Problem 94 - Calculus for Biology and Medicine

Question

Suppose \(X\) and \(Y\) are independent and Poisson with mean \(\lambda\). Given that \(X+Y=n\), find the probability that \(X=k\) for \(k=0,1,2, \ldots, n\)

Step-by-Step Solution

Answer

The probability is \( \frac{1}{2^n} \binom{n}{k} \).

Step 1: Recognize the Conditional Probability

We are tasked with finding the probability that \( X = k \) given that \( X + Y = n \). This setup indicates that we're looking for a conditional probability: \[P(X = k \mid X + Y = n)\]

Step 2: Apply Conditional Probability Formula

The conditional probability formula is given by: \[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]Here, \( A \) is the event \( X = k \) and \( B \) is the event \( X + Y = n \). Therefore: \[P(X = k \mid X + Y = n) = \frac{P(X = k \cap X + Y = n)}{P(X + Y = n)}\]

Step 3: Express Joint Probability

We know that the event \( X = k \cap X + Y = n \) is equivalent to \( X = k \) and \( Y = n - k \). Since \( X \) and \( Y \) are independent Poisson random variables, the joint probability is the product of their individual probabilities:\[P(X = k) \cdot P(Y = n - k)\]

Step 4: Evaluate Poisson Probabilities

The probability that a Poisson random variable with mean \( \lambda \) takes the value \( k \) is:\[P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\]Thus, we have:\[P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \quad \text{and} \quad P(Y = n-k) = \frac{\lambda^{n-k} e^{-\lambda}}{(n-k)!}\]

Step 5: Complete Joint Probability

Substitute the Poisson probabilities into the expression for the joint probability:\[P(X = k \cap X + Y = n) = \frac{\lambda^k e^{-\lambda}}{k!} \cdot \frac{\lambda^{n-k} e^{-\lambda}}{(n-k)!} = \frac{\lambda^n e^{-2\lambda}}{k!(n-k)!}\]

Step 6: Calculate the Denominator

We need to calculate \( P(X + Y = n) \). Since \( X + Y \) is the sum of two independent Poisson random variables both with mean \( \lambda \), \( X + Y \) follows a Poisson distribution with mean \( 2\lambda \):\[P(X + Y = n) = \frac{(2\lambda)^n e^{-2\lambda}}{n!}\]

Step 7: Formulate the Conditional Probability

Insert the joint probability and the denominator into the conditional probability expression:\[P(X = k \mid X + Y = n) = \frac{\frac{\lambda^n e^{-2\lambda}}{k!(n-k)!}}{\frac{(2\lambda)^n e^{-2\lambda}}{n!}}\]This simplifies to:\[P(X = k \mid X + Y = n) = \frac{\binom{n}{k} \lambda^k \lambda^{n-k}}{(2\lambda)^n}\]Which further reduces to:\[P(X = k \mid X + Y = n) = \frac{1}{2^n} \binom{n}{k}\]

Step 8: Interpret the Result

The result \( P(X = k \mid X + Y = n) = \frac{1}{2^n} \binom{n}{k} \) represents the probability of a Binomial distribution with parameters \( n \) and \( p = \frac{1}{2} \). This fits the scenario since \( X \) and \( Y \) have equal contributions to the total sum.

Key Concepts

conditional probability

Conditional probability is a fascinating concept in probability theory that deals with finding the probability of an event occurring given that another event has already happened. In the context of the problem, we are interested in finding the probability of the event that the random variable \(X\) equals \(k\), given that the sum of \(X+Y\) equals \(n\). This is expressed using conditional probability notation as \(P(X = k \mid X + Y = n)\).

The formula for conditional probability is:

• \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)

Here, \(A\) is the event that \(X = k\), and \(B\) is the event that \(X + Y = n\). This formula tells us that the conditional probability is found by dividing the joint probability of \(A\) and \(B\) happening together by the probability of \(B\) occurring. It's like finding a tiny part of the likelihood pie that's already been sliced based on the known condition event \(B\).

Using this probability concept is quite handy, especially in scenarios where events are not entirely independent and some conditions are already known, helping to refine our prediction of outcomes.

independent random variables

In probability, two random variables are independent if the occurrence of one does not impact the occurrence of the other. This independence simplifies many calculations and is crucial to the method we use in the original exercise. The problem gives us \(X\) and \(Y\) which are independent Poisson random variables. This independence is key when calculating the joint probability \(P(X = k \cap X + Y = n)\).

Because \(X\) and \(Y\) are independent and each follows a Poisson distribution with the same mean \(\lambda\), we can calculate this joint probability as the product of their individual probabilities:

• \(P(X = k \cap Y = n-k) = P(X=k) \cdot P(Y=n-k)\)

Understanding independence allows us to separate the variables and calculate probabilities of these individual outcomes more straightforwardly. When dealing with independent random variables, always look for opportunities to break down complex joint probabilities into simpler calculations.

binomial distribution

The binomial distribution is another fundamental concept in probability, often used to model the number of successes in a fixed number of trials in an experiment where each trial has two possible outcomes. Such experiments are known as Bernoulli trials. Based on the exercise, we deduced that this distribution comes into play characterized by parameters \( n \) and \( p \), where \( n \) is the number of trials, and \( p \) is the probability of success in each trial.

In the solution, we found that the conditional probability \(P(X = k \mid X + Y = n)\) simplifies to a binomial distribution, specifically:

• \(P(X = k \mid X + Y = n) = \frac{1}{2^n} \binom{n}{k}\)

This result reveals that given \(X+Y=n\), \(X\) behaves like a binomial random variable with parameters \(n\) and \(p = \frac{1}{2}\). The binomial coefficient \(\binom{n}{k}\) represents the number of ways \(k\) successes can occur in \(n\) trials. This simplification beautifully ties together the scenarios, showing that a sum of two independent Poisson with equal parameters distributes binomially under such a condition. But mainly, checking for a binomial distribution under certain summed scenarios can provide elegance and simplicity to a solution.