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Chapter 3: Problem 101

a. We observe a sequence of independent identical trials with two possible outcomes on each trial, \(S\) and \(F,\) and with \(P(S)=p .\) The number of the trial on which we observe the fifth success, \(Y\), has a negative binomial distribution with parameters \(r=5\) and \(p\). Suppose that we observe the fifth success on the eleventh trial. Find the value of \(p\) that maximizes \(P(Y=11\) ). b. Generalize the result from part (a) to find the value of \(p\) that maximizes \(P\left(Y=y_{0}\right)\) when \(Y\) has a negative binomial distribution with parameters \(r\) (known) and \(p\)

Short Answer

Expert verified
The value of \( p \) that maximizes \( P(Y=11) \) is \( \frac{5}{11} \). The general maximizer for \( P(Y=y_0) \) is \( \frac{r}{y_0} \).

Step by step solution

01

Understanding the Problem

The problem involves finding the probability that the fifth success occurs on the 11th trial in a sequence of trials with two possible outcomes. This probability follows a negative binomial distribution with parameters \( r = 5 \) (number of successes) and \( p \) (probability of success). We need to find the value of \( p \) that maximizes \( P(Y=11) \).
02

Probability of Y = 11 for a Negative Binomial Distribution

For a negative binomial distribution, the probability of having the \( r \)-th success on the \( y \)-th trial can be given by the formula: \[ P(Y=y) = \binom{y-1}{r-1} p^{r} (1-p)^{y-r} \].For \( Y = 11 \), \( r = 5 \), this becomes: \[ P(Y=11) = \binom{10}{4} p^{5} (1-p)^{6} \].
03

Identify the Expression to Maximize

We need to find the value of \( p \) that maximizes \[ P(Y=11) = \binom{10}{4} p^{5} (1-p)^{6} \].This requires taking the derivative of this probability function with respect to \( p \) and finding the critical points.
04

Differentiate the Expression with Respect to p

To find the value of \( p \) that maximizes this probability, differentiate \[ P(Y=11) = \binom{10}{4} p^{5} (1-p)^{6} \]with respect to \( p \). Use the product rule to differentiate:\[ \frac{d}{dp} \left( p^5 (1-p)^6 \right) = p^5 \cdot \frac{d}{dp} (1-p)^6 + (1-p)^6 \cdot \frac{d}{dp} p^5 \].Then find when this derivative equals zero.
05

Simplify and Solve for Critical Points

Set the derivative equal to zero to find critical points:\[ 5p^4(1-p)^6 - 6p^5(1-p)^5 = 0 \].This simplifies to \[ p(5(1-p) - 6p) = 0 \].Solving \( 5(1-p) - 6p = 0 \) gives \( 5 - 11p = 0 \) or \( p = \frac{5}{11} \).
06

Verify Maximization Condition

Check the second derivative or examine the behavior around \( p = \frac{5}{11} \) to ensure that it is a maximum. In this case, testing nearby points in the derivative shows \( p = \frac{5}{11} \) indeed maximizes \( P(Y=11) \).
07

Generalize the Maximization Result

To generalize for any \( y_0 \), the probability becomes \[ P(Y=y_0) = \binom{y_0-1}{r-1} p^r (1-p)^{y_0-r} \].Following similar differentiation, the critical point remains \( p = \frac{r}{y_0} \), providing a generalized formula for maximizing \( P(Y = y_0) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Maximization Problem
In the context of a negative binomial distribution, a maximization problem involves finding the probability of success rate \( p \) that optimizes a particular outcome.
Specifically, when dealing with a scenario where you wish to determine the likelihood of achieving the \( r \)-th success on the \( y \)-th trial, using known parameters, you are essentially addressing a maximization problem.
The process generally involves identifying the expression for probability, deriving it with respect to \( p \), and finding values that provide critical points.
By solving for these critical points, you can determine which value of \( p \) results in the highest probability for your given conditions.
For instance, in a problem where you seek to maximize \( P(Y=11) \) with \( r=5 \), using derivative calculus helps evaluate the expression of probability and find the optimal value of \( p \).
This step-by-step approach to maximizing the probability supports understanding how particular configurations of \( p \) influence the outcome within the framework of a negative binomial distribution.
Probability of Success
The probability of success, represented by \( p \), is a key parameter in trials involving binary outcomes.
In any negative binomial distribution, it defines the likelihood of achieving the desired success in repeated trials.
Here, scenarios are depicted where each trial can either result in success \( S \) or failure \( F \).
Understanding \( p \) is crucial, as it directly influences the distribution function and determines the statistical characteristics of the experiment conducted.
In exercises often seen in textbooks, estimating \( p \) particularly matters when exploring the probability associated with achieving success after a specific number of attempts, denoted typically by \( y \).
For maximizing \( P(Y=y) \), finding the right value of \( p \)鈥攅specially when seeking critical points鈥攊s essential since it determines how frequently a specific outcome manifests over a set number of iterations.
  • The value of \( p \) impacts the skewness and variance of the distribution.
  • The likelihood or odds ratios are adjusted proportionally depending on \( p \).
Grasping this concept helps students predict outcomes effectively when given data mimicking real-world processes.
Critical Points
Critical points in the context of mathematical optimization are values that indicate where maxima or minima occur for a function.
In the scope of maximizing a probability function, identifying these points is pivotal.
After deriving the probability function鈥攕uch as \( P(Y=11) \)鈥攚ith respect to the probability of success \( p \), setting the derivative equal to zero enables finding these critical points.
Solving for \( p \) not only reveals the potential peaks but also addresses where changes in inclines or declines within the function happen.
For example, in a problem with \( P(Y=11) \), simplification after this derivation leads to knowing \( p = \frac{5}{11} \) as a critical point.
Such points, when checked using the second derivative test, confirm whether they are indeed points of maximum probability.
Understanding this provides insights on how optimization in a mathematical function serves a broader statistical analysis useful in experiments and repeated trials.

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