证明无差异曲线方程图像为圆并确定圆心与半径

Let's work through this problem step by step. We start with the given utility function, where ( u ) is a fixed value (since we're analyzing a single indifference curve):
[ u = \alpha \mu - \beta \sigma^2 - \beta \mu^2 ]

First, rearrange the equation to group variable terms on one side and constants on the other:
[ \beta \sigma^2 + \beta \mu^2 - \alpha \mu = -u ]

Factor out ( \beta ) from the ( \mu )-related terms to simplify completing the square later:
[ \beta \sigma^2 + \beta\left(\mu^2 - \frac{\alpha}{\beta}\mu\right) = -u ]

Divide every term by ( \beta ) (since ( \beta > 0 ), this doesn't alter the equation's validity):
[ \sigma^2 + \mu^2 - \frac{\alpha}{\beta}\mu = -\frac{u}{\beta} ]

Now we'll complete the square for the ( \mu )-terms:

1. Take the coefficient of ( \mu ): ( -\frac{\alpha}{\beta} )
2. Divide it by 2: ( -\frac{\alpha}{2\beta} )
3. Square the result: ( \frac{\alpha^2}{4\beta2} )

Add this squared value to both sides to keep the equation balanced:
[ \sigma^2 + \left(\mu^2 - \frac{\alpha}{\beta}\mu + \frac{\alpha^2}{4\beta2}\right) = -\frac{u}{\beta} + \frac{\alpha^2}{4\beta2} ]

The grouped ( \mu )-terms now form a perfect square, and we can rewrite the right-hand side with a common denominator:
[ \sigma^2 + \left(\mu - \frac{\alpha}{2\beta}\right)^2 = \frac{\alpha^2 - 4\beta u}{4\beta^2} ]

Key Interpretations: #

This matches the standard equation of a circle: ( (x - h)^2 + (y - k)^2 = r^2 ). Here's how it maps to our variables:

• Our horizontal axis is ( \sigma ) (equivalent to ( x ) in the standard form), so the x-coordinate of the center ( h = 0 )
• Our vertical axis is ( \mu ) (equivalent to ( y )), so the y-coordinate of the center ( k = \frac{\alpha}{2\beta} )
• The radius squared is ( \frac{\alpha^2 - 4\beta u}{4\beta^2} ), so the positive radius is:
  [ r = \frac{\sqrt{\alpha^2 - 4\beta u}}{2\beta} ]

For this to represent a real circle, the expression under the square root must be positive: ( \alpha^2 - 4\beta u > 0 ), or ( u < \frac{\alpha^2}{4\beta} ). This makes sense for a valid indifference curve, as ( u ) can't be so large that the curve becomes imaginary.

Final Results: #

• The indifference curve is a circle centered at ( (\sigma, \mu) = \left(0, \frac{\alpha}{2\beta}\right) )
• Its radius is ( \frac{\sqrt{\alpha^2 - 4\beta u}}{2\beta} )

内容的提问来源于stack exchange，提问作者TAJ