Integrals of Compositions - Data 89 Course Notes

In Section 3.2 we introduced the idea of a function composition.

We’ve seen a variety of distributions that involve a composition, most importantly, the normal distribution, whose PDF has the form:

\text{PDF}(x) \propto e^{-\frac{1}{2} x^2}.

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The simplest function compositions use a linear function for $h$ :

f(x) = g(h(x)) = g((x - s)/\sigma), \quad h(x) = (x - s)/\sigma

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These compositions translate the function by $s$ , and dilates it by $\sigma$ (see Section 3.2). Many distribution families are defined by first picking some family of functions $g$ , then allowing arbitrary linear $h$ . In this case the parameters $s$ and $\sigma$ of $h$ act as location and scale parameters that can be adjusted to translate and dilate the base model, $g$ .

Integration by Substitution (Change of Variables)¶

Consider an integral of a function composition:

\int_{x = a}^b g(h(x)) dx.

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In Section 7.1 we worked out rules for integrating sums and products of functions. What about nested functions?

The procedure used to integrate a function composition is often called “ $u$ ” substitution after the convention $u = h(x)$ . There is no reason to call the output of the inner function $u$ . So, we won’t call this rule “ $u$ ” substitution. Instead, we’ll name it based on the action that justifies the rule: change of variables

Integration by Change of Variables (Version I)

Suppose that $f(x) = g \circ h(x) = g(h(x))$ and that $h(x)$ is a monotonic function.

Let $u = h(x)$ and $du(x) = h'(x) dx$ . Then:

\int g(h(x)) h'(x) dx = \int g(u) du.

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Let $G$ denote the indefinite integral of $g$ . Then:

\int_{a}^b g(h(x)) h'(x) dx = G(u) \Big|_{h(a)}^{h(b)} = G(h(b)) - G(h(a)).

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This procedure is equivalent to changing variables by replacing $x$ with $u = h(x)$ .

Proof

The proof follows from the fundamental theorem of calculus and the chain rule for derivatives.

First, let $F(x) = G \circ h(x) = G(h(x))$ where $G$ is the antiderivative of $g$ , $\frac{d}{dx} G(x) = G'(x) = g(x)$ . Then, by the product rule:

F'(x) = \frac{d}{dx} F(x) = G'(h(x)) h'(x) = g(h(x)) h'(x).

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So, by the Fundamental Theorem of Calculus:

F(x) = \int F'(x) dx = \int g(h(x)) h'(x) dx.

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Then:

G(h(b)) - G(h(a)) = F(b) - F(a) = \int_{x = a}^b F'(x) dx = \int_{x = a}^b g(h(x)) h'(x) dx.

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Example

Suppose that $X \sim \text{Normal}(0,1)$ . Then, since $X$ is a standard normal random variable, $\mathbb{E}[X] = 0$ and $\text{SD}[X] = 1$ . Let’s find the mean absolute deviation in $X$ (see Section 4.3).

First, write down the expectation as a weighted average:

\begin{aligned} \text{MAD}[X] & = \mathbb{E}[|X - \mathbb{E}[X]|] = \mathbb{E}[|X - 0|] \\ & = \int_{x = -\infty}^{\infty} |x| \text{PDF}(x) dx = \frac{1}{\sqrt{2 \pi}} \int_{x = -\infty}^{\infty} |x| e^{-\frac{1}{2} x^2} dx. \end{aligned}

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Next, let’s think a bit about symmetries. The normal density $\text{PDF}(x) \propto e^{-\frac{1}{2} x^2}$ is an even function. The function $|x|$ is also even since $|-x| = |x|$ . The product of two even functions is an even function. Therefore:

\text{MAD}[X] = \frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} x e^{-\frac{1}{2} x^2} dx.

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Notice that $e^{-\frac{1}{2} x^2}$ is a composition of the functions $g(x) = e^{-x}$ and the quadratic function $h(x) = \frac{1}{2} x^2$ .

So, let’s try integrating by changing variables.

