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 $Y$:

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.

16.2.1Change 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 $X$.

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,

16.2.1.1The Formula¶

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

16.2.2Understanding 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.

16.2.3Applying 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

So for $y > 0$,

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

16.2.4Change 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

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.

16.2.4.1Reciprocal 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

That is, for $v > 1$,

So

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.