10 Transformations

In Chapter 8, we defined random variables and described them by their PMFs. However, we are often interested in more complicated random variables that are functions of simpler random variables. In this chapter, we describe how random variables behave under transformations. That is, if $X$ is a random variable, what is the distribution of $Y = g(X)$?

10.1 Functions of a Random Variable

To build some intuition about transformations, we revisit the roulette payoffs from Example 8.2.

Example 10.1 (Roulette and Transformations) Let $S$ be the payoff from a $1 bet on the number 23. We showed in Example 8.2 that the PMF of $S$ is

$x$	$-1$	$35$
$f_{S}(x)$	$37/38$	$1/38$

What if $Y$ represents the payoff from a $10 bet on the same number? We can express $Y$ as a transformation $g(S)$; in particular, \[ Y = g(S) = 10 S. \tag{10.1}\]

The possible outcomes are $-10$ and $350$, so the PMF of $Y$ is

$x$	$-10$	$350$
$f_{Y}(x)$	$37/38$	$1/38$

In fact, $S$ itself can be written as a transformation of a simpler random variable $I$, which is the indicator of the ball landing on the number 23. That is, $I$ is a $\text{Bernoulli}(p=1/38)$ random variable (Definition 8.5), whose PMF is

$x$	$0$	$1$
$f_{I}(x)$	$37/38$	$1/38$

Convince yourself that \[ S = 36 I - 1 \tag{10.2}\] by plugging in $I = 0$ and $I = 1$.

Notice that we can do algebra on random variables. For example, we can derive a formula for $Y$ as a transformation of $I$: \[ Y = 10S = 10(36 I - 1) = 360 I - 10. \]

The next example illustrates two ways to derive the PMF of a transformed random variable.

Example 10.2 (Transformation of a binomial) Suppose $X$ is $\text{Binomial}(n,p)$ random variable. Define a new random variable $Y$ by $Y = n - X$. In other words, $Y = g(X)$, where $g(x) = n - x$.

By doing algebra on the random variables, we can see that the PMFs of $X$ and $Y$ are related by

\[ \begin{align} f_Y(k) &= P(Y = k) \\ &= P(n - X = k) \\ &= P(X = n - k) \\ &= f_X(n - k). \end{align} \]

Substituting $n - k$ into the binomial PMF, we obtain: \[ f_Y(k) = f_X(n-k) = \binom{n}{n-k} p^{n-k} (1-p)^{k} = \binom{n}{k} (1-p)^k (1 - (1-p))^{n-k}, \] which is the PMF of a binomial distribution with parameters $n$ and $1-p$!

We can reach the same conclusion by interpreting the random variables. Recall from Section 8.4 that a binomial random variable $X$ represents the number of heads in $n$ coin tosses, each with probability $p$ of coming up heads. In this analogy, $Y = n-X$ would correspond to the number of tails. Each tail happens with probability $1-p$ and the $n$ trials are still independent, so $Y \sim \text{Binomial}(n,1-p)$.

In Examples 10.1 and 10.2, only the possible values $x$ changed, while the probabilities stayed the same. It turns out that the probabilities can also change, as the next example illustrates.

Example 10.3 (Random walk)

A drunk man staggers out of a bar. Each step he takes is equally likely to be either 1 foot to the left or 1 foot to the right, independently of the other steps. How far is he from the bar entrance after 5 steps?

Let $L$ be the number of steps he takes to the left. Then $L \sim \text{Binomial}(n=5, p=1/2)$. The number of steps to the right, per Example 10.2, is $R = 5 - L$, which is also $\text{Binomial}(n=5, p=1/2)$. The PMF of $L$ (and $R$) is \[ f_L(x) = {5 \choose x} \left( \frac{1}{2} \right)^x \left( \frac{1}{2} \right)^{5-x} = \frac{5 \choose x}{32},\] which can be written in tabular form as

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f_{L}(x)$	$1/32$	$5/32$	$10/32$	$10/32$	$5/32$	$1/32$

Let $X$ be his position after 5 steps, relative to the bar entrance. Then, \[ X = R - L = (5 - L) - L = 5 - 2L. \tag{10.3}\]

Since $L$ is an integer from $0$ to $5$, $X$ is an odd integer from $-5$ to $5$. The PMF of $X$ has the same probabilities but distributed over different values.

$x$	$-5$	$-3$	$-1$	$1$	$3$	$5$
$f_{X}(x)$	$1/32$	$5/32$	$10/32$	$10/32$	$5/32$	$1/32$

We can write it as the following formula \[ \begin{align} f_X(x) = P(X = x) &= P(5 - 2L = x) \\ &= P(L = (5-x)/2) \\ &= f_L((5-x)/2) \\ &= \frac{{5 \choose (5-x)/2}}{32}, \end{align} \] as long as we are careful to only evaluate this function at $x = -5, -3, -1, 1, 3, 5$.

Let $D$ be his distance from the bar entrance. Notice that this distance can be expressed as a transformation of $X$ (or $L$). \[ D = |X| = |5 - 2L|. \tag{10.4}\]

The possible values of $D$ are $1$, $3$, and $5$. Each of these values corresponds to two values of $X$. To calculate the PMF of $D$, we will need to add these probabilities:

$P(D = 1) = P(X = 1) + P(X = -1) = \frac{10}{32} + \frac{10}{32} = \frac{20}{32}$.
$P(D = 3) = P(X = 3) + P(X = -3) = \frac{5}{32} + \frac{5}{32} = \frac{10}{32}$.
$P(D = 5) = P(X = 5) + P(X = -5) = \frac{1}{32} + \frac{1}{32} = \frac{2}{32}$.

