Question

Suppose that the joint probability mass function of \(X\) and \(Y\) is $$ P(X=i, Y=j)=\left(\begin{array}{l} j \\ i \end{array}\right) e^{-2 \lambda} \lambda^{i} / j !, \quad 0 \leq i \leq j $$ (a) Find the probability mass function of \(Y\). (b) Find the probability mass function of \(X\). (c) Find the probability mass function of \(Y-X\).

Short answer

The probability mass functions for the given problem are as follows: (a) The probability mass function of Y is: $$ P_Y(j) = e^{-2 \lambda} \frac{1}{j!} (\lambda+1)^j, \quad j = 0, 1, 2, ... $$ (b) The probability mass function of X is: $$ P_X(i) = \frac{\lambda^i}{i!} (e^{-\lambda} \sum_{j=0}^{i-1} \frac{(2\lambda)^j}{j!}), \quad i = 0, 1, 2, ... $$ (c) The probability mass function of Y-X is: $$ P_{Y-X}(k) = e^{-2 \lambda} \sum_{i=0}^{\infty} \frac{\lambda^i}{i!} (e^{2 \lambda} - \sum_{j=0}^{i+k-1} \frac{(2\lambda)^j}{j!}), \quad k = 0, 1, 2, ... $$

Step-by-step solution

Problem (a) - Probability mass function of Y #

To find P_Y(j), we need to sum the joint probability mass function over all possible values of i for each value of j. Therefore, $$ P_Y(j) = \sum_{i=0}^j P(X=i, Y=j) $$ By plugging the given joint probability mass function, we get: $$ P_Y(j) = \sum_{i=0}^j \left(\begin{array}{l} j \\\ i \end{array}\right) e^{-2 \lambda} \lambda^{i} / j ! $$ Now, we simplify the above expression: $$ P_Y(j) = e^{-2 \lambda} \frac{1}{j!} \sum_{i=0}^j \left(\begin{array}{l} j \\\ i \end{array}\right) \lambda^{i} $$ Using the binomial theorem, we know that: $$ \sum_{i=0}^j \left(\begin{array}{l} j \\\ i \end{array}\right) \lambda^{i} = (\lambda+1)^j $$ So, the probability mass function of Y is: $$ P_Y(j) = e^{-2 \lambda} \frac{1}{j!} (\lambda+1)^j, \quad j = 0, 1, 2, ... $$

Problem (b) - Probability mass function of X #

To find P_X(i), we need to sum the joint probability mass function over all possible values of j for i being fixed. Therefore, $$ P_X(i) = \sum_{j=i}^{\infty} P(X=i, Y=j) $$ By plugging the given joint probability mass function, we get: $$ P_X(i) = \sum_{j=i}^{\infty} \left(\begin{array}{l} j \\\ i \end{array}\right) e^{-2 \lambda} \lambda^{i} / j ! $$ Now, we simplify the above expression: $$ P_X(i) = e^{-2 \lambda} \frac{\lambda^i}{i!} \sum_{j=i}^{\infty} \left(\begin{array}{l} j \\\ i \end{array}\right) \frac{1}{(j-i)!} $$ Using the identity for falling factorials: $$ \sum_{j=i}^{\infty} \left(\begin{array}{l} j \\\ i \end{array}\right) \frac{1}{(j-i)!} = 1^i + 2^i + \cdots \\ = \frac{e^{2 \lambda} - \sum_{j=0}^{i-1} j!}{i!} $$ So, the probability mass function of X is: $$ P_X(i) = \frac{\lambda^i}{i!} (e^{-\lambda} \sum_{j=0}^{i-1} \frac{(2\lambda)^j}{j!}), \quad i = 0, 1, 2, ... $$

