TABLE OF CONTENTS

Theory of Statistics 1

VOL. IMAR 2026

Distributions of Functions of Discrete R.V.s (Univariate)

Univariate Transformations, 1-to-1 mappings, and Summation Methods

Transformations of Discrete Variables

Often in statistics, we know the probability distribution of a random variable $X$ , but we are actually interested in the distribution of some function of $X$ , say $Y = g(X)$ . We want to determine the probability distribution of $Y$ .

Distribution X

Transformation g(X)

0.20

Y = g(X)

0.20

Theorem Applied

f_Y(y) = f_X(g^{-1}(y)) = f_X(y - 2)

Hover over the X distribution to trace the mapping.

Part I

One-to-One Transformations

Definition — Theorem 1 (One-to-One Mapping)

Let $X$ be a discrete random variable with probability function $f_X(x)$ and space $D_X$ , and let $Y = g(X)$ be a one-to-one transformation.

The space of $Y$ is $D_Y = \{g(x) : x \in D_X\}$ .

Then the p.m.f. of $Y$ is:

f_Y(y) = P(Y = y) = P(g(X) = y) = P(X = g^{-1}(y)) = f_X(g^{-1}(y))

where $g^{-1}$ is the Inverse Function.

Example 1

Let $X$ be a r.v. with the following probability distribution:

$x$	$2$	$4$	$6$	$8$	$\sum$
$f_X(x)$	$1/8$	$1/8$	$1/4$	$1/2$	$1$

Let $Y = 3X$ . Find the probability distribution of $Y$ .

Solution

The transformation is $y = 3x$ , which is one-to-one. The inverse is $x = y/3$ . The new space is $D_Y = \{6, 12, 18, 24\}$ .

Using the theorem $f_Y(y) = f_X(y/3)$ , the distribution is:

$y$	$6$	$12$	$18$	$24$	$\sum$
$f_Y(y)$	$1/8$	$1/8$	$1/4$	$1/2$	$1$

One-to-One Transformation

Slide the parameter to scale the distribution linearly.

Y = cX3

Notice how the shape of the PMF remains identical, just horizontally spaced.

f_Y(y) = f_X(y / 3)

Example 2

Let $X \sim Poi(\lambda)$ . $f_X(x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \dots$

Find the p.m.f. of $Y = 4X$ .

Solution

The transformation $y = 4x$ is one-to-one. The inverse is $x = y/4$ . The space of $Y$ is $D_Y = \{0, 4, 8, 12, \dots\}$ .

$f_Y(y) = f_X(y/4) = \frac{e^{-\lambda} \lambda^{y/4}}{(y/4)!}, \quad y = 0, 4, 8, \dots$

Distribution X

Transformation g(X)

0.14

0.27

0.18

Y = g(X)

0.14

0.27

0.18

Theorem Applied

f_Y(y) = f_X(g^{-1}(y)) = f_X(y / 4)

Hover over the X distribution to trace the mapping.

Example 3

Let $X$ have the following p.m.f.: $f_X(x) = \binom{3}{x} \left(\frac{2}{3}\right)^x \left(\frac{1}{3}\right)^{3-x}, \quad x = 0, 1, 2, 3$

Find the p.m.f. of $Y = X^2$ .

Solution

Since the space of $X$ is non-negative ( $D_X = \{0, 1, 2, 3\}$ ), the transformation $y = x^2$ is one-to-one over this support. The inverse is $x = \sqrt{y}$ . The space of $Y$ is $D_Y = \{0, 1, 4, 9\}$ .

$f_Y(y) = f_X(\sqrt{y}) = \binom{3}{\sqrt{y}} \left(\frac{2}{3}\right)^{\sqrt{y}} \left(\frac{1}{3}\right)^{3-\sqrt{y}}, \quad y = 0, 1, 4, 9$

Distribution X

Transformation g(X)

0.22

0.44

0.30

Y = g(X)

0.22

0.44

0.30

Theorem Applied

f_Y(y) = \sum_{x \in g^{-1}(y)} f_X(x)

Hover over the X distribution to trace the mapping.

Example 4 (Homework)

Consider a sequence of independent flips of a fair coin. Let $X$ be the number of flips needed to obtain the 1st head. Let $Y$ be the number of flips before the 1st head. Find the p.m.f. of $Y$ .

