# 22.3 - Two-to-One Functions

22.3 - 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 22-3

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 $$X^2\le y$$, 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 $$X^2\le y$$, 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 $$X$$ be a continuous random variable with probability density function $$f(x)$$ for $$c_1<x<c_2$$.

Let $$Y=u(X)$$ be a continuous two-to-one function of $$X$$, which can be “broken up” into two one-to-one invertible functions with:

$$X_1=v_1(Y)$$ and $$X_2=v_2(Y)$$

1. Then, the probability density function for the two-to-one portion of $$Y$$ is:

$$f_Y(y)=f_X(v_1(y))\cdot |v'_1(y)|+f_X(v_2(y))\cdot |v'_2(y)|$$

for the “appropriate support” for $$y$$. That is, you have to add the one-to-one portions together.

2. And, the probability density function for the one-to-one portion of $$Y$$ is, as always:

$$f_Y(y)=f_X(v_2(y))\cdot |v'_2(y)|$$

for the “appropriate support” for $$y$$.

## Example 22-4

Suppose $$X$$ is a continuous random variable with that follows the standard normal distribution with, of course, $$-\infty<x<\infty$$. 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 $$-\infty<x<\infty$$:

That is, when $$-\infty<x<0$$, we have:

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

and when $$0<x<\infty$$, 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 $$0<y<\infty$$:

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

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 $$0<y<\infty$$, 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!

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