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Chapter 3: Problem 97

Use the conditional variance formula to find the variance of a geometric random variable.

Short Answer

Expert verified
The variance of a geometric random variable with probability of success \(p\) can be found using the conditional variance formula. First, find the expected value \(E(X) = \frac{1}{p}\) and the second moment \(E(X^2) = \frac{2}{p^2}\). Then, use the variance formula: \(Var(X) = E(X^2) - [E(X)]^2\), which gives \(Var(X) = \frac{1}{p^2}\).

Step by step solution

01

1. Understanding Geometric Random Variables and Variance Formula

A geometric random variable, denoted as \(X\), represents the number of trials needed to get the first success in independent Bernoulli trials. Let \(p\) be the probability of success in each trial. The probability mass function (PMF) of a geometric random variable is given by: \[ P(X = k) = (1 - p)^{k-1} p \] where \(k\) is the number of trials. The variance formula for a random variable is given by: \[ Var(X) = E(X^2) - [E(X)]^2 \] Our goal is to derive the expressions for the expected value \(E(X)\) and the second moment \(E(X^2)\), and then use the formula to find the variance.
02

2. Find the Expected Value (E(X)) of a Geometric Random Variable

The expected value of a geometric random variable can be found with: \[ E(X) = \sum_{k=1}^{\infty} kP(X = k) \] Substituting the PMF in the formula above, we get: \[ E(X) = \sum_{k=1}^{\infty} k (1 - p)^{k-1} p \] Let's evaluate the sum: \[E(X) = p\sum_{k=1}^\infty k(1-p)^{k-1}\] \[E(X) = p(1-p)^0 + 2p(1-p)^1 + 3p(1-p)^2 + \cdots\] Now, let's consider \((1 - p)E(X)\): \[(1 - p) E(X) = 0 + p(1-p)^0 + 2p(1-p)^1 + 3p(1-p)^2 + \cdots\] We can now subtract \((1 - p)E(X)\) from \(E(X)\) to obtain: \[pE(X) = p + (1 - p)^1 + (1-p)^2 + (1-p)^3 + \cdots\] This is the geometric series with a common ratio of \((1 - p)\): \[pE(X) = \frac{p}{1 - (1 - p)} = \frac{p}{p} = 1\] Therefore, \(E(X) = \frac{1}{p}\).
03

3. Find the Second Moment (E(X^2)) of a Geometric Random Variable

To find the second moment, we consider: \[E(X^2) = \sum_{k=1}^{\infty} k^2(1 - p)^{k-1} p \] This is slightly trickier to obtain directly. However, we can use the technique of differentiation. Recall that for \(|r| < 1\), \[\sum_{k=0}^{\infty} r^k = \frac{1}{1 - r} \] Now, differentiate both sides of the above equation with respect to r: \[\sum_{k=1}^{\infty} kr^{k-1} = \frac{1}{(1 - r)^2} \] Differentiate once more: \[\sum_{k=2}^{\infty} k(k-1)r^{k-2} = \frac{2}{(1 - r)^3} \] Substitute \(r = 1 - p\), and multiply both sides by \(p^2\): \[E(X^2) = \frac{2p^2}{(2p)^3} = \frac{2}{p^2}\]
04

4. Calculate the Variance of a Geometric Random Variable

Now that we have the expected value and the second moment, we can compute the variance of a geometric random variable using the formula: \[ Var(X) = E(X^2) - [E(X)]^2 \] Substitute the values: \[ Var(X) = \frac{2}{p^2} - \left(\frac{1}{p}\right)^2 \] Finally, \[ Var(X) = \frac{2}{p^2} - \frac{1}{p^2} = \frac{1}{p^2} \] So, the variance of a geometric random variable is \(\frac{1}{p^2}\).

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Key Concepts

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

Conditional Variance
Conditional variance is a concept used in probability and statistics to describe how much a random variable deviates from its expected value, given certain conditions. For a geometric random variable, we are interested in understanding the variance based on specific conditions that are dependent on other variables, such as previous outcomes.The formula for variance, \[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]is key in finding the conditional variance because it helps us understand how the expected outcome changes when new information is given. In essence, the conditional variance provides deeper insights into the variability of the data when influenced by changing conditions or additional information.In this problem, we are particularly looking at the variance using the geometric distribution's properties to establish this variance under certain conditions.
Expected Value
The expected value, often denoted as \(E(X)\), is a fundamental concept in probability that represents the average outcome of a random variable. For a geometric random variable, which is defined as the number of trials to achieve the first success, the expected value gives us an intuition about the average number of trials needed to succeed. In mathematical terms, the expected value of a geometric random variable \(X\) with probability \(p\) of success is represented by:\[ E(X) = \frac{1}{p} \]This formula stems from the probability mass function of the geometric distribution and tells us that, on average, if the success probability in each trial is \(p\), it will take \(\frac{1}{p}\) trials to get one success.
Probability Mass Function
The probability mass function (PMF) is a crucial tool when dealing with discrete random variables like a geometric random variable. It provides the probability that a discrete random variable is exactly equal to some value. For geometric random variables, the PMF is given by:\[ P(X = k) = (1 - p)^{k-1} p \]Here, \(k\) represents the number of trials needed to get the first success, and \(p\) is the probability of success on each trial. This formula reflects how the probability decreases exponentially with the number of trials until the first success, making it a quintessential geometric property. Understanding the PMF helps in evaluating both the expected value and variance of the geometric distribution.
Independent Bernoulli Trials
Independent Bernoulli trials are the foundation of the geometric distribution. In these trials, each trial is an independent event resulting in either a success or a failure, with a constant probability \(p\) of success and \(1-p\) of failure. This concept of independence is crucial as it ensures that the outcome of one trial does not influence another, which is a major assumption in the geometric distribution. When calculating the variance or expected value of a geometric random variable, it is essential to remember that these are built on the idea of independent Bernoulli trials, where the trials continue until the first success is observed. Keeping the trials independent assures that the mathematical properties derived from them, such as PMF, expected value, and variance, hold true consistently across calculations.

