# 16.2. Monotone Functions¶

If $$X$$ is discrete, and $$g$$ is a monotone function, then finding the distribution of the random variable $$Y = g(X)$$ is straightforward: convert each possible value of $$X$$ by applying $$g$$ and leave the probabilities alone.

For example, if $$X$$ is uniform on the three values $$0, 1, 2$$, and $$Y = e^X$$, then the you can just augment the distribution table of $$X$$ by the possible values of $$Y$$:

$$y$$

$$1$$

$$e$$

$$e^2$$

$$x$$

$$0$$

$$1$$

$$2$$

chance

$$\frac{1}{3}$$

$$\frac{1}{3}$$

$$\frac{1}{3}$$

The new random variable $$Y$$ is uniform on the three values $$1, e, e^2$$.

But if $$X$$ has a density then both $$X$$ and $$Y$$ have a continuum of values and we have to be more careful.

The method we developed in the previous section for finding the density of a linear function of a random variable can be extended to non-linear functions.

## 16.2.1. Change of Variable Formula for Density: Increasing Function¶

The function $$F^{-1}$$ is differentiable and increasing. We will now develop a general method for finding the density of such a function applied to any random variable that has a density.

Let $$X$$ have density $$f_X$$. Let $$g$$ be a smooth (that is, differentiable) increasing function, and let $$Y = g(X)$$. Examples of such functions $$g$$ are:

• $$g(x) = ax + b$$ for some $$a > 0$$. This case was covered in the previous section.

• $$g(x) = e^x$$

• $$g(x) = \sqrt{x}$$ on positive values of $$x$$

To develop a formula for the density of $$Y$$ in terms of $$f_X$$ and $$g$$, we will start with the cdf as we did above.

Let $$g$$ be smooth and increasing, and let $$Y = g(X)$$. We want a formula for $$f_Y$$. We will start by finding a formula for the cdf $$F_Y$$ of $$Y$$ in terms of $$g$$ and the cdf $$F_X$$ of $$X$$.

\begin{split} \begin{align*} F_Y(y) ~ & = ~ P(Y \le y) \\ &= ~ P(g(X) \le y) \\ &= ~ P(X \le g^{-1}(y)) ~~~~ \text{because } g \text{ is increasing} \\ &= ~ F_X(g^{-1}(y)) \end{align*} \end{split}

Now we can differentiate to find the density of $$Y$$. By the chain rule and the fact that the derivative of an inverse is the reciprocal of the derivative,

\begin{split} \begin{align*} f_Y(y) ~ &= ~ f_X(g^{-1}(y)) \frac{d}{dy} g^{-1}(y) \\ &= ~ f_X(x) \frac{1}{g'(x)} ~~~~~ \text{at } x = g^{-1}(y) \end{align*} \end{split}

### 16.2.1.1. The Formula¶

Let $$g$$ be a differentiable, increasing function. The density of $$Y = g(X)$$ is given by

$f_Y(y) ~ = ~ \frac{f_X(x)}{g'(x)} ~~~ \text{at } x = g^{-1}(y)$

## 16.2.2. Understanding the Formula¶

To see what is going on in the calculation, we will follow the same process as we used for linear functions in an earlier section.

• For $$Y$$ to be $$y$$, $$X$$ has to be $$g^{-1}(y)$$.

• Since $$g$$ need not be linear, the tranformation by $$g$$ won’t necessarily stretch the horizontal axis by a constant factor. Instead, the factor has different values at each $$x$$. If $$g'$$ denotes the derivative of $$g$$, then the stretch factor at $$x$$ is $$g'(x)$$, the rate of change of $$g$$ at $$x$$. To make the total area under the density equal to 1, we have to compensate by dividing by $$g'(x)$$. This is valid because $$g$$ is increasing and hence $$g'$$ is positive.

This gives us an intuitive justification for the formula.

## 16.2.3. Applying the Formula¶

Let $$X$$ have the exponential (1/2) density and let $$Y = \sqrt{X}$$. We can take the square root because $$X$$ is a positive random variable.