Let $u = h(x) = \frac{1}{2} x^2$ . Then $du = h'(x) dx = \frac{2}{2} x dx = x dx$ . So:

\begin{aligned} \text{MAD}[X] & = \frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} x e^{-\frac{1}{2} x^2} dx \\ & = \frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} e^{-\frac{1}{2} x^2} x dx \\ & = \frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} g(h(x)) h'(x) dx \\ & = \frac{2}{\sqrt{2 \pi}} \int_{u = h(0)}^{h(\infty)} g(u) du \\ & = \frac{2}{\sqrt{2 \pi}} \int_{u = 0}^{\infty} e^{-u} du. \end{aligned}

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Pay careful attention to the bounds of integration. In the end, we still integrate from 0 to infinity since $h(0) = \frac{1}{2} 0^2 = 0$ and $\lim_{x \rightarrow \infty} h(x) = \frac{1}{2} \lim_{x \rightarrow \infty} x^2 = \infty$ .

Integrating:

\int_{u = 0}^{\infty} e^{-u} du = -e^{-u} \Big|_{0}^{\infty} = -(0 - 1) = 1.

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Therefore:

\text{MAD}[X] = \frac{2}{\sqrt{2 \pi}} = \sqrt{\frac{2}{\pi}} \approx \sqrt{\frac{2}{3}} \approx 0.80.

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Notice that, as Jensen guarantees, $\text{MAD}[X] \approx \sqrt{2/3} < \sqrt{1} = 1 = \text{SD}[X]$ .

Often, we want to just take an integral of a composition, e.g. $\int_{x = a}^b g(h(x)) dx$ . In this case we are missing the $h'(x) dx$ term needed to replace $g(h(x)) dx$ with $g(u) du$ .

So, we will add it in ourselves by multiplying the integrand with $1 = h'(x)/h'(x)$ :

g(h(x)) dx = g(h(x)) \frac{h'(x)}{h'(x)} dx = \frac{g(h(x))}{h'(x)} h'(x) dx.

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Now, to write everything in terms of $u$ , we can substitute $h(x) = u$ and $h'(x) dx = du$ . However, we get stuck with $h'(x)$ . This is still a function of $x$ , not $u$ . We want to write everything in terms of $u$ .

If $h$ is monotonic, then $h$ is invertible over its range, so we can define an inverse function $h^{-1}$ such that $h^{-1}(h(x)) = x$ and $h(h^{-1}(u)) = u$ . Then, given $u = h(x)$ , $x = h^{-1}(u)$ . For example, if $h(x) = e^x$ , then $h^{-1}(x) = \log(x)$ . Then we can replace $h'(x)$ with $h'(h^{-1}(u))$ .

Now:

g(h(x)) dx = \frac{g(h(x))}{h'(x)} h'(x) dx = \frac{g(u)}{h'(h^{-1}(u))} du.

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Integrating returns the standard form for integration by change of variables:

\int g(h(x)) dx = \int \frac{g(u)}{h'(h^{-1}(u))} du.

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Then:

\int_{x = a}^b g(h(x)) dx = \int_{u = h(a)}^{h(b)} \frac{g(u)}{h'(h^{-1}(u))} du.

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Let $\tilde{g}(u) = g(u)/h'(h^{-1}(u))$ . Then let $\tilde{G}$ represent the anti-derivative (indefinite integral) of $\tilde{g}$ . Then:

\int_{x = a}^b g(h(x)) dx = \int_{u = h(a)}^{h(b)} \tilde{g}(u) du = \tilde{G}(h(b)) - \tilde{G}(h(a)).

(19)

Integration by Change of Variables (Version II)

Suppose that $f(x) = g \circ h(x) = g(h(x))$ and that $h(x)$ is a monotonic function.

Let $u = h(x)$ and $du(x) = h'(x) dx$ . Let $h^{-1}$ denote the inverse function that recovers $x$ from $u$ , $x = h^{-1}(u)$ .

Then:

\int g(h(x)) dx = \int \frac{g(u)}{h'(h^{-1}(u))} du = \int \tilde{g}(u) du.

