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Consider \(n\) independent trials, each of which results in one of the outcomes \(1, \ldots, k\) with respective probabilities \(p_{1}, \ldots, p_{k}, \quad \sum_{i=1}^{k} p_{i}=1 .\) Show that if all the \(p_{i}\) are small, then the probability that no trial outcome occurs more than once is approximately equal to \(\exp (-n(n-1)\) \(\left.\sum_{i} p_{i}^{2} / 2\right)\).

Short Answer

Expert verified
In this problem, we are asked to find the probability of no trial outcome occurring more than once when all the \(p_i\) values are small. We calculated this probability directly using combinations and the binomial approximation for large values. We then approximated the probability by using a Taylor expansion and factoring out common terms. Finally, we obtained the approximate probability as \(P \approx \exp\left(-n(n-1)\sum_{i=1}^k p_i^2 / 2\right)\), which confirms the result given in the question.

Step by step solution

01

Calculate the probability of no trial outcome occurring more than once

To find the probability of each outcome occurring only once, we will use combinations and multiply the probabilities of each outcome. Let \(N_i\) denote the number of occurrences of the outcome \(i\). Since we want no trial outcome to occur more than once, the only possible combinations of \(N_i\) are \(N_i=0\) or \(N_i=1\). For each outcome \(i\), there are \(C(n,N_i)\) different ways of picking \(N_i\) out of \(n\) trials. So, given a combination of \(N_i\), the probability of that combination occurring is, \[P(N_i) = {C(n,N_i)}p_i^1(1-p_i)^{n-1}\] We need to sum this over all possible combinations of \(N_i\).
02

Simplify the probability expression under the assumption of small \(p_i\)

Since we are asked to show this under the assumption of small \(p_i\), we will replace all the combinations \(C(n,N_i)\) by approximate values, knowing that when \(p_i\) is small, \(n\) is large, and \(n-N_i\) is also large. Using the binomial approximation for large values, we have \[C(n,N_i)\approx\frac{n^1}{1!} \quad\text{for}\quad N_i=1.\] Now, let's simplify the probability expression: \[P(N_i) \approx \frac{n^1}{1!}p_i^1(1-p_i)^{n-1}\] And for all possible combinations of \(N_i\), \[P = \sum_{i=1}^k P(N_i).\]
03

Approximate the probability further using Taylor expansion

To approximate the expression even further, we can use a Taylor expansion for the term \((1-p_i)^{n-1}\): \[(1-p_i)^{n-1}\approx e^{-(n-1)p_i},\] since for small values of \(p_i\), we have \(1 + x \approx e^x\). Replacing this approximation back into the probability expression, we get \[P(N_i) \approx \frac{n^1}{1!}p_i^1e^{-(n-1)p_i}\] Now, summing over all possible combinations of \(N_i\): \[P = \sum_{i=1}^k \frac{n^1}{1!}p_i^1e^{-(n-1)p_i}\]
04

Factor out common terms and simplify the summation expression

Now, we can factor out the common term \(ne^{-n+1}\) from the summation: \[P = ne^{-n+1}\sum_{i=1}^k p_i e^{-np_i}\] Now, we'll expand the exponential term in summation using Taylor series up to the first two terms: \[e^{-np_i} \approx 1 - np_i\] Plugging it back to our probability expression: \[P \approx ne^{-n+1} \sum_{i=1}^k p_i (1 - np_i)\] \[P \approx ne^{-n+1}\left(\sum_{i=1}^k p_i - n\sum_{i=1}^k p_i^2 \right)\] Since \(\sum_{i=1}^k p_i = 1\): \[P \approx ne^{-n+1}\left(1 - n\sum_{i=1}^k p_i^2 \right)\]
05

Calculate the approximate probability as given in the exercise

Finally, let's rewrite our probability expression in the form mentioned in the exercise: \[P \approx e^{-n+1}e^{-n(n-1)\sum_{i=1}^k p_i^2}\] \[P \approx \exp\left(-n(n-1)\sum_{i=1}^k p_i^2 / 2\right)\] This completes our derivation of the approximate probability expression of no trial outcome occurring more than once when all \(p_i\) values are small.

