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Suppose that \(Y\) has a binomial distribution with parameters \(n\) and \(p\) but that \(p\) varies from day to day according to a beta distribution with parameters \(\alpha\) and \(\beta\). Show that $$ \begin{array}{l} \text { a. } E(Y)=n \alpha /(\alpha+\beta) \\ \text { b. } V(Y)=\frac{n \alpha \beta(\alpha+\beta+n)}{(\alpha+\beta)^{2}(\alpha+\beta+1)} \end{array} $$

Short Answer

Expert verified
E(Y) = \(\frac{n\alpha}{\alpha+\beta}\), V(Y) = \(\frac{n\alpha\beta(\alpha+\beta+n)}{(\alpha+\beta)^2(\alpha+\beta+1)}\)."

Step by step solution

01

Understand the Distribution of Y

The random variable \(Y\) follows a binomial distribution with parameters \(n\) (number of trials) and \(p\) (success probability), where \(p\) follows a beta distribution with parameters \(\alpha\) and \(\beta\). This implies that the probability of success \(p\) is a random variable as well.
02

Calculate the Expected Value of Y

The expected value of a binomially distributed random variable \(Y\) with a variable success probability \(p\) is given by \(E(Y) = E(E(Y \mid p))\). Since \(E(Y \mid p) = np\), it follows that \(E(Y) = E(np) = nE(p)\). For a beta distribution with parameters \(\alpha\) and \(\beta\), the expectation is \(E(p) = \frac{\alpha}{\alpha+\beta}\). Thus, \(E(Y) = n \cdot \frac{\alpha}{\alpha+\beta} = \frac{n\alpha}{\alpha+\beta}\).
03

Calculate the Variance of Y

The variance of \(Y\) is given by \(V(Y) = E(V(Y \mid p)) + V(E(Y \mid p))\). First, calculate \(V(Y \mid p) = np(1-p)\), so \(E(V(Y \mid p)) = nE(p - p^2)\). With \(p\) following a beta distribution, we have \(E(p) = \frac{\alpha}{\alpha+\beta}\) and \(E(p^2) = \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)}\), leading to \(E(V(Y \mid p)) = n\left(\frac{\alpha}{\alpha+\beta} - \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)}\right)\). Also, \(V(E(Y \mid p)) = n^2V(p)\), where \(V(p) = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\). Substitute these values to obtain \(V(Y) = \frac{n\alpha\beta(\alpha+\beta+n)}{(\alpha+\beta)^2(\alpha+\beta+1)}\).

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Key Concepts

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

Beta Distribution
The beta distribution is a crucial concept in probability and statistics, often applied when dealing with random variables whose values are constrained to lie within a certain interval, specifically between 0 and 1. This makes it especially useful for modeling probabilities or proportions. It is characterized by two shape parameters, \( \alpha \) and \( \beta \), which determine the distribution's behavior and shape.

The beta distribution is flexible and can take various forms depending on its parameters. For example:
  • When \( \alpha = \beta = 1 \), the beta distribution becomes uniform, meaning every outcome is equally likely.
  • If \( \alpha > 1 \) and \( \beta > 1 \), the distribution will be unimodal, having a peak indicating more frequent outcomes around the middle.
  • When \( \alpha < 1 \) or \( \beta < 1 \), the distribution will be U-shaped or skewed towards one of the bounds.
The beta distribution is widely used in contexts where the probability of an event is not fixed but rather a random variable itself. This concept is perfectly illustrated in the given exercise, where the probability \( p \) of success in a binomial distribution is described by a beta distribution. Such usage allows for modeling uncertainties in the success probabilities, which is common in real-world scenarios where outcomes are unpredictable.
Expected Value
The expected value is a foundational concept in probability and statistics, representing the long-run average of outcomes from random variables. It is essentially a weighted average, considering all possible outcomes and their probabilities.

For a random variable \( Y \) with a particular distribution, the expected value \( E(Y) \) gives a central measure of the distribution's location. When \( Y \) follows a binomial distribution with parameters \( n \) and \( p \), and \( p \) itself varies according to a beta distribution, the expected value involves additional calculations.

To find \( E(Y) \):
  • First, recognize that \( Y \) is conditionally expected to have \( np \) successes, given \( p \).
  • Since \( p \) is a random variable following a beta distribution with parameters \( \alpha \) and \( \beta \), its expected value is \( E(p) = \frac{\alpha}{\alpha + \beta} \).
  • Thus, \( E(Y) = nE(p) = \frac{n\alpha}{\alpha + \beta} \).
The expected value in this case adjusts for the variability in \( p \), effectively integrating the uncertainty of the beta distribution into the calculation of the mean success count. This shows the versatility and applicability of expected value in predicting likely outcomes even when faced with uncertain probabilities.
Variance
Variance is another critical statistical metric that measures the spread or dispersion of a set of values. While the expected value provides a central location, variance indicates how much the values spread out from the mean. In probability terms, it shows how much the outcomes of a random variable deviate from its expected value on average.