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Let $\tilde{G}$ denote the indefinite integral of $\tilde{g}(x) = g(u)/h'(h^{-1}(u))$ . Then:

\int_{a}^b g(h(x)) dx = \tilde{G}(u) \Big|_{h(a)}^{h(b)} = \tilde{G}(h(b)) - \tilde{G}(h(a)).

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This form is a bit messy as a single formula. It’s easier to remember as a procedure:

Identify a term in the integrand we would like to replace (e.g. a function of $x$ , $h(x)$ , that we will use to define a new variable, $u$ ).
Replace $h(x)$ with $u$ everywhere inside the integral.
Find $h'(x) = \frac{d}{dx} h(x)$ .
Replace $dx$ with $du/h'(x)$ inside the integral.
Solve for $x$ in terms of $u$ , $x = h^{-1}(u)$ .
Replace all remaining $x$ ’s inside the integral with $h^{-1}(u)$ .
Update the bounds of integration from $x \in [a,b]$ to $u \in [h(a),h(b)]$ .
Integrate.

Example

Suppose that $X \sim \text{Normal}(0,1)$ . Find $\mathbb{E}[|X|^3]$ .

As before, start by writing the expectation as a weighted average:

\mathbb{E}[|X|^3] = \frac{1}{\sqrt{2 \pi}} \int_{x = -\infty}^{\infty} |x|^3 e^{-\frac{1}{2} x^2} dx.

(22)

Then, since $|x|^3$ is the composition of an odd function, $x^3$ , with an even function, $|x|$ , the function $|x|^3$ is even. The normal density is also even, so $|x|^3 e^{-\frac{1}{2} x^2}$ is an even function. So, using the same symmetry argument we used before:

\mathbb{E}[|X|^3] = \frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} x^3 e^{-\frac{1}{2} x^2} dx.

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We’d like to set $u = h(x) = \frac{1}{2} x^2$ and $g(x) = e^{-x}$ . However, this time the term outside $g(h(x))$ is $x^3 \neq h'(x) = \frac{d}{dx} \frac{1}{2} x^2 = x$ . So we need to use the second version of integration by change of variables.

$u = h(x) = \frac{1}{2} x^2$ .
Update the integral:
$\frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} x^3 e^{-u} dx$
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$h'(x) = \frac{d}{dx} \frac{1}{2} x^2 = x$ . So, $du = x dx$ .
Replace $dx$ with $du/h'(x)$ inside the integral:
$\frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} x^3 e^{-u} \frac{x dx}{x} = \frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} x^3 e^{-u} \frac{du}{x}$
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Simplifying:
$\frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} x^2 e^{-u} du$
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Solve for $x$ in terms of $u$ .
$u = h(x) = \frac{1}{2} x^2 \Rightarrow x = \pm \sqrt{2 u}$
(27)
Since we are only integrating over positive $x$ , we can assume that $x = \sqrt{2 u}$ so $h^{-1}(u) = \sqrt{2 u}$ .
Replace all remaining $x$ ’s inside the integral with $h^{-1}(u)$ .
$\frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} (\sqrt{2 u})^2 e^{u} du = \frac{2}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} 2 u e^{-u} du$
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Simplifying again:
$\frac{4}{\sqrt{2 \pi}} \int_{x = 0}^{\infty} u e^{-u} du.$
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Update the bounds of integration from $x \in [a,b]$ to $u \in [h(a),h(b)] = [\frac{1}{2} 0^2, \frac{1}{2} \infty^2] = [0,\infty]$ .
$\frac{4}{\sqrt{2 \pi}} \int_{u = 0}^{\infty} u e^{-u} du.$
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Integrate. Notice that the integrand is a product, $u \times e^{-u}$ . So, we’ll use integration by parts (see Section 7.1):
$\begin{aligned} \int_{u = 0}^{\infty} u e^{-u} du & = -ue^{-u} |_{0}^{\infty} - \int_{u = 0}^{\infty}(-e^{-u}) du \\ & = (0 - 0) + \int_{u=0}^{\infty} e^{-u} du \\ & = -e^{-u} |_{0}^1 = -(0-1) = 1. \end{aligned}$
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Therefore:

\mathbb{E}[|X|^3] = \frac{4}{\sqrt{2 \pi}} = 2 \sqrt{\frac{2}{\pi}}.