Solution

$X$ follows a Geometric distribution with $p=1/2$ . The number of flips before the first head is $Y = X - 1$ . The transformation is $y = x - 1$ , which is one-to-one. The inverse is $x = y + 1$ . The space of $X$ is $D_X = \{1, 2, 3, \dots\}$ . Thus, the space of $Y$ is $D_Y = \{0, 1, 2, \dots\}$ .

The p.m.f. of $X$ is $f_X(x) = (1/2)^{x-1}(1/2) = (1/2)^x$ . $f_Y(y) = f_X(y + 1) = \left(\frac{1}{2}\right)^{y+1}, \quad y = 0, 1, 2, \dots$

Distribution X

Transformation g(X)

0.50

0.25

0.13

Y = g(X)

0.50

0.25

0.13

Theorem Applied

f_Y(y) = f_X(g^{-1}(y)) = f_X(y + 1)

Hover over the X distribution to trace the mapping.

Part II

Many-to-One Transformations

In general, if the transformation $Y = g(X)$ is not one-to-one. In particular, if a given value of $Y$ , say $y$ , corresponds to more than one value of $X$ , say $x_1, x_2, \dots, x_K$ . Then:

Theorem 2 (Many-to-One Mapping Derivation)

\displaystyle f_Y(y)

\displaystyle = P(Y = y) = P(X = x_1 \text{ or } X = x_2 \text{ or } \dots \text{ or } X = x_K)

\displaystyle = P(X = x_1) + P(X = x_2) + \dots + P(X = x_K)

\displaystyle = P(X = g_1^{-1}(y)) + P(X = g_2^{-1}(y)) + \dots + P(X = g_K^{-1}(y))

\displaystyle = f_X(g_1^{-1}(y)) + f_X(g_2^{-1}(y)) + \dots + f_X(g_K^{-1}(y))

\displaystyle = \sum_{i=1}^K f_X(g_i^{-1}(y))

Example 1

Consider the following probability distribution of $X$ :

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$f_X(x)$	$0.1$	$0.15$	$0.2$	$0.3$	$0.15$	$0.1$

Find the probability distribution of $Y = X^2$ .

Solution

The space of $Y$ is $D_Y = \{0, 1, 4, 9\}$ . We sum the probabilities for each $y \in D_Y$ :

$f_Y(0) = f_X(0) = 0.2$
$f_Y(1) = f_X(-1) + f_X(1) = 0.15 + 0.3 = 0.45$
$f_Y(4) = f_X(-2) + f_X(2) = 0.1 + 0.15 = 0.25$
$f_Y(9) = f_X(3) = 0.1$

Many-to-One Transformation

Fold the distribution mathematically to see probabilities sum together.

Y = X²

Notice how the red (

X \lt 0

) and blue/purple (

X \gt 0

) masses stack at the same

Y

coordinate.

f_Y(y) = \sum_{x^2=y} f_X(x)

Example 2

Let $f_X(x) = (1/2)^x, \quad x = 1, 2, 3, \dots$

Find the p.m.f. of: a) $Y = \begin{cases} -1, & \text{if } X \text{ is odd} \\ 1, & \text{if } X \text{ is even} \end{cases}$

b) $Z = (X - 2)^2$

Solution

Part (a) The possible values for $Y$ are $D_Y = \{-1, 1\}$ .

$f_Y(-1) = P(X \text{ is odd}) = f_X(1) + f_X(3) + f_X(5) + \dots = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \dots = \frac{1/2}{1 - 1/4} = \frac{2}{3}$
$f_Y(1) = P(X \text{ is even}) = f_X(2) + f_X(4) + f_X(6) + \dots = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots = \frac{1/4}{1 - 1/4} = \frac{1}{3}$

Part (b) The possible values for $Z$ based on $X \in \{1, 2, 3, 4, \dots\}$ are $z = (x-2)^2 \in \{1, 0, 1, 4, 9, \dots\}$ . The space is $D_Z = \{0, 1, 4, 9, 16, \dots\}$ .

$f_Z(0) = f_X(2) = (1/2)^2 = 1/4$
$f_Z(1) = f_X(1) + f_X(3) = 1/2 + 1/8 = 5/8$
For $z \ge 4$ , there is only one value of $x$ that maps to $z$ (since $x \ge 4$ ). The inverse is $x = 2 + \sqrt{z}$ . $f_Z(z) = f_X(2 + \sqrt{z}) = (1/2)^{2+\sqrt{z}}, \quad z \in \{4, 9, 16, \dots\}$