For the random variable \( Y \) in our exercise, variance steps in as a tool to grasp the unpredictability in the experimental outcomes. Since \( Y \) follows a binomial distribution with a variable \( p \), calculating its variance involves understanding both the variability within trials and across different probabilities:
  • First, calculate conditional variance \( V(Y \mid p) = np(1-p) \), representing variability in outcomes with a fixed \( p \).
  • Next, derive \( E(V(Y \mid p)) \), which is the expected value of this variance based on a random \( p \), using the properties of the beta distribution.
  • The variance \( V(p) = \frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)} \) influences \( V(E(Y \mid p)) = n^2 V(p) \).
  • Combine these to find the total variance \( V(Y) = \frac{n \alpha \beta (\alpha + \beta + n)}{(\alpha + \beta)^2 (\alpha + \beta + 1)} \).
Variance captures the essence of uncertainty by considering possible variations in both the individual trials and different probability settings. By doing so, it provides a comprehensive measure of unpredictability in scenarios where the success probability itself is not fixed.
Probability Distribution
A probability distribution is a mathematical function describing all possible values and outcomes for a random variable along with their respective probabilities. It serves as the backbone for statistical analysis, enabling us to model real-world phenomena and draw conclusions about data.

For any random variable, including \( Y \) in our exercise, understanding its probability distribution helps determine the likelihood of different outcomes. The binomial distribution, a common probability distribution, describes the number of successes in a fixed number of independent Bernoulli trials at a constant probability of success \( p \).

However, when \( p \) is not constant but follows another distribution, such as the beta distribution, it combines to create a more complex probability model. In such cases:
  • The binomial distribution function becomes a compound distribution, integrating over all possible values of \( p \).
  • The combined distribution allows for modeling scenarios where the success probability itself is subject to random variability.
  • This approach provides a more realistic framework for expressing probabilities when facing uncertainty around \( p \), often encountered in fields like quality control, economics, and risk assessment.
By comprehending probability distributions and their interactions, we enrich our ability to predict and reason about random events, enhancing decision-making and statistical inferences in uncertain conditions.

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Most popular questions from this chapter

The length of life \(Y\) for fuses of a certain type is modeled by the exponential distribution, with $$f(y)=\left\\{\begin{array}{ll}(1 / 3) e^{-y / 3}, & y>0 \\\0, & \text { elsewhere }\end{array}\right.$$ (The measurements are in hundreds of hours.) a. If two such fuses have independent lengths of life \(Y_{1}\) and \(Y_{2}\), find the joint probability density function for \(Y_{1}\) and \(Y_{2}\). b. One fuse in part (a) is in a primary system, and the other is in a backup system that comes into use only if the primary system fails. The total effective length of life of the two fuses is then \(Y_{1}+Y_{2} .\) Find \(P\left(Y_{1}+Y_{2} \leq 1\right)\).

A firm purchases two types of industrial chemicals. Type I chemical costs \(\$ 3\) per gallon, whereas type II costs \(\$ 5\) per gallon. The mean and variance for the number of gallons of type I chemical purchased, \(Y_{1},\) are 40 and \(4,\) respectively. The amount of type II chemical purchased, \(Y_{2}\), has \(E\left(Y_{2}\right)=65\) gallons and \(V\left(Y_{2}\right)=8 .\) Assume that \(Y_{1}\) and \(Y_{2}\) are independent and find the mean and variance of the total amount of money spent per week on the two chemicals.

In Exercise 5.12 we were given the following joint probability density function for the random variables \(Y_{1}\) and \(Y_{2},\) which were the proportions of two components in a sample from a mixture of insecticide: $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 2, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1,0 \leq y_{1}+y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find \(P\left(Y_{1} \geq 1 / 2 | Y_{2} \leq 1 / 4\right)\) b. Find \(P\left(Y_{1} \geq 1 / 2 | Y_{2}=1 / 4\right)\)

The management at a fast-food outlet is interested in the joint behavior of the random variables \(Y_{1},\) defined as the total time between a customer's arrival at the store and departure from the service window, and \(Y_{2}\), the time a customer waits in line before reaching the service window. Because \(Y_{1}\) includes the time a customer waits in line, we must have \(Y_{1} \geq Y_{2}\). The relative frequency distribution of observed values of \(Y_{1}\) and \(Y_{2}\) can be modeled by the probability density function $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} e^{-y_{1}}, & 0 \leq y_{2} \leq y_{1}<\infty \\ 0, & \text { elsewhere } \end{array}\right.$$ with time measured in minutes. Find a. \(P\left(Y_{1}<2, Y_{2}>1\right)\). b. \(P\left(Y_{1} \geq 2 Y_{2}\right)\). c. \(P\left(Y_{1}-Y_{2} \geq 1\right)\). (Notice that \(Y_{1}-Y_{2}\) denotes the time spent at the service window.)

Refer to Exercises 5.6,5.24 , and \(5.50 .\) Suppose that a radioactive particle is randomly located in a square with sides of unit length. A reasonable model for the joint density function for \(Y_{1}\) and \(Y_{2}\) is $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 1, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. What is \(E\left(Y_{1}-Y_{2}\right) ?\) b. What is \(E\left(Y_{1} Y_{2}\right) ?\) c. What is \(E\left(Y_{1}^{2}+Y_{2}^{2}\right) ?\) d. What is \(V\left(Y_{1} Y_{2}\right) ?\)

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