To summarize, the PMF of $D$ is given by:

$x$	$1$	$3$	$5$
$f_{D}(x)$	$20/32$	$10/32$	$2/32$

Notice that these probabilities are different from the probabilities in the PMF of $X$ because multiple values of $X$ mapped to the same value of $D$. In mathematical language, the transformation $g(x) = |x|$ is not one-to-one.

When the transformation $Y = g(X)$ is not one-to-one, then the PMFs of $X$ and $Y$ may not have the same probabilities. But it is straightforward to determine the PMF of $Y$ by the following recipe:

Identify the possible values of $Y$, say $y_1, y_2, \dots$.
For each possible value $y_i$, identify all the $x$ values such that $g(x) = y_i$ and add their probabilities:

\[ f_Y(y_i) = \sum_{x: g(x) = y_i} f_X(x). \tag{10.5}\]

The next example illustrates how to apply this recipe.

Example 10.4 (Call option) In finance, a call option is a contract that allows the holder to buy a certain share at a pre-determined price (called the “strike price”) at a pre-determined time.

For example, suppose you hold a call option that allows you to buy a stock at a price of $55 in one week. The price of the stock next week is a random variable $X$.

If $X = 53$, then the option is worth nothing because there is no point in exercising the option for $55 when you could just buy the stock for $53.
But if $X = 60$, then the option is worth $5 because $60 - 55 = 5$. (In other words, you could make $5 by exercising the option to buy the stock for $55 and flipping it for $60.)

In other words, the value of this option $Y$ is a transformation of the stock price next week $X$, namely \[ Y = \max(X - 55, 0). \] This function is not one-to-one because any value of $X$ below 55 will map to $Y = 0$.

Suppose the PMF of $X$ is

$x$	$50$	$53$	$57$	$60$
$f_X(x)$	$1/8$	$2/8$	$4/8$	$1/8$

The possible values of $Y$ are $0$, $2$, and $5$. To determine $f_Y(0)$, we need to sum the probabilities of the values of $X$ that map to $Y = 0$:

$f_Y(0) = f_X(50) + f_X(53) = 1/8 + 2/8 = 3/8$.

The other two probabilities are more straightforward:

$f_Y(2) = f_X(57) = 4/8$.
$f_Y(5) = f_X(60) = 1/8$.

To summarize, the value of the option $Y$ has PMF

$y$	$0$	$2$	$5$
$f_Y(y)$	$3/8$	$4/8$	$1/8$

10.2 Functions of Multiple Variables

A random variable can also be a transformation of more than one random variable.

Example 10.5 (Higher of two dice rolls) Consider a game where two dice are rolled and the higher of the two numbers is taken. If the higher number is 1, 2, 3, or 4, then Player 1 wins; if the higher number is 5 or 6, then Player 2 wins. Would you rather be Player 1 or Player 2?

At first, it might seem like Player 1 has the upper hand, since there are four outcomes where they win, compared with two outcomes where Player 2 wins. The flaw in this reasoning is that the outcomes are not equally likely.

We can answer this question by defining a random variable $H$ to be the higher of the numbers on the two dice. If we denote by $X$ and $Y$ the numbers on the first and second die, respectively, then $H$ is a function of both $X$ and $Y$: \[ H = \max(X, Y). \tag{10.6}\]

We can determine the PMF of $H$ by listing the 36 possible combinations of $X$ and $Y$, along with the corresponding value of $H$.

$X$	$Y$	$H = \max(X, Y)$
$1$	$1$	$1$
$1$	$2$	$2$
$1$	$3$	$3$
$1$	$4$	$4$
$1$	$5$	$5$
$\dots$	$\dots$	$\dots$
$6$	$2$	$6$
$6$	$3$	$6$
$6$	$4$	$6$
$6$	$5$	$6$
$6$	$6$	$6$

We can obtain the PMF of $H$ by simply counting the outcomes.

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$f_H(x)$	$1/36$	$3/36$	$5/36$	$7/36$	$9/36$	$11/36$

Using this PMF, we can calculate the probability that Player 1 wins as \[ P(H \leq 4) = 1/36 + 3/36 + 5/36 + 7/36 = 16/36, \] which is less than $1/2$. It is better to be Player 2!

This example is also explained in the following video.

There are also cases where we can reason about the distribution of a transformation without doing any calculations.

Example 10.6 (Sum of independent Bernoullis) Let $X_1, X_2, \dots, X_n$ be $n$ independent $\text{Bernoulli}(p)$ random variables. What is the distribution of $Y = X_1 + X_2 + \cdots + X_n$?

In Section 8.4, we described a Bernoulli random variable $X_i$ as the outcome of a single toss of a coin with probability $p$ of landing heads. That is, $X_i = 1$ if the coin lands heads, and $X_i = 0$ otherwise.

Now, if we toss this coin $n$ times, we can let $X_i$ represent the outcome of the $i$th toss. Now, \[ Y = X_1 + X_2 + \dots + X_n \] represents the total number of heads across all $n$ tosses, which was the description of a $\text{Binomial}(n,p)$ random variable.

However, a full discussion of transformations of multiple random variables will have to wait until we discuss how to describe the distribution of multiple random variables in Chapter 13.

\(X\)	\(Y\)	\(H = \max(X, Y)\)
\(1\)	\(1\)	\(1\)
\(1\)	\(2\)	\(2\)
\(1\)	\(3\)	\(3\)
\(1\)	\(4\)	\(4\)
\(1\)	\(5\)	\(5\)
\(\dots\)	\(\dots\)	\(\dots\)
\(6\)	\(2\)	\(6\)
\(6\)	\(3\)	\(6\)
\(6\)	\(4\)	\(6\)
\(6\)	\(5\)	\(6\)
\(6\)	\(6\)	\(6\)