# 22.2 - Change-of-Variable Technique

22.2 - Change-of-Variable Technique

On the last page, we used the distribution function technique in two different examples. In the first example, the transformation of $$X$$ involved an increasing function, while in the second example, the transformation of $$X$$ involved a decreasing function. On this page, we'll generalize what we did there first for an increasing function and then for a decreasing function. The generalizations lead to what is called the change-of-variable technique.

## Generalization for an Increasing Function

Let $$X$$ be a continuous random variable with a generic p.d.f. $$f(x)$$ defined over the support $$c_1<x<c_2$$. And, let $$Y=u(X)$$ be a continuous, increasing function of $$X$$ with inverse function $$X=v(Y)$$. Here's a picture of what the continuous, increasing function might look like:

The blue curve, of course, represents the continuous and increasing function $$Y=u(X)$$. If you put an $$x$$-value, such as $$c_1$$ and $$c_2$$, into the function $$Y=u(X)$$, you get a $$y$$-value, such as $$u(c_1)$$ and $$u(c_2)$$. But, because the function is continuous and increasing, an inverse function $$X=v(Y)$$ exists. In that case, if you put a $$y$$-value into the function $$X=v(Y)$$, you get an $$x$$-value, such as $$v(y)$$.

Okay, now that we have described the scenario, let's derive the distribution function of $$Y$$. It is:

$$F_Y(y)=P(Y\leq y)=P(u(X)\leq y)=P(X\leq v(y))=\int_{c_1}^{v(y)} f(x)dx$$

for $$d_1=u(c_1)<y<u(c_2)=d_2$$. The first equality holds from the definition of the cumulative distribution function of $$Y$$. The second equality holds because $$Y=u(X)$$. The third equality holds because, as shown in red on the following graph, for the portion of the function for which $$u(X)\le y$$, it is also true that $$X\le v(Y)$$:

And, the last equality holds from the definition of probability for a continuous random variable $$X$$. Now, we just have to take the derivative of $$F_Y(y)$$, the cumulative distribution function of $$Y$$, to get $$f_Y(y)$$, the probability density function of $$Y$$. The Fundamental Theorem of Calculus, in conjunction with the Chain Rule, tells us that the derivative is:

$$f_Y(y)=F'_Y(y)=f_x (v(y))\cdot v'(y)$$

for $$d_1=u(c_1)<y<u(c_2)=d_2$$.

## Generalization for a Decreasing Function

Let $$X$$ be a continuous random variable with a generic p.d.f. $$f(x)$$ defined over the support $$c_1<x<c_2$$. And, let $$Y=u(X)$$ be a continuous, decreasing function of $$X$$ with inverse function $$X=v(Y)$$. Here's a picture of what the continuous, decreasing function might look like:

The blue curve, of course, represents the continuous and decreasing function $$Y=u(X)$$. Again, if you put an $$x$$-value, such as $$c_1$$ and $$c_2$$, into the function $$Y=u(X)$$, you get a $$y$$-value, such as $$u(c_1)$$ and $$u(c_2)$$. But, because the function is continuous and decreasing, an inverse function $$X=v(Y)$$ exists. In that case, if you put a $$y$$-value into the function $$X=v(Y)$$, you get an x-value, such as $$v(y)$$.

That said, the distribution function of $$Y$$ is then:

$$F_Y(y)=P(Y\leq y)=P(u(X)\leq y)=P(X\geq v(y))=1-P(X\leq v(y))=1-\int_{c_1}^{v(y)} f(x)dx$$

for $$d_2=u(c_2)<y<u(c_1)=d_1$$. The first equality holds from the definition of the cumulative distribution function of $$Y$$. The second equality holds because $$Y=u(X)$$. The third equality holds because, as shown in red on the following graph, for the portion of the function for which $$u(X)\le y$$, it is also true that $$X\ge v(Y)$$:

The fourth equality holds from the rule of complementary events. And, the last equality holds from the definition of probability for a continuous random variable $$X$$. Now, we just have to take the derivative of $$F_Y(y)$$, the cumulative distribution function of $$Y$$, to get $$f_Y(y)$$, the probability density function of $$Y$$. Again, the Fundamental Theorem of Calculus, in conjunction with the Chain Rule, tells us that the derivative is:

$$f_Y(y)=F'_Y(y)=-f_x (v(y))\cdot v'(y)$$

for $$d_1=u(c_2)<y<u(c_1)=d_1$$. You might be alarmed in that it seems that the p.d.f. $$f(y)$$ is negative, but note that the derivative of $$v(y)$$ is negative, because $$X=v(Y)$$ is a decreasing function in $$Y$$. Therefore, the two negatives cancel each other out, and therefore make $$f(y)$$ positive.

Phew! We have now derived what is called the change-of-variable technique first for an increasing function and then for a decreasing function. But, continuous, increasing functions and continuous, decreasing functions, by their one-to-one nature, are both invertible functions. Let's, once and for all, then write the change-of-variable technique for any generic invertible function.

Definition. Let $$X$$ be a continuous random variable with generic probability density function $$f(x)$$ defined over the support $$c_1<x<c_2$$. And, let $$Y=u(X)$$ be an invertible function of $$X$$ with inverse function $$X=v(Y)$$. Then, using the change-of-variable technique, the probability density function of $$Y$$ is:

$$f_Y(y)=f_X(v(y))\times |v'(y)|$$

defined over the support $$u(c_1)<y<u(c_2)$$.

Having summarized the change-of-variable technique, once and for all, let's revisit an example.

## Example 22-1 Continued

Let's return to our example in which $$X$$ is a continuous random variable with the following probability density function:

$$f(x)=3x^2$$

for $$0<x<1$$. Use the change-of-variable technique to find the probability density function of $$Y=X^2$$.

#### Solution

Note that the function:

$$Y=X^2$$

defined over the interval $$0<x<1$$ is an invertible function. The inverse function is:

$$x=v(y)=\sqrt{y}=y^{1/2}$$

for $$0<y<1$$. (That range is because, when $$x=1, y=0$$; and when $$x=1, y=1$$). Now, taking the derivative of $$v(y)$$, we get:

$$v'(y)=\dfrac{1}{2} y^{-1/2}$$

Therefore, the change-of-variable technique:

$$f_Y(y)=f_X(v(y))\times |v'(y)|$$

tells us that the probability density function of $$Y$$ is:

$$f_Y(y)=3[y^{1/2}]^2\cdot \dfrac{1}{2} y^{-1/2}$$

And, simplifying we get that the probability density function of $$Y$$ is:

$$f_Y(y)=\dfrac{3}{2} y^{1/2}$$

for $$0<y<1$$. We shouldn't be surprised by this result, as it is the same result that we obtained using the distribution function technique.

## Example 22-2 continued

Let's return to our example in which $$X$$ is a continuous random variable with the following probability density function:

$$f(x)=3(1-x)^2$$

for $$0<x<1$$. Use the change-of-variable technique to find the probability density function of $$Y=(1-X)^3$$.

#### Solution

Note that the function:

$$Y=(1-X)^3$$

defined over the interval $$0<x<1$$ is an invertible function. The inverse function is:

$$x=v(y)=1-y^{1/3}$$

for $$0<y<1$$. (That range is because, when $$x=1, y=1$$; and when $$x=1, y=0$$). Now, taking the derivative of $$v(y)$$, we get:

$$v'(y)=-\dfrac{1}{3} y^{-2/3}$$

Therefore, the change-of-variable technique:

$$f_Y(y)=f_X(v(y))\times |v'(y)|$$

tells us that the probability density function of $$Y$$ is:

$$f_Y(y)=3[1-(1-y^{1/3})]^2\cdot |-\dfrac{1}{3} y^{-2/3}|=3y^{2/3}\cdot \dfrac{1}{3} y^{-2/3}$$

And, simplifying we get that the probability density function of Y is:

$$f_Y(y)=1$$

for $$0<y<1$$. Again, we shouldn't be surprised by this result, as it is the same result that we obtained using the distribution function technique.

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