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Change of Density¶

The rule we just worked out for changing variables inside of an integral provides a general rule for updating the density of a continuous random variable after transforming the random variable.

Suppose that $X$ is a continuous random variable, with density function $f_X(x)$ . Then to find the probability that $X$ lands in an interval, we would evaluate an integral over the density:

\text{Pr}(X \in [a,b]) = \int_{x = a}^b f_X(x) dx.

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Similarly, to find the CDF we would integrate:

F_X(x) = \text{Pr}(X \in (-\infty,x]) = \int_{s = -\infty}^x f_X(s) ds.

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Suppose now, that $Y = h(X)$ for some monotonically increasing, differentiable function $h$ . What is the density of $Y$ , $f_Y(y)$ ?

Well, just like $X$ , the chance that $Y$ lands in an interval is related to its density by an integral:

\text{Pr}(Y \in [c,d]) = \int_{x = c}^d f_Y(y) dy.

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The CDF of $Y$ is also related to its density by an integral:

F_Y(y) = \text{Pr}(Y \in (-\infty,y]) = \int_{s = -\infty}^y f_Y(s) ds.

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There are now two ways to find the density of $Y$ :

Start from integrals involving $x$ . Use integration by change of variables to replace $x$ with $y = h(x)$ .
Start from integrals involving $y$ . Try to re-express them in terms of $x$ . Then, match sides to solve for the density of $y$ . Check that integrating over $y$ gives the same answer as integrating over $x$ .

We will adopt the second approach.

Consider the CDF of $Y$ . As always, if we know the CDF, then we can recover the density, and chances on intervals. So, if we can work out the CDF of $Y$ , then we have recovered it’s distribution.

In particular, if we know the CDF of $Y$ , then we can find its density since:

f_Y(y) = \frac{d}{dy} F_Y(y).

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The CDF of $Y$ is:

F_Y(y) = \text{Pr}(Y \leq y) = \text{Pr}(h(X) \leq y).

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Recall that, if $h(x)$ is monotonically increasing, then $h(x) \leq y$ if and only if $x \leq h^{-1}(y)$ . Therefore:

F_Y(y) = \text{Pr}(X \leq h^{-1}(y)) = F_X(h^{-1}(y)).

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So, recalling the product rule, and that $\frac{d}{dx} F_X(x) = f_X(x)$ ,

f_Y(y) = \frac{d}{dy} F_X(h^{-1}(y)) = f_X(h^{-1}(y)) \frac{d}{dy} h^{-1}(y).

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So:

Change of Density Formula

If $X$ is a continuous random variable with density $f_X$ , and $Y = h(X)$ for some differentiable, monotonically increasing function $h$ , then:

f_Y(y) = f_X(h^{-1}(y)) \frac{d}{dy} h^{-1}(y).

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It is often helpful to write this result in terms of $x$ . Let $h'(x) = \frac{d}{dx} h(x)$ . Then,

\frac{d}{dy} h^{-1}(y) = \frac{1}{h'(x)} \text{ at } x = h^{-1}(y)

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So:

f_Y(y) = f_X(x) \frac{1}{h'(x)} \text{ at } x = h^{-1}(y).

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If $h$ is differentiable and invertible, but decreasing, then:

f_Y(y) = f_X(x) \frac{1}{|h'(x)|} \text{ at } x = h^{-1}(y).

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This formula is the change of density formula.

The absolute value enters since densities are nonnegative and the $1/h'(x) dx$ term represents the length of an interval. Lengths can never be negative. We didn’t need the absolute value in the increasing case since, if $h$ is increasing, then $h'(x) > 0$ so $|h'(x)| = h'(x)$ .

Derivatives of Inverses

Why is $\frac{d}{dy} h^{-1}(y)$ equal to $1/h'(x)$ at $x = h^{-1}(y)$ ?

Well:

h(h^{-1}(y)) = y

(45)

so:

\frac{d}{dy} h(h^{-1}(y)) = \frac{d}{dy} y = 1.

