# Two-to-One Functions

You might have noticed that all of the examples we have looked at so far involved monotonic functions that, because of their one-to-one nature, could therefore be inverted. The question naturally arises then as to how we modify the change-of-variable technique in the situation in which the transformation is not monotonic, and therefore not one-to-one. That's what we'll explore on this page! We'll start with an example in which the transformation is two-to-one. We'll use the distribution function technique to find the p.d.f of the transformed random variable. In so doing, we'll take note of how the change-of-variable technique must be modified to handle the two-to-one portion of the transformation. After summarizing the necessary modification to the change-of-variable technique, we'll take a look at another example using the change-of-variable technique.

**Example**

Suppose *X* is a continuous random variable with probability density function:

\(f(x)=\dfrac{x^2}{3}\)

for −1 < *x* < 2. What is the p.d.f. of \(Y=X^2\)?

**Solution.** First, note that the transformation:

\(Y=X^2\)

is not one-to-one over the interval −1 < *x* < 2:

For example, in the interval −1 < *x* < 1, if we take the inverse of \(Y=X^2\), we get:

\(X_1=-\sqrt{Y}=v_1(Y)\)

for −1 < *x* < 0, and:

\(X_2=+\sqrt{Y}=v_2(Y)\)

for 0 < *x* < 1.

As the graph suggests, the transformation is two-to-one between when 0 < *y* < 1, and one-to-one when 1 < *y* < 4. So, let's use the distribution function technique, separately, over each of these ranges. First, consider when **0 < ****y**** < 1**. In that case:

\(F_Y(y)=P(Y\leq y)=P(X^2 \leq y)=P(-\sqrt{y}\leq X \leq \sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})\)

The first equality holds by the definition of the cumulative distribution function. The second equality holds because the transformation of interest is *Y* = *X*^{2}. The third equality holds, because when *X*^{2} ≤ *y*, the random variable *X* is between the positive and negative square roots of *y*. And, the last equality holds again by the definition of the cumulative distribution function. Now, taking the derivative of the cumulative distribution function *F*(*y*), we get (from the Fundamental Theorem of Calculus and the Chain Rule) the probability density function *f*(*y*):

\(f_Y(y)=F'_Y(y)=f_X(\sqrt{y})\cdot \dfrac{1}{2} y^{-1/2} + f_X(-\sqrt{y})\cdot \dfrac{1}{2} y^{-1/2}\)

Using what we know about the probability density function of *X*:

\(f(x)=\dfrac{x^2}{3}\)

we get:

\(f_Y(y)=\dfrac{(\sqrt{y})^2}{3} \cdot \dfrac{1}{2} y^{-1/2}+\dfrac{(-\sqrt{y})^2}{3} \cdot \dfrac{1}{2} y^{-1/2}\)

And, simplifying, we get:

\(f_Y(y)=\dfrac{1}{6}y^{1/2}+\dfrac{1}{6}y^{1/2}=\dfrac{\sqrt{y}}{3}\)

for 0 < *y* < 1. Note that it readily becomes apparent that in the case of a two-to-one transformation, we need to sum two terms, each of which arises from a one-to-one transformation.

So, we've found the p.d.f. of *Y* when 0 < *y* < 1. Now, we have to find the p.d.f. of *Y* when **1 < y < 4**. In that case:

\(F_Y(y)=P(Y\leq y)=P(X^2 \leq y)=P(X\leq \sqrt{y})=F_X(\sqrt{y})\)

The first equality holds by the definition of the cumulative distribution function. The second equality holds because *Y* = *X*^{2}. The third equality holds, because when *X*^{2} ≤ *y*, the random variable \(X \le \sqrt{y}\). And, the last equality holds again by the definition of the cumulative distribution function. Now, taking the derivative of the cumulative distribution function *F*(*y*), we get (from the Fundamental Theorem of Calculus and the Chain Rule) the probability density function *f*(*y*):

\(f_Y(y)=F'_Y(y)=f_X(\sqrt{y})\cdot \dfrac{1}{2} y^{-1/2}\)

Again, using what we know about the probability density function of *X*, and simplifying, we get:

\(f_Y(y)=\dfrac{(\sqrt{y})^2}{3} \cdot \dfrac{1}{2} y^{-1/2}=\dfrac{\sqrt{y}}{6}\)

for 1 < *y* < 4.

Now that we've seen how the distribution function technique works when we have a two-to-one function, we should now be able to summarize the necessary modifications to the change-of-variable technique.

## Generalization.
Let \(X_1=v_1(Y)\) and \(X_2=v_2(Y)\) (1) Then, the probability density function for the two-to-one portion of
for the “appropriate support” for (2) And, the probability density function for the one-to-one portion of
for the “appropriate support” for |

*Y*=

*u*(

*X*) be a continuous two-to-one function of

*X*, which can be “broken up” into two one-to-one invertible functions with:

### Example

Suppose *X* is a continuous random variable with that follows the standard normal distribution with, of course, −∞ < *x* < ∞. Use the change-of-variable technique to show that the p.d.f. of \(Y=X^2\) is the chi-square distribution with 1 degree of freedom.

**Solution.** The transformation \(Y=X^2\) is two-to-one over the entire support −∞ < *x* < ∞:

That is, when −∞ < *x* < 0, we have:

\(X_1=-\sqrt{Y}=v_1(Y)\)

and when 0 < *x* < ∞, we have:

\(X_2=+\sqrt{Y}=v_2(Y)\)

Then, the change of variable technique tells us that, over the two-to-one portion of the transformation, that is, when 0 < *y* < ∞:

\(f_Y(y)=f_X(\sqrt{y})\cdot \left |\dfrac{1}{2} y^{-1/2}\right|+f_X(-\sqrt{y})\cdot \left|-\dfrac{1}{2} y^{-1/2}\right|\)

Recalling the p.d.f. of the standard normal distribution:

\(f_X(x)=\dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{x^2}{2}\right]\)

the p.d.f. of *Y* is then:

\(f_Y(y)=\dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{(\sqrt{y})^2}{2}\right]\cdot \left|\dfrac{1}{2} y^{-1/2}\right|+\dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{(\sqrt{y})^2}{2}\right]\cdot \left|-\dfrac{1}{2} y^{-1/2}\right|\)

Adding the terms together, and simplifying a bit, we get:

\(f_Y(y)=2 \dfrac{1}{\sqrt{2\pi}} \text{exp}\left[-\dfrac{y}{2}\right]\cdot \dfrac{1}{2} y^{-1/2}\)

Crossing out the 2s, recalling that \(\Gamma(1/2)=\sqrt{\pi}\), and rewriting things just a bit, we should be able to recognize that, with 0 < *y* < ∞, the probability density function of *Y*:

\(f_Y(y)=\dfrac{1}{\Gamma(1/2) 2^{1/2}} e^{-y/2} y^{-1/2}\)

is indeed the p.d.f. of a chi-square random variable with 1 degree of freedom!