Problem 14 - Calculus for Biology and Medicine

Question

Geometric Distribution In Example 2, we tossed a coin repeatedly until the first heads showed up. Assume that the probability of heads is \(p\), where \(p \in(0,1)\). Let \(Y\) be a random variable that counts the number of trials until the first heads shows up. (a) Show that \(P(Y=1)=p, P(Y=2)=(1-p) p\), and \(P(Y=3)=(1-p)^{2} p\) (b) Explain why $$ P(Y=j)=(1-p)^{j-1} p $$ for \(j=1,2, \ldots\) This equation is called the geometric distribution. (c) Prove that $$ \sum_{j \geq 1} P(Y=j)=1 $$ as follows: (i) For \(0 \leq q<1\), define $$ S_{n}=1+q+q^{2}+\cdots+q^{n} $$ Show that $$ S_{n}-q S_{n}=1-q^{n+1} $$ and conclude from this equation that $$ S_{n}=\frac{1-q^{n+1}}{1-q} $$ (ii) Show that $$ P(Y \leq k)=\sum_{j=1}^{k} P(Y=j)=p \sum_{j=1}^{k}(1-p)^{j-1} $$ Use your results in (i) to show that this formula simplifies to $$ 1-(1-p)^{k} $$ and conclude from this equation that $$ \lim _{k \rightarrow \infty} P(Y \leq k)=1 $$ which is equivalent to $$ \sum_{j \geq 1} P(Y=j)=1 $$

Step-by-Step Solution

Answer

Part (a) shows the pattern of the geometric distribution probabilities. Part (b) derives the formula \( P(Y=j) = (1-p)^{j-1} p \). Part (c) proves \( \sum P(Y=j) = 1 \) using series manipulation.

Step 1: Understanding Geometric Probabilities

The probability \( P(Y=1) \) is the probability of getting heads on the first trial, which is simply \( p \) because the coin toss is independent. For \( P(Y=2) \), heads appear on the second trial: the first trial was tails (\( 1-p \)) and the second trial was heads (\( p \)). Thus, \( P(Y=2) = (1-p)p \). Similarly, for \( P(Y=3) \), tails appear in the first two trials (both having probability \( 1-p \)) and heads on the third (\( p \)), so \( P(Y=3) = (1-p)^2 p \).

Step 2: Formula for the Geometric Distribution

The pattern from Step 1 shows that for any \( j \), to get heads on the \( j^{th} \) trial, there must be \( j-1 \) tails first. Thus, \( P(Y=j) = (1-p)^{j-1} p \). This formula fits the observations from part (a).

Step 3: Define and Manipulate the Series Sum

We define \( S_{n} = 1 + q + q^2 + \, \ldots \, + q^n \). To manipulate this, multiply the series by \( q \) to get \( qS_{n} = q + q^2 + \, \ldots \, + q^{n+1} \). Then, subtract \( qS_{n} \) from \( S_{n} \) to find that \( S_{n} - qS_{n} = 1 - q^{n+1} \). Simplifying, we find \( S_{n} = \frac{1 - q^{n+1}}{1 - q} \).

Step 4: Sum of Probabilities for Geometric Distribution

Using the result from Step 3 with \( q = 1-p \), we can compute \( S_k = \sum_{j=1}^{k} (1-p)^{j-1} = \frac{1 - (1-p)^k}{p} \). Thus, \( P(Y \leq k) = pS_k = 1 - (1-p)^k \). To find the limit as \( k \rightarrow \infty \), note that \( (1-p)^k \rightarrow 0 \) because \( 0 < 1-p < 1 \). Hence, \( \lim_{k \rightarrow \infty} P(Y \leq k) = 1 \), confirming that \( \sum_{j \geq 1} P(Y = j) = 1 \).

Key Concepts

Probability

Probability is the measure of how likely an event is to occur. In a simple coin toss, the probability of getting heads is denoted by \( p \), where \( 0 < p < 1 \). Each side of the coin is equally likely to occur in a fair toss.

For example, if \( p = 0.5 \), there is an equal probability of getting heads or tails.

In the context of a geometric distribution, probability helps us find the likelihood of a first success on the \( j^{th} \) trial, like getting the first heads after a number of tosses.

We can calculate these probabilities using the general formula \( P(Y=j) = (1-p)^{j-1}p \), where \( (1-p)^{j-1} \) represents the probability of getting \( j-1 \) tails before a head is finally observed.

This approach emphasizes the likelihood of independent, successive events leading up to a single successful outcome.

Random Variable

A random variable is a numerical representation of the outcomes of a random process. In the problem of geometric distribution, the random variable \( Y \) counts how many trials are needed to get the first 'heads'.

\( Y \) is known as a discrete random variable because it can take on countable values, such as 1, 2, 3, and so forth, depending on the number of coin tosses.

It's important to differentiate between the random variable itself, which represents the process, and the probability distribution, which describes the probabilities associated with each potential outcome.

Understanding how \( Y \) functions helps us conceptualize the problem at hand, recognizing that each value of \( Y \) corresponds to a specific probability, like \( P(Y=2) = (1-p)p \), where the second toss gives us our first head.

Series Sum

The concept of a series sum becomes crucial when analyzing the probabilities in a geometric distribution. In mathematics, a series is the sum of the terms of a sequence.

To find the total probability of all possible outcomes, we often need to calculate a sum of a geometric series. The formula \( S_{n} = 1 + q + q^2 + \ldots + q^n \) helps us accomplish this by managing series with a common ratio \( q \).

In our problem, specifically, \( q \) represents \( 1-p \) and is used for calculating the sum of probabilities:

• Start with \( S_{n} \), the sum of powers of \( q \).
• Subtract a shifted series: \( S_{n} - q S_{n} = 1 - q^{n+1} \).
• Simplify to get \( S_{n} = \frac{1-q^{n+1}}{1-q} \).

This series sum reveals how the probabilities sum to 1 as the terms extend infinitely.

Probability Distribution

Probability distribution describes how the probabilities are distributed over different potential outcomes of a random variable. In a geometric distribution, this relates to a series of independent coin flips and getting the first heads.

For each trial \( j \), the probability distribution follows \( P(Y=j) = (1-p)^{j-1} p \). This formula represents the nature of a geometric distribution where the probability of a success decreases exponentially with each additional failure before success.

To ensure the probabilities are valid, they must sum to 1, meaning:

• Each potential outcome contributes a part to the whole probability.
• Using the series sum techniques, we see \( \lim_{k \rightarrow \infty} P(Y \leq k) = 1 \).

This implies that if you continue flipping the coin indefinitely, you are guaranteed to eventually get heads, hence the sum of all probabilities is 1.

Understanding a probability distribution assists in grasping the long-term behavior of random processes, which is critical in fields such as statistics and risk assessment.