Problem (c) - Probability mass function of Y-X #

To find P_{Y-X}(k), we need to sum the joint probability mass function over all possible values of i and j that satisfy Y-X=k while keeping k fixed. Therefore, $$ P_{Y-X}(k) = \sum_{i=0}^{\infty} \sum_{j=i+k}^{\infty} P(X=i, Y=j) $$ By plugging the given joint probability mass function, we get: $$ P_{Y-X}(k) = \sum_{i=0}^{\infty} \sum_{j=i+k}^{\infty} \left(\begin{array}{l} j \\\ i \end{array}\right) e^{-2 \lambda} \lambda^{i} / j ! $$ Now, we simplify the above expression: $$ P_{Y-X}(k) = e^{-2 \lambda} \sum_{i=0}^{\infty} \frac{\lambda^i}{i!} \sum_{j=i+k}^{\infty} \left(\begin{array}{l} j \\\ i \end{array}\right) \frac{1}{(j-i)!} $$ Notice that the inner summand follows a similar pattern to the one we encountered while calculating the PMF of X. The pattern observed was: $$ \sum_{j=i+k}^{\infty} \left(\begin{array}{l} j \\\ i \end{array}\right) \frac{1}{(j-i)!} = \frac{e^{2 \lambda} - \sum_{j=0}^{i+k-1} j!}{i!} $$ So, the probability mass function of Y-X is: $$ P_{Y-X}(k) = e^{-2 \lambda} \sum_{i=0}^{\infty} \frac{\lambda^i}{i!} (e^{2 \lambda} - \sum_{j=0}^{i+k-1} \frac{(2\lambda)^j}{j!}), \quad k = 0, 1, 2, ... $$

Key concepts

Joint Probability Mass Function

Understanding the joint probability mass function (PMF) is critical for grasping more complex probability scenarios, as it provides a way to describe the likelihood of two discrete random variables occurring simultaneously. In probability theory, a PMF outlines the probability that a discrete random variable takes on a specific value, and the joint PMF takes this concept to two dimensions, where we now focus on pairs of outcomes.

Returning to the exercise, we recall that the joint PMF, denoted as \(P(X=i, Y=j)\), is given as a function of two variables, where \(X\) and \(Y\) are two discrete random variables, and \(i\) and \(j\) represent specific outcomes. The problem presents a joint PMF that incorporates the exponential function, factorials, and the binomial coefficient - which is a number of ways to choose \(i\) elements out of a set of \(j\), commonly written as \(\left(\begin{array}{l} j \ i \end{array}\right)\).

This notation is significant as it also sets the stage for utilizing the binomial theorem in simplifying the function during the solution process. To find the individual PMFs for \(Y\), \(X\), and \(Y-X\), we apply the concept of marginalization, which means summing over the joint PMF for all possible values of the other variable. This process allows students to peel apart the combined behavior of \(X\) and \(Y\), to explore the individual characteristics of each random variable.

Probability Theory

Probability theory serves as the foundation of statistical analysis, encompassing the methods to quantify uncertainty. At its core, probability theory enables us to predict the chances of various outcomes and to understand random phenomena. In educational contexts, it's imperative to start by defining the basic terms such as the probability of an event, random variables, and particularly, the probability mass function (PMF) for discrete random variables.

In the provided exercise, we delve into probability theory by examining discrete random variables \(X\) and \(Y\). The PMFs we seek to find represent the foundational blocks upon which more complex probability distributions are built. Whether it's the PMF of a single variable or a combination of variables like \(Y-X\), a deep comprehension of these basics paves the path for tackling more advanced topics, such as joint distributions, conditional probability, and expected values.

By tackling the PMFs through a step-by-step process, as illustrated in the exercise, we gain not only the numerical answers but also the conceptual knowledge of how individual and joint behavior of variables can be depicted through probability distributions. This step-by-step walkthrough shows the calculation of PMFs through marginalization and demonstrates the powerful connections between various probability concepts.

Binomial Theorem

The binomial theorem is a cornerstone of algebra and probability theory due to its vast spectrum of applications. It comes into play when expanding expressions that are raised to a power, usually presented in the form \((a+b)^n\). This theorem shows how to represent this expression as a sum of terms \(C(n, k) \times a^{n-k} \times b^k\), where \(C(n, k)\) is the binomial coefficient, and \(k\) varies from 0 to \(n\).

In the context of the exercise, the binomial theorem simplifies the PMF of \(Y\) by allowing us to sum the probability for each \(i\) from 0 to \(j\). The result of such a summation yields a neatly raised binomial expression, \(( lambda+1)^j\), demonstrating a clear connection between combinatorials used in the joint PMF and the expansion formulas stemming from the binomial theorem.

Moreover, students often find the abstraction of the binomial theorem easier to grasp when it's applied to concrete problems, such as calculating a PMF. Seeing the theorem in action reinforces its importance, not just as an algebraic tool, but as a bridge linking various areas of mathematics and making complex probabilities computations manageable.