Let’s find the density of $$Y$$ by applying the formula we have derived above. We will organize our calculation in four preliminary steps, and then plug into the formula.

• The density of the original random variable: The density of $$X$$ is $$f_X(x) = (1/2)e^{-(1/2)x}$$ for $$x > 0$$.

• The function being applied to the original random variable: Take $$g(x) = \sqrt{x}$$. Then $$g$$ is increasing and its possible values are $$(0, \infty)$$.

• The inverse function: Let $$y = g(x) = \sqrt{x}$$. We will now write $$x$$ in terms of $$y$$, to get $$x = y^2$$.

• The derviative: The derivative of $$g$$ is given by $$g'(x) = 1/(2\sqrt{x})$$.

We are ready to plug this into our formula. Keep in mind that the possible values of $$Y$$ are $$(0, \infty)$$. For $$y > 0$$ the formula says

$f_Y(y) ~ = ~ \frac{f_X(x)} {g'(x)} ~~~ \text{at } x = g^{-1}(y)$

So for $$y > 0$$,

\begin{split} \begin{align*} f_Y(y) ~ &= ~ \frac{(1/2)e^{-\frac{1}{2}x}}{1/(2\sqrt{x})} ~~~~ \mbox{at } x = y^2 \\ &= ~ \sqrt{x} e^{-\frac{1}{2}x} ~~~~ \mbox{at } x = y^2 \\ &= ~ \sqrt{y^2} e^{-\frac{1}{2}y^2} \\ &= ~ y e^{-\frac{1}{2}y^2} \end{align*} \end{split}

This is called the Rayleigh density. Its graph is shown below. Quick Check

$$V$$ has density $$f_V(v) = 2v$$ for $$v \in (0, 1)$$ and 0 elsewhere. Let $$W = -\log(V)$$. What are the possible values of $$W$$? What is the density of $$W$$?

## 16.2.4. Change of Variable Formula for Density: Monotone Function¶

Let $$g$$ be smooth and monotone (that is, either increasing or decreasing). The density of $$Y = g(X)$$ is given by

$f_Y(y) ~ = ~ \frac{f_X(x)}{\lvert g'(x) \rvert} ~~~ \text{at } x = g^{-1}(y)$

We have proved the result for increasing $$g$$. When $$g$$ is decreasing, the proof is analogous to proof in the linear case and accounts for $$g'$$ being negative. We won’t take the time to write it out.

## 16.2.5. Reciprocal of a Uniform Variable¶

Let $$U$$ be uniform on $$(0, 1)$$ and let $$V = 1/U$$. The distribution of $$V$$ is called the inverse uniform but the word “inverse” is confusing in the context of change of variable. So we will simply call $$V$$ the reciprocal of $$U$$.

To find the density of $$V$$, start by noticing that the possible values of $$V$$ are in $$(1, \infty)$$ as the possible values of $$U$$ are in $$(0, 1)$$.

The components of the change of variable formula for densities:

• The original density: $$f_U(u) = 1$$ for $$0 < u < 1$$.

• The function: Define $$g(u) = 1/u$$.

• The inverse function: Let $$v = g(u) = 1/u$$. Then $$u = g^{-1}(v) = 1/v$$.

• The derivative: Then $$g'(u) = -u^{-2}$$.

By the formula, for $$v > 1$$ we have

$f_V(v) ~ = ~ \frac{f_U(u)}{\lvert g'(u) \rvert} ~~~ \text{at } u = g^{-1}(v)$

That is, for $$v > 1$$,

$f_V(v) ~ = ~ 1 \cdot u^2 ~~~ \text{at } u = 1/v$

So

$f_V(v) ~ = ~ \frac{1}{v^2}, ~~~ v > 1$

You should check that $$f_V$$ is indeed a density, that is, it integrates to 1. You should also check that the expectation of $$V$$ is infinite.

The density $$f_V$$ belongs to the Pareto family of densities, much used in economics. 