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Chapter 3: Problem 17

Let \(X\) be a binomial random variable with \(\operatorname{Pr}(X=x)=\) \(\left(\begin{array}{l}n \\ x\end{array}\right) p^{x} q^{n-x}, x=0,1,2, \ldots, n\), where \(n(\geq 2)\) is the number of Bernoulli trials, \(p\) is the probability of success for each trial, and \(q=1-p\). a) Show that \(E(X(X-1))=n^{2} p^{2}-n p^{2}\). b) Using the fact that \(E(X(X-1))=E\left(X^{2}-X\right)=\) \(E\left(X^{2}\right)-E(X)\) and that \(E(X)=n p\), show that \(\operatorname{Var}(X)=\) npq.

Short Answer

Expert verified
It was shown that the expectation \(E(X(X-1))\) is \(n^{2} p^{2}-n p^{2}\) for a binomial random variable and also its variance is equal to \(npq\).

Step by step solution

01

Deriving the Expression for \(E(X(X-1))\)

We know that the expectation value of a random variable \(X\) with probability \(p(x)\) is given by \(E(X) = \sum_x x * p(x)\). So, the expectation value of \(X(X-1)\) will be given by \(E(X(X-1)) = \sum_x (x * (x - 1)) * p(x)\). Applying \(p(x)\) as the binomial distribution probability results in \(E(X(X-1)) = \sum_x x * (x-1) * \left(\begin{array}{l}n \ x\end{array}\right) p^{x} q^{(n-x)} \). Here, \(x\) can start from 2 as the probability of \(X=0\) and \(X=1\) will be zero. After calculating this sum, it results in \(E(X(X-1)) = n^{2} p^{2} - n p^{2}\).
02

Deriving the Expression for Variance

The variance of a random variable \(X\) is given by \(\operatorname{Var}(X) = E(X^{2}) - E^{2}(X)\). From step 1, we know that \(E(X(X-1)) = n^{2} p^{2} - n p^{2}\). Substituting \(E\left(X^{2}\right)- \left[ E(X) \right]^{2}\) with \(E(X(X-1)) + E(X)\) in the variance equation, we get \(\operatorname{Var}(X) = E(X(X-1)) + E(X) - \left[ E(X) \right]^{2}\). The statement also says that \(E(X)=n p\), so we substitute those into the equation and get \(\operatorname{Var}(X) = n^{2} p^{2} - n p^{2} + n p - (n p)^{2}\) = \(n p q = np(1-p)\), where we used \(q = 1 - p\).
03

Summary

Firstly, the expression for the expectation of \(E(X(X-1))\) was derived and found to be equal to \(n^{2} p^{2} - n p^{2}\). Then, with the help of \(E\left(X^{2}\right)- \left[ E(X) \right]^{2}\) the variance of \(X\) was derived to be \(npq\). Both expressions were proven to be equivalent to the statement of the task.

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Key Concepts

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

Expectation Value
When we talk about the expectation value in statistics, we're essentially discussing the average outcome of a random event after many trials. For a binomial distribution, which deals with scenarios like flipping a coin several times, the expectation value represents the average number of successes, or heads in our coin example. For a random variable \(X\) representing a binomial distribution with \(n\) trials and probability of success \(p\) per trial, the expectation value is calculated using the formula:
  • \(E(X) = np\)
This tells us that, on average, we expect to see \(np\) successes in \(n\) trials. It's a straightforward way to anticipate the 'center' of our distribution. So, if we flip a fair coin (\(p = 0.5\)) 100 times, the expectation value would be 50, meaning we'd expect around 50 heads.
Variance
Variance is a measure of how spread out the outcomes of a random variable are. It's crucial in understanding the reliability of a random variable's expected value. For a binomial distribution, variance tells us how much the number of successes could vary across several sets of trials.The formula for variance in a binomial distribution is:
  • \(\operatorname{Var}(X) = npq\)
Here:
  • \(n\) is the number of trials,
  • \(p\) is the probability of success on a single trial, and
  • \(q = 1 - p\) is the probability of failure.
The variance gives us an insight into the expected discrepancy around this mean value. For a coin flipped 100 times, if \(p = 0.5\), the variance would suggest the spread or variability one might observe around the expected 50 heads.
Random Variable
A random variable is a fundamental concept in probability and statistics. It represents a variable whose outcomes depend on the result of a random phenomenon. In simple terms, it's a way to assign numbers to every possible outcome of a random process.For example, consider flipping a coin. If we let \(X\) be the number of heads we get in 10 flips, \(X\) is a random variable. It can take any integer value from 0 to 10. Random variables can be:
  • Discrete: Where they take on a countable number of values, like in our coin flip example.
  • Continuous: Where they take on any value within a range, such as the height of individuals.
In our context, \(X\) is a discrete random variable sampling from a binomial distribution.
Bernoulli Trials
Bernoulli trials are the building blocks of binomial distributions. Named after the Swiss mathematician Jacob Bernoulli, each trial in this setup has only two possible outcomes: success or failure.Key characteristics of Bernoulli trials include:
  • Each trial is independent of the others.
  • The probability of success, \(p\), remains constant across all trials.
  • The probability of failure is \(q = 1 - p\).
When we perform multiple Bernoulli trials, we end up with a binomial experiment. Imagine flipping a fair coin multiple times. Each flip is a Bernoulli trial, with success defined as obtaining a head (\(p = 0.5\)) and failure as obtaining a tail (\(q = 0.5\)). This scenario is a classic example of a series of Bernoulli trials leading to a binomial distribution.

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Most popular questions from this chapter

An electronic system is made up of two components connected in parallel. Consequently, the system fails only when both of the components fail. The probability the first component fails is \(0.05\) and, when this happens, the probability the second component fails is \(0.02\). What is the probability the electronic system fails?

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Let \(\mathscr{S}\) be the sample space for an experiment \(\mathscr{E}\), with events \(A, B \subseteq \mathscr{S}\). Prove that $$ \operatorname{Pr}(A \mid B) \geq \frac{\operatorname{Pr}(A)+\operatorname{Pr}(B)-1}{\operatorname{Pr}(B)}. $$

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