(46)

Now, applying the product rule:

\frac{d}{dy} h(h^{-1}(y)) = h'(h^{-1}(y)) \left[ \frac{d}{dy} h^{-1}(y) \right].

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So, matching terms:

h'(h^{-1}(y)) \left[ \frac{d}{dy} h^{-1}(y) \right] = 1

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or:

\frac{d}{dy} h^{-1}(y) = \frac{1}{h'(h^{-1}(y))}.

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Geometrically, $h^{-1}$ is the reflection of $h$ across the line $y = x$ . So, since the slope of $h$ is change in $y = h(x)$ over change in $x$ , the slope of $h^{-1}$ is the change in $x$ over the change in $y$ , which is one over the slope of $h$ .

We’ve already studied a special case.

Suppose that $h(x)$ is linear, $h(x) = \sigma x + s$ for some $\sigma > 0$ . Then $h'(x) = \sigma$ , so, by the change of density formula:

f_Y(y) \propto f_X(h^{-1}(y)) = f_X\left( \frac{x - s}{\sigma} \right).

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It follows that $Y = h(X)$ has density proportional to the density of $X$ translated by $s$ and dilated by $\sigma$ . If we dilate a distribution by a factor of $\sigma$ , then we must divide its height by $\sigma$ so that it integrates to one. As before, if we double the width of a rectangle, we have to half its height to keep its area fixed.

By that logic, we should have:

f_Y(y) = \frac{1}{\sigma} f_X\left( \frac{x - s}{\sigma} \right)

(51)

Using the exact change of density formula gives the same result since $h'(x) = \sigma$ for all $x$ :

f_Y(y) = f_X(h^{-1}(y)) \frac{1}{|h'(h^{-1}(y))|} = f_X\left( \frac{x - s}{\sigma} \right) \frac{1}{|\sigma|} .

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This is a sensible rule. Replacing $X$ with $2X$ doubles the distance between any sampled values of $X$ , so should halve its density.

Let’s check that the change of density formula actually returns the correct density for $Y$ . To confirm, we’ll use integration by substitution. We will check the case when $h$ is monotonically increasing. To check the general case, break the range of $x$ into segments where $h$ is monotonic then work one segment at a time.

As long as $h$ is increasing:

\text{Pr}(X \in [a,b]) = \text{Pr}(Y \in [h(a),h(b)]).

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So, integrating over the density of $Y$ :

\text{Pr}(Y \in [h(a),h(b)]) = \int_{y = h(a)}^h(b) f_Y(y) dy.

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Using the change of density formula:

\int_{y = h(a)}^{h(b)} f_Y(y) dy = \int_{y = h(a)}^{h(b)} f_X(h^{-1}(y)) \left[\frac{d}{dy} h^{-1}(y) \right] dy.

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That looks like a mess, but we have all the parts we need to change of variables.

Let $x = h^{-1}(y)$ . Then:

$f_X(h^{-1}(y)) = f_X(x)$ ,
$dx = \frac{d}{dy} h^{-1}(y) dy$
$h^{-1}(h(a)) = a$ and $h^{-1}(h(b)) = b$ .

So:

\begin{aligned} \text{Pr}(X \in [a,b]) & = \int_{y = h(a)}^{h(b)} f_X(h^{-1}(y)) \left[\frac{d}{dy} h^{-1}(y) \right] dy \\ & = \int_{x = a}^b f_X(x) dx = \text{Pr}(X \in [a,b]). \end{aligned}

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To visualize this change of density, open the code cell below. Set the X distribution to “Beta” and use parameters $\alpha = \beta = 3$ . Set $g(x)$ to “Quadratic” and use coefficients $a = 1$ and $b = 0$ .

Set the number of samples to 1,000 and click “Draw Samples.” Then click “Transform” to push the same set of samples through the function $g(x)$ . Finally, click “Show Density.”

from utils import show_change_of_density

show_change_of_density()

Notice that the histogram for $X$ is symmetric about $X = 0.5$ , while the histogram for $Y$ is skewed so that its mode is closer to $Y = 0$ . Why?

