# 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*) ≤ *y*, it is also true that *X* ≤ *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*) ≤ *y*, it is also true that *X* ≥ *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*_{2} = *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.

\(f_Y(y)=f_X(v(y))\times |v'(y)|\) defined over the support |

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

### Example

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* = 0, *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

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* = 0, *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.