3
$\begingroup$

I wonder if someone can help me interpret the vertical bar notation used in the picture. From the graph, it is apparent that the consumer will consume only good $x_1$, since the indifference curve is steeper than the budget line at $x^*$. I assume that is what eq. C.5 is expressing? I'm hoping that C.6 and C.7 will become apparent if I first understand C.5.

Appreciate any help!

enter image description here

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The vertical bar notation is used to denote conditions/restrictions applied to the expression to its left. For example, $\frac{\mathrm dx_2}{\mathrm dx_1}\bigg\vert_{\text{$u$ constant}}$ reads:

The derivative of $x_2$ with respect to $x_1$, holding $u$ constant.

This apparently refers to the slope of the indifference curve, holding utility constant at $I_1$. Similarly, the RHS of [C.5] refers to the slope of the budget line. The inequality holds at $x^*$, where the slope of the indifference curve is smaller (i.e. more negative) than the slope of the budget line.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks - this makes sense. The one confusing thing is that the inequality has the same dx2/dx1 fraction on both sides. But, if I understand you correctly, it is implicit that the RHS fraction refers to the budget line because the bar notation refers to the budget line? $\endgroup$
    – Tomas R
    Jul 4 at 15:47
  • 1
    $\begingroup$ @TomasR: Yes, that's my educated guess based on the snapshot you provide. The RHS has a restriction on income $M$ which usually occurs in the budget equation (e.g. $M=p_1x_1+p_2x_2$), while the LHS has a restriction on utility level, which naturally brings to mind indifference curves. Given utility function $u(x_1,x_2)$, the slope of the an indifference curve is given by $\frac{\mathrm dx_2}{\mathrm dx_1}=-\frac{\partial u(\cdot)/\partial x_1}{\partial u(\cdot)/\partial x_2}$ according to the implicit function theorem. $\endgroup$
    – Herr K.
    Jul 4 at 16:25

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.