Think about how the slope of $g(x)$ affects the spacing between samples. When $g(x)$ has a shallow slope, then distant samples are mapped near each other, so the density of $Y$ increases. In contrast, where the slope is steep, nearby samples get spread apart, so $Y$ is less dense. This explains the $1/|\text{slope of transform}|$ term in the change of density formula.

Try varying the transform and distribution for $X$ . You’ll see that, where the transform increases slowly, $Y$ is denser, and where it increases quickly, $Y$ is less dense.

Here are some more examples. You can use the same demonstration to test each case.

Example

Suppose that $X \sim \text{Exp}(\lambda)$ and $Y = \sqrt{x}$ .

Let $h(x) = x^{1/2}$ . It follows that:

$h'(x) = \frac{1}{2} x^{-1/2}$
$h^{-1}(y) = y^2$

So:

f_Y(y) = \frac{1}{|h'(x)|} f_X(x) = 2 x^{1/2} \lambda e^{-\lambda x} \text{ at } x = h^{-1}(y) = y^2.

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Plugging in:

f_Y(y) = 2 \lambda y e^{-\lambda y^2}.

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The support of $Y$ is $[0,\infty)$ since the smallest possible value of $\sqrt{X}$ is zero, and the largest is infinity.

Example

Suppose that $X \sim \text{Exponential}(0,1)$ and $Y = 1/x$ .

Let $h(x) = x^{-1}$ . It follows that:

$|h'(x)| = |-x^{-2}| = x^{-2}$
$h^{-1}(y) = 1/y = y^{-1}$ .

So:

f_Y(y) = \frac{1}{|h'(x)|} f_X(x) = x^2 \lambda e^{-\lambda x} \text{ at } x = h^{-1}(y) = \frac{1}{y}.

(64)

Plugging in:

f_Y(y) = \frac{\lambda}{y^2} e^{-\frac{\lambda}{y}}.

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The support of $Y$ is $[0,\infty)$ since the smallest possible value of $1/X$ is zero ( $X \rightarrow \infty$ ), and the largest is infinity ( $X \rightarrow 0$ ).

Example

Suppose that $X \sim \text{Uniform}(0,1)$ and $Y = -\log(X)$ .

Let $h(x) = -\log(x)$ . It follows that:

$|h'(x)| = |-x^{-1}| = x^{-1}$
$h^{-1}(y) = e^{-y}$ .

So:

f_Y(y) = \frac{1}{|h'(x)|} f_X(x) = x \times 1 = x \text{ at } x = h^{-1}(y) = e^{-y}.

(66)

Plugging in:

f_Y(y) = e^{-y}.

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The support of $Y$ is $[0,\infty)$ since the smallest possible value of $-\log(X)$ is zero ( $X = 1$ ), and the largest is infinity ( $X \rightarrow 0$ ).

So, $Y \sim \text{Exponential}(1)$ .

Example

Suppose that $X \sim \text{Normal}(0,1)$ and $Y = X^2$ .

Let $h(x) = x^2$ . Now we have a problem since $h$ is not monotonic. However, since the density of $X$ is an even function, and $h$ is an even function, the density of $Y$ would not change had we conditioned on $X > 0$ . Restricted to positive $x$ , $x^2$ is monotonic. So, we can use our familiar rule, replacing $f_{X}(x)$ with $f_{X|X > 0}(x)$ .

Then:

$|h'(x)| = |2x| = 2x$
$h^{-1}(y) = \sqrt{y}$ .

So:

f_Y(y) = \frac{1}{|h'(x)|} f_{X|X > 0}(x) = \frac{1}{2 x} \times \frac{2}{\sqrt{2 \pi}} e^{-\frac{1}{2} x^2} \text{ at } x = h^{-1}(y) = \sqrt{y}.

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Plugging in:

f_Y(y) = \frac{1}{\sqrt{2 \pi}} y^{-1/2} e^{- \frac{1}{2} y}.

(69)

The support of $Y$ is $[0,\infty)$ since the smallest possible value of $X^2$ is zero (at $X = 0$ ), and the largest is infinity (as $X \rightarrow \infty$ ).

7.2 Integrals of Compositions

Integration by Substitution (Change of Variables)¶

Change of Density¶