Is this equation a parabola or a hyperbola?

Question

In a 1972 paper by Robert Merton, the following equation is derived:

$$\sigma(\mu;A,B,C,D)=\sqrt{\frac{A \mu^2-2B\mu+C}{D}}$$

This is known as the Markowitz frontier in finance. When this is plotted, the convention is to measure $\mu$ along the y-axis and $\sigma$ along the x-axis. Regardless of which variable is assigned to which axis, the result is a parabola (or maybe a hyperbola?)-looking thing.

In his paper Merton calls it a parabola. However, I recently saw an online lecture given at Yale university wherein Robert Shiller calls it a hyperbola.

Both Merton and Shiller are Nobel laureates. Which one is correct?

EDIT: Ah! To be fair to Merton, in his paper he doesn't call the above equation a parabola. He does, however, call the square of the above equation a parabola (pp. 1854). Does the above equation become a parabola if you square it?

Answer 1

Assuming that $B^2 < AC$ and $AD>0$ such that the expression under the square root sign is always positive, this produces (one branch of) a hyperbola.

To see it's not a parabola, simply note that for large $\mu$ the graph asymptotically approaches $\sigma=\left|\mu-\frac BA\right|\sqrt{A/D}$, which is not how a parabola behaves.

On the other hand, it's clearly a second-degree curve (square both sides to get a quadratic equation in $\sigma$ and $\mu$), and the only conic section that has asymptotes is a hyperbola.

The expression under the square root sign (and thus $\sigma^2$ as a function of $\mu$) does indeed produce a parabola.

Answer 2

Assuming the radicand is positive for all $\mu$ (i.e., that $A/D > 0$ and $B^{2} - AC < 0$), it's (one branch of) a hyperbola, with asymptotes $\sigma = \pm \sqrt{A/D} \mu$.

Answer 3

A hyperbola can be defined such that for the general quadratic equation

$$ax^2+2bxy+cy^2+dx+ey+f=0$$

we have that $ac-b^2<0$. Here, we have

$$A\mu^2-D\sigma^2-2B\mu+C=0$$

Then, if $-AD<0$, then the quadratic is a hyperbola and Shiller was correct.

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NOTE:

If the dependent variable is considered to be the variance $\sigma^2$, then quadratic is a parabola and Merton was correct.

The conclusion is that the answer depends on whether one considers the dependent variable to be the standard deviation $\sigma$, in which case the expression is hyperbolic, or one considers the dependent variable to be the variance $\sigma^2$, in which case the expression is parabolic.