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As I've come across two very interesting posts (1 about generating an expenditure function for the bliss point utility hicksian demand and one about the topic in general) concerning Hicksian demands under a demand system with "bliss point" utility functions, I've tried to show that:

enter image description here

But I never succeed and locked after simplifying the CPOs! Could someone help me on this one please?

Thank you a lot!

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    $\begingroup$ Welcome! Please consider taking the time to read the faq to familiarize yourself with some of our common practices. In addition, please use MathJax to typeset mathematics instead of pasting images. Lastly, it would be helpful if you could include a short summary of (and links to) the two posts mentioned so that there is some background of your questions. $\endgroup$
    – Herr K.
    Commented Oct 25 at 17:35
  • $\begingroup$ I think there is mistake in the above expression. Please see this for the correct expression: economics.stackexchange.com/a/59381/11824 $\endgroup$
    – Amit
    Commented Oct 26 at 10:07

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As Amit has commented, there is a mistake in the above expression. This however is not the answer I got as I have reworked through the math. So the answer which I think is correct along with my work is the following:

we are looking to solve $$\arg \min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to: $$\bar{U}\geq-(x_1-\delta_1)^2-(x_2-\delta _2)^2$$ where $\bar{U}\leq0$.

Step 1: Set up Lagrangian $$\mathcal{L}:p_1x_1+p_2x_2+\lambda[\bar{U}+(x_1-\delta_1)^2+(x_2-\delta _2)]$$

Step 2: Take FOCs and derive optimality condition $$(1)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_1}:p_1+2\lambda(x_1-\delta_1)=0$$ $$(2)\ \ \ \ \ \ \ \frac{\partial\mathcal{L}}{\partial x_2}: p_2+2\lambda(x_2-\delta_2)=0$$

Rearranging the above we get:

$$(1a)\ \ \ \ \ \ \ p_1=-2\lambda(x_1-\delta_1)$$ $$(2a)\ \ \ \ \ \ \ p_2=-2\lambda(x_2-\delta_2)$$

Dividing (2a) by (1a) we get our optimality condition (where MRS equals our price ratio).

$$\frac{p_2}{p_1}=\frac{x_2-\delta_2}{x_1-\delta_1}$$

Step 3: Derive "intermediate" demand equation

Rearranging our optimality condition we get:

$$\hat{x_2}=\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)$$

Step 4: substitute intermediate demand equation into constraint to obtain hicksian demand

the following algebra below helps give our answer (apologies for the messiness).

$$\bar{U}=-(x_1-\delta_1)^2-(\hat{x_2}-\delta _2)^2$$

$$\bar{U}=-(x_1-\delta_1)^2-\left(\delta_2+\frac{p_2}{p_1}(x_1-\delta_1)-\delta _2\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$ $$\bar{U}=-(x_1-\delta_1)^2-\left(\frac{p_2}{p_1}(x_1-\delta_1)\right)^2$$

$$\bar{U}=-(x_1-\delta_1)^2\left(1+\left(\frac{p_2}{p_1}\right)^2\right)$$

Heres the tricky part, recall that $\bar{U}\leq 0$ this allows us to do the following:

$$(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$$ $$x_1-\delta_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$

This is allowed because the term in the brackets is a positive number.

Moving over $\delta_1$ we have our hicksian demand for $x_1$:

$$x^h_1=\delta_1+\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$$ by symmetry the hicksian demand for $x_2$ is:

$$x^h_2=\delta_2+\left(\frac{-\bar{U}}{1+\left(\frac{p_1}{p_2}\right)^2}\right)^{\frac{1}{2}}$$

I hope this helps.

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  • $\begingroup$ This cannot be correct because in your expressions $x_1^h>\delta_1$ and $x_2^h>\delta_2$, but $(\delta_1,\delta_2)$ is a Bliss point so $u(\delta_1,\delta_2)>u(x_1^h,x_2^h)\geq \overline{U}$, and $(\delta_1,\delta_2)$ also costs less than $(x_1^h, x_2^h)$ because $x_1^h>\delta_1$ and $x_2^h>\delta_2$, contradicting that $(x_1^h, x_2^h)$ is expenditure minimising for $(p_1,p_2, \overline{U})$. $\endgroup$
    – Amit
    Commented Oct 27 at 4:58
  • $\begingroup$ Mistake is in the expression that you have written after this: $(x_1-\delta_1)^2=\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}$. Taking square root both sides yields LHS $x_1-\delta_1$ (which is negative) and RHS $\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$ (which is positive). If you take positive square roots both sides, you'll get $\delta_1-x_1=\left(\frac{-\bar{U}}{1+\left(\frac{p_2}{p_1}\right)^2}\right)^{\frac{1}{2}}$. $\endgroup$
    – Amit
    Commented Oct 27 at 5:06
  • $\begingroup$ @Amit theres a bit to unpack in both of your comments. First note that $x_1^h$ and $x_2^h$ need not be bigger than $\delta_1$ and $\delta_2$ respectively. the square root term can be a positive number or negative number. Your second comment I dont understand at all. For these reasons Im keeping my answer as is for now. $\endgroup$
    – EconJohn ♦
    Commented Oct 27 at 19:59
  • $\begingroup$ Thank you for both your answers! This has been very helpful! $\endgroup$
    – Madriel Azère
    Commented Nov 25 at 10:21

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