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Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent random variables, each with a beta distribution, with \(\alpha=\beta=2\) Find a. the probability distribution function of \(Y_{(n)}=\max \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) b. the density function of \(Y_{(n)}\) c. \(E\left(Y_{(n)}\right)\) when \(n=2\)

Short Answer

Expert verified
The CDF is \([x^2(3-2x)]^n\), the PDF is its derivative, and \(E(Y_{(n)})\) is calculated using integration for \(n=2\).

Step by step solution

01

Understanding the Random Variables

Each random variable \(Y_i\) follows a beta distribution with parameters \(\alpha=\beta=2\). The probability density function (PDF) for \(Y_i\) is given by \( f(y) = 6y(1-y) \) for \(0 \leq y \leq 1\).
02

Finding the CDF of \(Y_{(n)}\)

The cumulative distribution function (CDF) of \(Y_{(n)} = \max(Y_1, Y_2, \ldots, Y_n)\) is given by \( F_{Y_{(n)}}(x) = P(Y_{(n)} \leq x) = P(Y_1 \leq x, Y_2 \leq x, \ldots, Y_n \leq x) = [F_Y(x)]^n \), where \( F_Y(x) = x^2(3-2x) \) is the CDF of the beta distribution \(\alpha=\beta=2\).
03

Finding the PDF of \(Y_{(n)}\)

The PDF of \(Y_{(n)}\) is obtained by differentiating its CDF: \( f_{Y_{(n)}}(x) = \frac{d}{dx} \left( F_{Y_{(n)}}(x) \right) = \frac{d}{dx} \left( [x^2 (3-2x)]^n \right) \). Use the chain rule and power rule to find the derivative.
04

Computing \(E(Y_{(n)})\) for \(n=2\)

The expectation of \(Y_{(n)}\), when \(n=2\), can be calculated using the integral \( E(Y_{(n)}) = \int_0^1 x \cdot f_{Y_{(n)}}(x) \,dx \). Substitute the PDF of \(Y_{(n)}\), \(f_{Y_{(n)}}(x)\), into the integral and solve.

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Key Concepts

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

Probability Density Function
When dealing with continuous random variables, the concept of a **Probability Density Function (PDF)** is essential. Imagine you have a smooth curve that plots likelihoods across different outcomes—a PDF is this curve. It tells us how probable specific points within a range are, relative to others. For a beta-distributed variable with parameters \(\alpha = 2\) and \(\beta = 2\), the density function is represented as:
\[ f(y) = 6y(1-y), \quad 0 \leq y \leq 1 \]
This formula gives more weight to outcomes around the middle of the possible range (0.5) and less to outcomes near the edges (0 or 1). You can think of it as how much 'density' there is at any spot, much like how we might describe how crowded a place feels.
The key aspects of a PDF to remember are:
  • The total area under the curve is always 1, reflecting the certainty that the random variable takes a value within the specified range.
  • The PDF itself is not a probability but describes where the probability is relatively more intense.
Understanding the PDF allows us to comprehend the likelihood of a random variable taking on various values.
Cumulative Distribution Function
To ascertain the probability that a random variable is less than or equal to a value, we use the **Cumulative Distribution Function (CDF)**. In simpler terms, the CDF is the sum of probabilities for all outcomes up to a point \(x\). It's like continuously adding the probabilities as you move from left to right along the number line. For a beta distribution with \(\alpha = \beta = 2\), it is expressed as:
\[ F_Y(x) = x^2(3-2x) \]
The CDF of \(Y_{(n)} = \max(Y_1, Y_2, \ldots, Y_n)\) can be calculated by raising the CDF of a single \(Y_i\) to the power \(n\):
\[ F_{Y_{(n)}}(x) = [F_Y(x)]^n \]
This represents the probability that all of the independent variables \(Y_i\) are less than or equal to \(x\). It's like stacking up these probabilities—only where they all match does \(Y_{(n)}\) achieve this condition which makes this quantity smaller and tighter around the higher values as \(n\) increases.
Remember, the CDF steadily increases from 0 to 1 as \(x\) goes from the smallest to the largest possible values of the variable. Once we differentiate this CDF, we return to find the PDF.
Expectation of Random Variables
The **Expectation** of a random variable, denoted by \(E\), is essentially the mean or average value that the variable takes on if we were to observe it an infinite number of times. For continuous variables, this expectation is determined through integration. Specifically, for \(Y_{(n)}\) in our example with \(n=2\), we compute it by the formula:
\[ E(Y_{(n)}) = \int_0^1 x \cdot f_{Y_{(n)}}(x) \,dx \]
Expectation helps us by summarizing the overall behavior of the random variable. It gives a single value that can concisely describe typical results from repeated sampling of \(Y_{(n)}\).
Key points are:
  • This calculation requires knowledge of the PDF of the distribution.
  • The expectation provides a central tendency figure for decision making or predictions.
  • It is the "balancing point" of the probability distribution where the outcomes weight each other out.
Thus, computing the expectation over a range takes into account not just the mid-point, but a full weighted average, considering how the distribution spreads across its domain.

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Most popular questions from this chapter

The opening prices per share \(Y_{1}\) and \(Y_{2}\) of two similar stocks are independent random variables, =ach with a density function given by $$f(y)=\left\\{\begin{array}{ll} (1 / 2) e^{-(1 / 2)(y-4)}, & y \geq 4 \\ 0, & \text { elsewhere } \end{array}\right.$$ On a given morning, an investor is going to buy shares of whichever stock is less expensive. Find the a. probability density function for the price per share that the investor will pay. b. expected cost per share that the investor will pay.

Let \(Y\) be a random variable with a density function given by $$f(y)=\left\\{\begin{array}{ll} (3 / 2) y^{2}, & -1 \leq y \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the density function of \(U_{1}=3 Y\) b. Find the density function of \(U_{2}=3-Y\) c. Find the density function of \(U_{3}=Y^{2}\)

The joint distribution for the length of life of two different types of components operating in a system was given in Exercise 5.18 by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} (1 / 8) y_{1} e^{-\left(y_{1}+y_{2}\right) / 2}, & y_{1}>0, y_{2}>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ The relative efficiency of the two types of components is measured by \(U=Y_{2} / Y_{1}\). Find the probability density function for \(U\).

The weight (in pounds) of "medium-size" watermelons is normally distributed with mean 15 and variance 4. A packing container for several melons has a nominal capacity of 140 pounds. What is the maximum number of melons that should be placed in a single packing container if the nominal weight limit is to be exceeded only \(5 \%\) of the time? Give reasons for your answer.

The length of time necessary to tune up a car is exponentially distributed with a mean of .5 hour. If two cars are waiting for a tune-up and the service times are independent, what is the probability that the total time for the two tune-ups will exceed 1.5 hours? [Hint: Recall the result of Example 6.12.]

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