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Key Concepts

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

Independent Trials
In probability theory, when we talk about independent trials, we're referring to a sequence of experiments or tests where the outcome of one trial does not affect the others. Imagine tossing a fair coin multiple times; each flip does not alter the chance of the next flip landing on heads or tails. This concept is crucial because it allows us to multiply individual probabilities to find the joint probability of multiple outcomes.
In the exercise above, each trial results in one of the outcomes labeled from 1 to k, with probabilities designated as \( p_i \). Since the trials are independent, the occurrence of any specific outcome in a trial is unaffected by the results of the previous trials.
This independence forms the backbone of calculating compounded probabilities, particularly when we aim to determine the likelihood of certain patterns occurring across multiple trials.
Probabilities
Probabilities represent the likelihood of different outcomes in a trial. They are expressed as values ranging from 0 to 1, where 0 signifies an impossible event, and 1 indicates a certain event. In the context of independent trials, each outcome has an associated probability \( p_i \), and their sum over all possible outcomes must equal 1: \( \sum_{i=1}^{k} p_{i} = 1 \).
In our exercise, probabilities are assigned to each outcome such that no outcome occurs more than once. The expression \( P(N_i) = C(n,N_i)p_i(1-p_i)^{n-1} \) helps compute the probability of any given outcome exactly once in n trials.
When probabilities \( p_i \) are small, it's generally due to a large number of possible outcomes (like a rare event in rolls of a die), and this becomes significant for approximations like the exponential approximation mentioned in the exercise.
Exponential Approximation
The exponential approximation is a common technique in probability when dealing with small probability values. It's based on the expression \((1-p_i)^{n-1}\), which is complex to manage directly. By applying an exponential approximation, we simplify this to \(e^{-(n-1)p_i}\), helping us handle these calculations more effectively.
This technique is particularly useful when \( p_i \) is small, making it feasible to replace \( (1 + x) \) with \( e^x \) for simplicity. In practice, this means that our complex binomial expressions converge to a simpler, more workable format or model that remains reasonably accurate, especially as a first-order approximation.
Taylor Expansion
Taylor expansion breaks down complex mathematical functions into simpler terms, essentially polynomials that can approximate values like exponentials more easily. In this context, we're using Taylor expansion to approximate exponential terms like \( e^{-np_i} \). For small \( p_i \), this translates to \( 1 - np_i \), which simplifes matters greatly.
When we have small probabilities involved, the Taylor expansion allows us to express functions in terms of their series representation, providing a practical and sufficiently accurate approximation.
  • This approximation streamlines calculations significantly, making what would otherwise be intricate computations straightforward enough to manage—particularly important in exercises involving probabilities and exponential functions.
  • It transforms series with countless terms into workable expressions by limiting to just the first few, often the leading terms in the series, which is usually enough to capture the essence of the function's behavior for small inputs.

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Most popular questions from this chapter

Each night different meteorologists give us the probability that it will rain the next day. To judge how well these people predict, we will score each of them as follows: If a meteorologist says that it will rain with probability \(p,\) then he or she will receive a score of \(1-(1-p)^{2} \quad\) if it does rain \(1-p^{2}\) if it does not rain We will then keep track of scores over a certain time span and conclude that the meteorologist with the highest average score is the best predictor of weather. Suppose now that a given meteorologist is aware of our scoring mechanism and wants to maximize his or her expected score. If this person truly believes that it will rain tomorrow with probability \(p^{*},\) what value of \(p\) should he or she assert so as to maximize the expected score?

Suppose that a biased coin that lands on heads with probability \(p\) is flipped 10 times. Given that a total of 6 heads results, find the conditional probability that the first 3 outcomes are (a) \(h, t, t\) (meaning that the first flip results in heads, the second in tails, and the third in tails); (b) \(t, h, t\)

\(A\) and \(B\) play the following game: \(A\) writes down either number 1 or number \(2,\) and \(B\) must guess which one. If the number that \(A\) has written down is \(i\) and \(B\) has guessed correctly, \(B\) receives \(i\) units from \(A\). If \(B\) makes a wrong guess, \(B\) pays \(\frac{3}{4}\) unit to \(A .\) If \(B\) randomizes his decision by guessing 1 with probability \(p\) and 2 with probability \(1-p,\) determine his expected gain if (a) \(A\) has written down number 1 and (b) \(A\) has written down number 2 What value of \(p\) maximizes the minimum possible value of \(B\) 's expected gain, and what is this maximin value? (Note that \(B\) 's expected gain depends not only on \(p,\) but also on what \(A\) does.) Consider now player \(A\). Suppose that she also randomizes her decision, writing down number 1 with probability q. What is \(A\) 's expected loss if (c) \(B\) chooses number 1 and (d) \(B\) chooses number \(2 ?\) What value of \(q\) minimizes \(A\) 's maximum expected loss? Show that the minimum of \(A\) 's maximum expected loss is equal to the maximum of \(B\) 's minimum expected gain. This result, known as the minimax theorem, was first established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. The common value is called the value of the game to player \(B\).

A person tosses a fair coin until a tail appears for the first time. If the tail appears on the \(n\) th flip, the person wins \(2^{n}\) dollars. Let \(X\) denote the player's winnings. Show that \(E[X]=+\infty .\) This problem is known as the St. Petersburg paradox. (a) Would you be willing to pay \(\$ 1\) million to play this game once? (b) Would you be willing to pay \(\$ 1\) million for each game if you could play for as long as you liked and only had to settle up when you stopped playing?

Suppose that a die is rolled twice. What are the possible values that the following random variables can take on: (a) the maximum value to appear in the two rolls; (b) the minimum value to appear in the two rolls; (c) the sum of the two rolls; (d) the value of the first roll minus the value of the second roll?

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