particular utility function using the Slutsky equation: $$ \frac{\partial D_i}{\partial p_i} = \frac{\partial H_i}{\partial p_i} - x_i^{*} \frac{\partial D_i}{\partial M} $$ This shows that the derivative of the Marshallian demand function with respect to price equals the derivative of the Hicksian demand function with respect to price minus the optimal $x_i^{*}$ times the derivative of the Marshallian demand function with respect to income. In this special case, the Marshallian demand function equals the Hicksian demand function, as $\frac{\partial D_i}{\partial M} = 0$.

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  • $\begingroup$ Ah, thanks, that makes a lot of sense! So, since the equivalent variation (EV) and compensated variation (CV) both measure the substitution effect, they have to offset each other, don't they? $\endgroup$ – Louki Nov 3 '16 at 13:23
  • $\begingroup$ As there is no income effect and only a single Hicksian demand function that equals the Marshallian demand function, this means that CV = EV. $\endgroup$ – Michael Pho Nov 3 '16 at 13:25

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