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Suppose I have the utility function $$U(x,y) = \sqrt{x} + y$$ subject to budget constraint $$p_x x + p_y y = m$$ Then $$x_M =\frac{p_y^2}{4 p_x^2}$$ $$y_M = \frac{m}{p_y} - \frac{p_y}{4 p_x}$$

where $M$ denotes Marshallian.

Now suppose I increase $p_x$ to $p_x'$.

Why is the income effect zero?

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    $\begingroup$ Could you perhaps define or post a link to the income effect? $\endgroup$
    – Giskard
    Commented Jun 3, 2015 at 5:18

2 Answers 2

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From the formula for $x_M$, we see it has no dependence on income $m$. So $$\frac{\partial x_M}{\partial m} =0$$ Thus, the Slutsky equation

$$\frac{\partial x_M}{\partial p_x} = \frac{\partial x_H}{\partial p_x} +-\frac{\partial x_M}{\partial m}x_M$$

implies

$$\frac{\partial x_M}{\partial p_x} = \frac{\partial x_H}{\partial p_x} +(0)x_M $$ $$\frac{\partial x_M}{\partial p_x} = \frac{\partial x_H}{\partial p_x} $$ Hence, and since $-\frac{\partial x_M}{\partial m}x_M$ is the income effect, this implies the income effect is zero and all the change is due to the substitution effect.

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Intuitively, the marginal utility of x falls faster than the marginal utility of y (which is actually constant), so with enough money all marginal funds go into y. Similarly, with enough money, an decrease in money only reduces the quantity of y, not x. But you should know this isn't true globally. If $$m < \frac{p_y^2}{4 p_x}$$ then the demand functions are: $$x_M =\frac{m}{p_x}$$ $$y_M = 0$$

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  • $\begingroup$ Is it a general statement that when $m < \frac{p_y^2}{4 p_x}$, all income will be allocated to $x_M$? What about the case when the marginal utility of x is smaller than the marginal utility of y (which is 1 in this case)? $x>0.25$ would satisfy this. Would it mean that the x will be consumed up to 0.25 unit, before all other income are allocated to $y$? $\endgroup$
    – Aqqqq
    Commented Nov 1, 2019 at 11:11
  • $\begingroup$ Why not ask this as a new question? Something like, "Building on question 5933, what are the demand functions for quasi-linear utility of the form..." $\endgroup$
    – BKay
    Commented Nov 1, 2019 at 19:30
  • $\begingroup$ Asked here: economics.stackexchange.com/questions/32559/… $\endgroup$
    – Aqqqq
    Commented Nov 2, 2019 at 8:17

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