# 16.1. Linear Transformations¶

Linear transformations are both simple and ubiquitous: every time you change units of measurement, for example to standard units, you are performing a linear transformation.

## 16.1.1. Linear Transformation: Exponential Density¶

Let $$T$$ have the exponential $$(\lambda)$$ distribution and let $$T_1 = \lambda T$$. Then $$T_1$$ is a linear transformation of $$T$$. Therefore

$E(T_1) = \lambda E(T) = 1 ~~~ \text{and} ~~~ SD(T_1) = \lambda SD(T) = 1$

The parameter $$\lambda$$ has disappeared in these results. Let’s see how that follows from the distribution of $$T_1$$. The cdf of $$T_1$$ is

$F_{T_1}(t) = P(T_1 \le t) = P(T \le t/\lambda) = 1 - e^{-\lambda (t/\lambda)} = 1 - e^{-t}$

That’s the cdf of the exponential $$(1)$$ distribution, consistent with the expectation and SD we found above.

To summarize, if $$T$$ has the exponential $$(\lambda)$$ distribution then the distribution of $$T_1 = \lambda T$$ is exponential $$(1)$$.

You can think of the exponential $$(1)$$ distribution as the fundamental member of the family of exponential distributions. All others in the family can be found by changing the scale of measurement, that is, by multiplying by a constant.

If $$T_1$$ has the exponential $$(1)$$ distribution, then $$T = \frac{1}{\lambda}T_1$$ has the exponential $$(\lambda)$$ distribution. The factor $$1/\lambda$$ is called the scale parameter.

Here are graphs of the densities of $$T_1$$ and $$T = \frac{1}{2}T_1$$. By the paragraph above, $$T$$ has the exponential $$(2)$$ distribution. The formulas for the two densities are

$f_{T_1} (s) = e^{-s} ~~~~~~~~~~~~~~ f_T(t) = 2e^{-2t}$

Let’s try to understand the relation between these two densities in a way that will help us generalize what we are seeing in this example.

The relation between the two random variables is $$T = \frac{1}{2}T_1$$.

• For any $$t$$, the chance that $$T$$ is near $$t$$ is the same as the chance that $$T_1$$ is near $$s = 2t$$. This explains the factor $$e^{-2t}$$ in the density of $$T$$.

• If we think of $$T_1$$ as a point on the horizontal axis, then to create $$T$$ you have to divide $$T_1$$ by $$2$$. So the transformation consists of halving all distances on the horizontal axis. The total area under the density of $$T$$ must equal $$1$$, so we have to compensate by doubling all distances on the vertical axis. This explains the factor $$2$$ in the density of $$T$$.

Quick Check

If $$T$$ is exponential $$(\lambda)$$ and $$c > 0$$, then $$S = cT$$ is exponential with one of the following rates. Which one?

$$c\lambda$$, $$\frac{c}{\lambda}$$, $$\frac{\lambda}{c}$$

## 16.1.2. Linear Change of Variable Formula for Densities¶

We use the same idea to find the density of a linear transformation of a random variable.

Let $$X$$ be a random variable with density $$f_X$$, and let $$Y = aX + b$$ for constants $$a \ne 0$$ and $$b$$. Let $$f_Y$$ be the density of $$Y$$. Then

$f_Y(y) ~ = ~ f_X\big{(} \frac{y-b}{a}\big{)} \frac{1}{\lvert a \rvert}$

Let’s take this formula in two pieces, as in the exponential example.

• For $$Y$$ to be $$y$$, $$X$$ has to be $$(y-b)/a$$.

• The linear function $$y = ax+b$$ involves multiplying distances along the horizontal axis by $$\lvert a \rvert$$; the sign of $$a$$ doesn’t affect distances. To get a density, we have to compensate by dividing all vertical distances by $$\lvert a \rvert$$.

This is a good way to understand the formula, and will help you understand the corresponding formula for non-linear transformations.

For a formal proof, start with the case $$a > 0$$.

$F_Y(y) = P(aX+b \le y) = P\big{(}X \le \frac{y-b}{a}\big{)} = F_X\big{(}\frac{y-b}{a}\big{)}$

By the chain rule of differentiation,

$f_Y(y) = f_X\big{(}\frac{y-b}{a}\big{)} \cdot \frac{1}{a}$

If $$a < 0$$ then division by $$a$$ causes the direction of the inequality to switch:

$F_Y(y) = P(aX+b \le y) = P\big{(}X \ge \frac{y-b}{a}\big{)} = 1 - F_X\big{(}\frac{y-b}{a}\big{)}$

Now the chain rule yields

$f_Y(y) ~ = ~ -f_X\big{(}\frac{y-b}{a}\big{)} \cdot \frac{1}{a} ~ = ~ f_X\big{(}\frac{y-b}{a}\big{)} \cdot \frac{1}{\lvert a \rvert}$

Quick Check

$$V$$ has density $$f$$ on the whole real line.

(a) Write the density of $$W = 5 + 3V$$ in terms of $$f$$.

(b) Write the density of $$W = 5 - 3V$$ in terms of $$f$$.

## 16.1.3. The Normal Densities¶

Let $$Z$$ have the standard normal density

$\phi(z) ~ = ~ \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z^2}, ~~~ -\infty < z < \infty$

Let $$X = \sigma Z + \mu$$ for constants $$\mu$$ and $$\sigma$$ with $$\sigma > 0$$. Then for any real number $$x$$, the density of $$X$$ is

\begin{split} \begin{align*} f_X(x) ~ &= ~ \phi\big{(} \frac{x-\mu}{\sigma} \big{)} \frac{1}{\sigma} \\ \\ &= ~ \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2} \big{(} \frac{x-\mu}{\sigma} \big{)}^2} \end{align*} \end{split}

Thus every normal random variable is a linear transformation of a standard normal variable.

Quick Check

Let $$X$$ have the normal distribution with mean $$0$$ and variance $$90$$. Write $$X$$ as a linear function of a standard normal random variable.

## 16.1.4. The Uniform Densities, Revisited¶

Let the distribution of $$U$$ be uniform on $$(0, 1)$$ and for constants $$b > a$$ let $$V = (b-a)U + a$$. In an earlier section we saw that $$V$$ has the uniform distribution on $$(a, b)$$. But let’s see what’s involved in confirming that result using our new formula.

First it is a good idea to be clear about the possible values of $$V$$. Since the possible values of $$U$$ are in $$(0, 1)$$, the possible values of $$V$$ are in $$(a, b)$$.

At $$v \in (a, b)$$, the density of $$V$$ is

$f_V(v) ~ = ~ f_U\big{(} \frac{v - a}{b-a} \big{)} \frac{1}{b-a} ~ = ~ 1 \cdot \frac{1}{b-a} ~ = ~ \frac{1}{b-a}$

That’s the uniform density on $$(a, b)$$.