Question

Let \(X\) and \(Y\) be two independent, nonnegative integer-valued, random variables whose distribution has the property $$ \operatorname{Pr}\\{X=x \mid X+Y=x+y\\}=\frac{\left(\begin{array}{l} m \\ x \end{array}\right)\left(\begin{array}{l} n \\ y \end{array}\right)}{\left(\begin{array}{l} m+n \\ x+y \end{array}\right)} $$ for all nonnegative integers \(x\) and \(y\) where \(m\) and \(n\) are given positive integers. Assume that \(\operatorname{Pr}\\{X=0\\}\) and \(\operatorname{Pr}\\{Y=0\\}\) are strictly positive. Show that both \(X\) and \(Y\) have binomial distributions with the same parameter \(p\), the other parameters being \(m\) and \(n\), respectively.

Short answer

Given the conditional probability property for random variables $X$ and $Y$, we can rewrite it using binomial coefficients and simplify the expression. Using the properties of conditional probabilities and the independence of $X$ and $Y$, we can find the probability mass functions of $X$ and $Y$. These expressions resemble binomial distributions. Considering the given strictly positive probabilities, we can conclude that both $X$ and $Y$ have binomial distributions with the same parameter $p$.

Step-by-step solution

Rewrite the given property using binomial coefficients

Rewrite the property using binomial coefficient notation: $$ \operatorname{Pr}\\{X=x \mid X+Y=x+y\\} = \frac{\binom{m}{x}\binom{n}{y}}{\binom{m+n}{x+y}} $$

Apply the binomial coefficient identity

Apply the identity \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\): $$ \operatorname{Pr}\\{X=x \mid X+Y=x+y\\} = \frac{\frac{m!}{x!(m-x)!}\frac{n!}{y!(n-y)!}}{\frac{(m+n)!}{(x+y)!(m+n-(x+y))!}} $$ Simplify the expression: $$ \operatorname{Pr}\\{X=x \mid X+Y=x+y\\} = \frac{m!n!(x+y)!}{x!y!(m-x)!(n-y)!(m+n)!} $$

Find the probability mass functions

First, use the fact that X and Y are independent, so \(\operatorname{Pr}\\{X=x, Y=y\\} = \operatorname{Pr}\\{X=x\\} \cdot \operatorname{Pr}\\{Y=y\\}\). Next, use the property of conditional probabilities: \(\operatorname{Pr}\\{X=x \mid X+Y=x+y\\} = \frac{\operatorname{Pr}\\{X=x, Y=y\\}}{\operatorname{Pr}\\{X+Y=x+y\\}}\). Plug in the expression of independence and the given property: $$ \frac{\operatorname{Pr}\\{X=x\\} \cdot \operatorname{Pr}\\{Y=y\\}}{\operatorname{Pr}\\{X+Y=x+y\\}} = \frac{m!n!(x+y)!}{x!y!(m-x)!(n-y)!(m+n)!} $$ Now, we can solve this equation for the probability mass functions of X and Y: $$ \operatorname{Pr}\\{X=x\\} = \frac{m!(x+y)!}{x!(m-x)!(m+n)!} \cdot \operatorname{Pr}\\{X+Y=x+y\\} $$ $$ \operatorname{Pr}\\{Y=y\\} = \frac{n!(x+y)!}{y!(n-y)!(m+n)!} \cdot \operatorname{Pr}\\{X+Y=x+y\\} $$

Show that X and Y have binomial distributions with parameter p

Observe that the expressions for \(\operatorname{Pr}\\{X=x\\}\) and \(\operatorname{Pr}\\{Y=y\\}\) have the same form as binomial distributions: $$ \operatorname{Pr}\\{X=x\\} = \binom{m}{x} p^x (1-p)^{m-x} $$ $$ \operatorname{Pr}\\{Y=y\\} = \binom{n}{y} p^y (1-p)^{n-y} $$ It's clear that both X and Y have binomial distributions, but we need to show that they have the same parameter p. From the expressions for \(\operatorname{Pr}\\{X=x\\}\) and \(\operatorname{Pr}\\{Y=y\\}\), we can observe that the terms \(\frac{(x+y)!}{(m+n)!}\) are both common. Moreover, we are given that the probabilities \(\operatorname{Pr}\\{X=0\\}\) and \(\operatorname{Pr}\\{Y=0\\}\) are strictly positive, which means that \(p \neq 0\) and \(1-p \neq 0\). Hence, we can conclude that X and Y have binomial distributions with the same parameter p.

Key concepts

Probability Mass Function

A probability mass function (PMF) characterizes the distribution of a discrete random variable and gives the probability that a random variable is exactly equal to some value. For a binomial distribution, the PMF is defined for a specific number of successes out of a fixed number of independent trials, each with the same probability of success.

For example, if you have a fair coin (with a 50% chance of landing on heads) and you flip it three times, the PMF would give you the probability of getting exactly 2 heads. In mathematical terms, if the random variable \(X\) represents the number of heads, its PMF is given by:
\[\operatorname{Pr}\{X = x\} = \binom{n}{x} p^x (1-p)^{n-x}\]
where \(\binom{n}{x}\) is the binomial coefficient representing the number of ways to choose \(x\) successes out of \(n\) trials, \(p\) is the probability of success on a single trial, and \((1-p)\) is the probability of failure.

Conditional Probability

Conditional probability is the probability of an event occurring, given that another event has already occurred. This concept is fundamental when dealing with dependent events in probability. The notation \(\operatorname{Pr}\{A \mid B\}\) signifies the probability of event \(A\) given event \(B\).

Continuing with the coin example, if you know that at least one flip resulted in heads, the conditional probability might be used to determine the likelihood that exactly two flips were heads out of three, assuming you know at least one was heads. If \(B\) is the event of getting at least one head and \(A\) is the event of getting exactly two heads, then:\[\operatorname{Pr}\{A \mid B\} = \frac{\operatorname{Pr}\{A \cap B\}}{\operatorname{Pr}\{B\}}\]
In problems involving independence, such as the example with independent random variables \(X\) and \(Y\), conditional probabilities can often simplify calculations and reasoning.

Independent Random Variables

Independent random variables are pivotal in probability theory because the outcome of one does not affect the outcome of the other. This means that knowing the value of one variable gives no information about the other. As a result, the probability of their joint occurrence is the product of their individual probabilities.

For two independent random variables \(X\) and \(Y\), the probability that \(X\) equals \(x\) and \(Y\) equals \(y\) can be expressed as:\[\operatorname{Pr}\{X=x, Y=y\} = \operatorname{Pr}\{X=x\} \cdot \operatorname{Pr}\{Y=y\}\]
This property is crucial when dealing with the distribution of sums of independent variables, like in the given problem where \(X+Y=x+y\). Independence simplifies the calculation of the PMF of the sum of the two variables.

Binomial Coefficient

The binomial coefficient is a term used in combinatorics to indicate the number of ways to choose a subset of items from a larger set, regardless of order. It's often denoted as \(\binom{n}{k}\), read as 'n choose k'. This concept features prominently in binomial distributions, where it represents the number of possible outcomes with a specific number of successes.

The binomial coefficient is calculated as:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
where \(n!\) is the factorial of \(n\) and represents the product of all positive integers up to \(n\), '!' denotes factorial, and \(0! = 1\) by definition. The binomial coefficient has symmetry properties such as \(\binom{n}{k} = \binom{n}{n-k}\), and it is essential for understanding binomial probabilities and distributions.