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Chapter 9: Problem 500

Let \(\mathrm{X}\) be a continuous random variable. We wish to find probabilities concerning \(\mathrm{X}\). These probabilities are determined by a density function. Find a density function such that the probability that \(\mathrm{X}\) falls in an interval \((\mathrm{a}, \mathrm{b})(0<\mathrm{a}\) \(<\mathrm{b}<1\) ) is proportional to the length of the interval \((\mathrm{a}, \mathrm{b})\). Check that this is a proper probability density function.

Short Answer

Expert verified
The probability density function (pdf) for a continuous random variable X, such that the probability of X falling in an interval (a, b) (0 < a < b < 1) is proportional to its length is given by: \(f(x) = 1\), for \(0 < x < 1\) We have shown that this function is a proper pdf, as it is nonnegative for all values of X, and the integral of the pdf over the entire range of possible values of X equals 1.

Step by step solution

01

Determine the proportionality constant

Since the question states that the probability of X falling in the interval (a, b) is proportional to its length, we can represent this as follows: Probability (X 鈭� (a, b)) 鈭� (b - a) Let's denote the proportionality constant as k. So we have: Probability (X 鈭� (a, b)) = k * (b - a)
02

Define the probability density function (pdf)

As we know, for a continuous random variable, the probability of X falling in the interval (a, b) can be found by integrating the pdf over the interval (a, b): Probability (X 鈭� (a, b)) = 鈭玔a, b] f(x) dx = k * (b - a) To find the pdf, we need to integrate the given equation over the interval: f(x) = k, for 0 < x < 1
03

Step 3a: Prove that the pdf is nonnegative for all values of X

To prove that our pdf is nonnegative for all values of X, we must show that: f(x) = k 鈮� 0, for 0 < x < 1 Since k is a proportionality constant and the probability is always nonnegative, we can conclude that k 鈮� 0. Thus, our pdf is nonnegative for all values of X.
04

Step 3b: Prove that the integral of the pdf over the entire range of possible values of X equals 1

Finally, we need to prove that the integral of our pdf equals 1 over the entire range of possible values of X, which is (0, 1): 鈭玔0, 1] f(x) dx = 1 Substitute f(x) with its expression: 鈭玔0, 1] k dx = 1 Now, we integrate: k * 鈭玔0, 1] dx = 1 k * (x |[0, 1]) = 1 k * (1 - 0) = 1 k = 1 Now that we have found the value of k, we can define our proper pdf as: f(x) = 1, for 0 < x < 1 Thus, we have found a probability density function that satisfies the given conditions, and demonstrated that it is a proper probability density function.

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Key Concepts

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

Continuous Random Variable
When dealing with the realm of probability, a continuous random variable is a cornerstone concept. In contrast to a discrete random variable, which can take on only specific isolated values, a continuous random variable can assume any value within a given range. Think of it like measuring height or weight鈥攜ou could get any value within the possible range, not just whole numbers.

In the exercise, X is such a variable, capable of embodying any real number in the interval between 0 and 1. When you want to calculate probabilities for continuous random variables, you don't count outcomes like you do for discrete variables; instead, you measure the area under a curve on a graph. This curve is described by the probability density function (pdf), which portrays how the total probability is distributed across the range of possible values.
Integrating Probability Density
The process of integrating the probability density function is an integral part (pun intended) of working with continuous random variables. This integration calculates the probability that a random variable X falls within a certain interval.

Mathematically, if you have a probability density function f(x) for a random variable X, and you want to know the probability that X is between a and b, you would integrate f(x) over the interval from a to b, denoted as \( 鈭玙{a}^{b} f(x) dx \). This integral computes the area under the curve of the pdf from a to b, which represents the probability of X falling within the specified range. It's this graphical representation that helps us visualize and thus better understand the distribution and likelihood of different outcomes.
Proportionality Constant
A proportionality constant is a fixed number that relates two variables that are directly proportional to each other. In our exercise, the probability that X falls within any interval (a, b) is directly proportional to the length of that interval. The constant of proportionality k serves as a scale factor that adjusts the linear relationship to the context of probability.

For the pdf to properly represent probability, two conditions must be met: the function must be nonnegative, and the total area under the curve (which equals the integral of the pdf over its entire range) must be equal to 1. These conditions ensure that the resultant values are valid probabilities. In solving for k, we found it to be 1, satisfying the conditions鈥�k adjusts the linear proportionality into a probability measure between 0 and 1, making our function f(x) = 1 a valid pdf. Remember, the value of the proportionality constant is dictated by the context of the probability and the conditions that must be met for the function to be a true pdf.

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Most popular questions from this chapter

Let \(X\) possess a Poisson distribution with mean \(\mu\), 1.e. $$ \mathrm{f}(\mathrm{X}, \mu)=\mathrm{e}^{-\mu}\left(\mu^{\mathrm{X}} / \mathrm{X} ;\right) $$ Suppose we want to test the null hypothesis \(\mathrm{H}_{0}: \mu=\mu_{0}\) against the alternative hypothesis, \(\mathrm{H}_{1}: \mu=\mu_{1}\), where \(\mu_{1}<\mu_{0}\). Find the best critical region for this test.

A reading test is given to an elementary school class that consists of 12 Anglo-American children and 10 MexicanAmerican children. The results of the test are \(-\) AngloAmerican children: \(\underline{\mathrm{X}}_{1}=74, \mathrm{~S}_{1}=8 ;\) Mexican-American children: \(\underline{X}_{2}=70, \mathrm{~S}_{2}=10 .\) Is the difference between the means of the two groups significant at the \(.05\) level of significance?

A pair of dice is thrown 120 times. What is the approximate probability of throwing at least 15 sevens? Assume that the rolls are independent and remember that the probability of rolling a seven on a single roll is \(6 / 36=1 / 6\).

Two continuous random variables \(\mathrm{X}\) and \(\mathrm{Y}\) may also be jointly distributed. Suppose \((X, Y)\) has a distribution which is uniform over a unit circle centered at \((0,0)\). Find the joint density of \((\mathrm{X}, \mathrm{Y})\) and the marginal densities of \(\mathrm{X}\) and \(\mathrm{Y}\). Are \(\mathrm{X}\) and \(\mathrm{Y}\) independent?

Find the expected value and variance of a random variable, $$ \mathrm{Y}=\mathrm{a}_{1} \mathrm{X}_{1}+\mathrm{a}_{2} \mathrm{X}_{2}+\ldots \ldots+\mathrm{a}_{\mathrm{n}} \mathrm{X}_{\mathrm{n}} $$ where the \(\mathrm{X}_{\mathrm{i}}\) are independent and each have mean \(\mu\) and variance \(\sigma^{2}\). The \(a_{i}\) are constants.
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