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Most popular questions from this chapter

Independent trials, resulting in one of the outcomes \(1,2,3\) with respective probabilities \(p_{1}, p_{2}, p_{3}, \sum_{i=1}^{3} p_{i}=1\), are performed. (a) Let \(N\) denote the number of trials needed until the initial outcome has occurred exactly 3 times. For instance, if the trial results are \(3,2,1,2,3,2,3\) then \(N=7\) Find \(E[N]\). (b) Find the expected number of trials needed until both outcome 1 and outcome 2 have occurred.

A deck of \(n\) cards, numbered 1 through \(n\), is randomly shuffled so that all \(n !\) possible permutations are equally likely. The cards are then turned over one at a time until card number 1 appears. These upturned cards constitute the first cycle. We now determine (by looking at the upturned cards) the lowest numbered card that has not yet appeared, and we continue to turn the cards face up until that card appears. This new set of cards represents the second cycle. We again determine the lowest numbered of the remaining cards and turn the cards until it appears, and so on until all cards have been turned over. Let \(m_{n}\) denote the mean number of cycles. (a) Derive a recursive formula for \(m_{n}\) in terms of \(m_{k}, k=1, \ldots, n-1\). (b) Starting with \(m_{0}=0\), use the recursion to find \(m_{1}, m_{2}, m_{3}\), and \(m_{4}\). (c) Conjecture a general formula for \(m_{n}\). (d) Prove your formula by induction on \(n\). That is, show it is valid for \(n=1\), then assume it is true for any of the values \(1, \ldots, n-1\) and show that this implies it is true for \(n\). (e) Let \(X_{i}\) equal 1 if one of the cycles ends with card \(i\), and let it equal 0 otherwise, \(i=1, \ldots, n\). Express the number of cycles in terms of these \(X_{i}\). (f) Use the representation in part (e) to determine \(m_{n}\). (g) Are the random variables \(X_{1}, \ldots, X_{n}\) independent? Explain. (h) Find the variance of the number of cycles.

Polya's urn model supposes that an urn initially contains \(r\) red and \(b\) blue balls. At each stage a ball is randomly selected from the urn and is then returned along with \(m\) other balls of the same color. Let \(X_{k}\) be the number of red balls drawn in the first \(k\) selections. (a) Find \(E\left[X_{1}\right]\) (b) Find \(E\left[X_{2}\right]\). (c) Find \(E\left[X_{3}\right]\). (d) Conjecture the value of \(E\left[X_{k}\right]\), and then verify your conjecture by a conditioning argument. (e) Give an intuitive proof for your conjecture. Hint: Number the initial \(r\) red and \(b\) blue balls, so the urn contains one type \(i\) red ball, for each \(i=1, \ldots, r ;\) as well as one type \(j\) blue ball, for each \(j=1, \ldots, b\). Now suppose that whenever a red ball is chosen it is returned along with \(m\) others of the same type, and similarly whenever a blue ball is chosen it is returned along with \(m\) others of the same type. Now, use a symmetry argument to determine the probability that any given selection is red.

Two players take turns shooting at a target, with each shot by player \(i\) hitting the target with probability \(p_{i}, i=1,2\). Shooting ends when two consecutive shots hit the target. Let \(\mu_{i}\) denote the mean number of shots taken when player \(i\) shoots first, \(i=1,2\) (a) Find \(\mu_{1}\) and \(\mu_{2}\). (b) Let \(h_{i}\) denote the mean number of times that the target is hit when player \(i\) shoots first, \(i=1,2\). Find \(h_{1}\) and \(h_{2}\).

\(A\) and \(B\) roll a pair of dice in turn, with \(A\) rolling first. A's objective is to obtain a sum of 6 , and \(B\) 's is to obtain a sum of 7 . The game ends when either player reaches his or her objective, and that player is declared the winner. (a) Find the probability that \(A\) is the winner. (b) Find the expected number of rolls of the dice. (c) Find the variance of the number of rolls of the dice.
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