TABLE OF CONTENTS

Theory of Statistics 1

VOL. IMAR 2026

Bivariate & Trivariate Discrete Transformations

Direct Summation, Auxiliary Variables, and Convolutions

Bivariate Transformations

We have 2 discrete random variables $(X_1, X_2)$ and we know its distribution $f_{X_1, X_2}(x_1, x_2)$ . We are interested in a random variable $Y$ which is some function of $X_1, X_2$ : $Y = g(X_1, X_2)$

We want to determine the probability distribution of $Y$ .

Example 1

Consider the following bivariate probability distribution of $X_1$ and $X_2$ :

$X_1 \backslash X_2$	$-2$	$1$	$2$
$-1$	$1/6$	$1/6$	$1/12$
$0$	$1/12$	$1/12$	$0$
$1$	$1/6$	$1/6$	$1/12$

Find the probability distribution of $Y = X_1 + X_2$ .

Bivariate Transformation: $Y = X_1 + X_2$

Hover over the joint table or the marginal table to trace the mappings.

Joint PMF

f_{X_1, X_2}(x_1, x_2)

$x_1 \backslash x_2$	$-2$	$1$	$2$
$-1$	0.167 Y=-3	0.167 Y=0	0.083 Y=1
$0$	0.083 Y=-2	0.083 Y=1	0.000 Y=2
$1$	0.167 Y=-1	0.167 Y=2	0.083 Y=3

Y = X_1 + X_2

→

Summation

Marginal PMF

f_Y(y)

$y$	$f_Y(y)$
-3	0.167
-2	0.083
-1	0.167
0	0.167
1	0.167
2	0.167
3	0.083

Select any cell or marginal value to see the underlying summation theorem applied.

Method 1

The Direct Method (Not 1-1 Transformation)

Direct Summation

We map each $(x_1, x_2)$ pair to $y$ :

$(x_1, x_2)$	$y$	$f_{X_1, X_2}(x_1, x_2)$
$(-1, -2)$	$-3$	$1/6$
$(-1, 1)$	$0$	$1/6$
$(-1, 2)$	$1$	$1/12$
$(0, -2)$	$-2$	$1/12$
$(0, 1)$	$1$	$1/12$
$(0, 2)$	$2$	$0$
$(1, -2)$	$-1$	$1/6$
$(1, 1)$	$2$	$1/6$
$(1, 2)$	$3$	$1/12$

By summing the probabilities for each unique $y$ , we get the distribution of $Y$ .

Method 2

The Auxiliary Variable Method (1-1 Transformation)

Auxiliary Transformation

We introduce an auxiliary variable $Y_2 = X_2$ , making the transformation $(Y_1 = X_1 + X_2, Y_2 = X_2)$ one-to-one.

$(x_1, x_2)$	$(y_1, y_2)$	$f_{Y_1, Y_2}(y_1, y_2) = f_{X_1, X_2}(x_1, x_2)$
$(-1, -2)$	$(-3, -2)$	$1/6$
$(-1, 1)$	$(0, 1)$	$1/6$
$(-1, 2)$	$(1, 2)$	$1/12$
$(0, -2)$	$(-2, -2)$	$1/12$
$(0, 1)$	$(1, 1)$	$1/12$
$(0, 2)$	$(2, 2)$	$0$
$(1, -2)$	$(-1, -2)$	$1/6$
$(1, 1)$	$(2, 1)$	$1/6$
$(1, 2)$	$(3, 2)$	$1/12$

From $f_{Y_1, Y_2}(y_1, y_2)$ , we obtain $f_{Y_1}(y_1)$ by summing over $y_2$ .

In general, let $(X_1, X_2)$ be a random vector with jpmf $f_{X_1, X_2}(x_1, x_2)$ . $S = \{(x_1, x_2) : f_{X_1, X_2}(x_1, x_2) > 0\}$ is the support of $(X_1, X_2)$ .

Let $Y_1 = g_1(X_1, X_2)$ and $Y_2 = g_2(X_1, X_2)$ define a one-to-one transformation that maps $S$ onto $T$ , where $T = \{(y_1, y_2) : f_{Y_1, Y_2}(y_1, y_2) > 0\}$ is the support of $(Y_1, Y_2)$ .

Then, the jpmf of $(Y_1, Y_2)$ is:

f_{Y_1, Y_2}(y_1, y_2) = \begin{cases} f_{X_1, X_2}(g_1^{-1}(y_1, y_2), g_2^{-1}(y_1, y_2)), & \text{for } (y_1, y_2) \in T \\ 0, & \text{otherwise} \end{cases}

Where $x_1 = g_1^{-1}(y_1, y_2), x_2 = g_2^{-1}(y_1, y_2)$ is the single-valued inverse of $y_1 = g_1(x_1, x_2), y_2 = g_2(x_1, x_2)$ .

From $f_{Y_1, Y_2}(y_1, y_2)$ , we may obtain $f_{Y_1}(y_1)$ by summing over $y_2$ and $f_{Y_2}(y_2)$ by summing over $y_1$ .

Example 2

Let $X_1$ and $X_2$ be 2 independent random variables having Poisson distributions with the parameters $\lambda_1$ and $\lambda_2$ respectively.

Find the probability distribution of the r.v. $Y = X_1 + X_2$ .

Solution: Convolution of Poisson

By definition, $Y$ takes values $y \in \{0, 1, 2, \dots\}$ .

Poisson Convolution Derivation

\displaystyle f_Y(y)

\displaystyle = P(X_1 + X_2 = y) = \sum_{x_1=0}^y P(X_1 = x_1, X_2 = y - x_1)

\displaystyle = \sum_{x_1=0}^y f_{X_1}(x_1) f_{X_2}(y - x_1) \quad \text{(since independent)}

\displaystyle = \sum_{x_1=0}^y \frac{e^{-\lambda_1} \lambda_1^{x_1}}{x_1!} \frac{e^{-\lambda_2} \lambda_2^{y - x_1}}{(y - x_1)!}

\displaystyle = \frac{e^{-(\lambda_1 + \lambda_2)}}{y!} \sum_{x_1=0}^y \frac{y!}{x_1!(y-x_1)!} \lambda_1^{x_1} \lambda_2^{y-x_1} \quad \text{(multiply \& divide by } y!)

\displaystyle = \frac{e^{-(\lambda_1 + \lambda_2)}}{y!} (\lambda_1 + \lambda_2)^y \quad \text{(via Binomial Theorem)}

This shows that $Y \sim Poi(\lambda_1 + \lambda_2)$ .

Part III

Trivariate Case

Example

Let $(X_1, X_2, X_3)$ have the joint discrete probability function given by:

$(x_1, x_2, x_3)$	$(0,0,0)$	$(0,0,1)$	$(0,1,1)$	$(1,0,1)$	$(1,1,0)$	$(1,1,1)$
$f(x_1, x_2, x_3)$	$1/8$	$3/8$	$1/8$	$1/8$	$1/8$	$1/8$

Find the joint probability function of $Y_1 = X_1 + X_2 + X_3$ & $Y_2 = X_3 - X_2$ .

Solution

First, compute $y_1$ and $y_2$ for each possible state:

$(x_1, x_2, x_3)$	$(0,0,0)$	$(0,0,1)$	$(0,1,1)$	$(1,0,1)$	$(1,1,0)$	$(1,1,1)$
$f(x_1, x_2, x_3)$	$1/8$	$3/8$	$1/8$	$1/8$	$1/8$	$1/8$
$y_1 = x_1 + x_2 + x_3$	$0$	$1$	$2$	$2$	$2$	$3$
$y_2 = x_3 - x_2$	$0$	$1$	$0$	$1$	$1$	$0$

The joint pdf of $(y_1, y_2)$ is compiled by summing the probabilities for matching pairs of $(y_1, y_2)$ :

$y_1 \backslash y_2$	$0$	$1$	$\sum$
$0$	$1/8$	$0$	$1/8$
$1$	$0$	$3/8$	$3/8$
$2$	$1/8$	$2/8$	$3/8$
$3$	$1/8$	$0$	$1/8$
$\sum$	$3/8$	$5/8$	$1$