/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 In Exercise 5.6 , we assumed tha... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise 5.6 , we assumed that if 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. Find the marginal density functions for \(Y_{1}\) and \(Y_{2}\) b. What is \(P\left(.3 < Y_{1} < .5\right) ? P\left(.3 < Y_{2} < .5\right) ?\) c. For what values of \(y_{2}\) is the conditional density \(f\left(y_{1} | y_{2}\right)\) defined? d. For any \(y_{2}, 0 \leq y_{2} \leq 1\) what is the conditional density function of \(Y_{1}\) given that \(Y_{2}=y_{2} ?\) e. Find \(P\left(.3 < Y_{1} < .5 | Y_{2}=.3\right)\) f. Find \(P\left(.3 < Y_{1} < .5 | Y_{2}=.5\right)\) g. Compare the answers that you obtained in parts \((\mathrm{a}),(\mathrm{d}),\) and \((\mathrm{e}) .\) For any \(y_{2}, 0

Short Answer

Expert verified
The probabilities are 0.2 for both the marginal and conditional cases, indicating independence.

Step by step solution

01

Marginal density functions

The joint density function is uniform over the unit square. To find the marginal density functions of \(Y_1\) and \(Y_2\), we integrate out the other variable. - For \(Y_1\): \[ f_{Y_1}(y_1) = \int_0^1 f(y_1, y_2) \, dy_2 = \int_0^1 1 \, dy_2 = 1. \]- For \(Y_2\): \[ f_{Y_2}(y_2) = \int_0^1 f(y_1, y_2) \, dy_1 = \int_0^1 1 \, dy_1 = 1. \]Thus, the marginal density functions are both 1 for \(0 \le y_i \le 1\) and 0 elsewhere.
02

Probability within interval for marginals

To find \(P(0.3 < Y_1 < 0.5)\) and \(P(0.3 < Y_2 < 0.5)\), we integrate the marginal densities.- \(P(0.3 < Y_1 < 0.5) = \int_{0.3}^{0.5} f_{Y_1}(y_1) \, dy_1 = \int_{0.3}^{0.5} 1 \, dy_1 = 0.5 - 0.3 = 0.2.\)- \(P(0.3 < Y_2 < 0.5) = \int_{0.3}^{0.5} f_{Y_2}(y_2) \, dy_2 = \int_{0.3}^{0.5} 1 \, dy_2 = 0.5 - 0.3 = 0.2.\)
03

Definition of conditional density

The conditional density \(f(y_1 | y_2)\) is only defined where the joint density \(f(y_1, y_2)\) is nonzero and \(f_{Y_2}(y_2)\) is non-zero. Since each marginal is 1 within the interval [0,1], the conditional density is defined for \(0 \leq y_2 \leq 1\).
04

Conditional density function

For any given \(y_2\), the conditional density \(f(y_1 | y_2)\) is \(\frac{f(y_1, y_2)}{f_{Y_2}(y_2)}\), which becomes:\[ f(y_1 | y_2) = \frac{1}{1} = 1, \quad \text{for } 0 \le y_1 \le 1. \] Therefore, \(f(y_1 | y_2) = 1\) for \(0 \le y_1 \le 1\).
05

Conditional probability with fixed \(Y_2\)

For \(P(0.3 < Y_1 < 0.5 | Y_2 = 0.3)\), we integrate the conditional density over the interval.\[ P(0.3 < Y_1 < 0.5 | Y_2 = 0.3) = \int_{0.3}^{0.5} 1 \, dy_1 = 0.2. \]The calculation is identical for \(Y_2 = 0.5\) due to the uniformity:\[ P(0.3 < Y_1 < 0.5 | Y_2 = 0.5) = \int_{0.3}^{0.5} 1 \, dy_1 = 0.2. \]
06

Comparison and analysis

The probabilities \(P(0.3 < Y_1 < 0.5)\) and \(P(0.3 < Y_1 < 0.5 | Y_2 = y_2)\) computed in Steps 2 and 5, respectively, are equal at 0.2. This equality occurs for any fixed \(y_2\) in the open interval \((0, 1)\), demonstrating that the marginal and conditional probabilities are the same in this scenario, implying that \(Y_1\) and \(Y_2\) are independent.

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Key Concepts

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

Understanding Marginal Density Functions
Marginal density functions help in understanding the behavior of any one random variable independently of the others in a given joint distribution. In essence, marginalization involves integrating out the variables you're not interested in.

For example, if you have a joint density function for two variables, like in our exercise, and you want to find the marginal density for one variable, you would integrate the joint density function over the entire range of the other variable. This process effectively "sums out" the irrelevant variables, allowing you to focus on the variable of interest.

In mathematical terms, if the joint density function is given by \( f(y_1, y_2) \), then the marginal density function for \( Y_1 \) is calculated by integrating over \( y_2 \):
  • \( f_{Y_1}(y_1) = \int_{-\infty}^{\infty} f(y_1, y_2) \, dy_2 \).
In our scenario, because the joint density was uniform, the integration resulted in a simple constant value indicating the uniform spread of the particle within the unit square.
Exploring Joint Density Functions
The joint density function provides a comprehensive view of how two random variables are distributed in relation to each other over a particular range. It's pivotal for understanding how two variables interact within a specific context, like locating a particle within a defined space.

When you have a joint density function, say \( f(y_1, y_2) \), it gives the density of probabilities over each point in the defined space for both variables. This is particularly useful to determine the likelihood of a set combination of values occurring together, which is crucial in multivariate studies.

A joint density function is essentially a generalization of a single-variable density function, allowing you to manage and analyze the intricacies of multiple interrelated variables. With uniform joint density functions, such as \( f(y_1, y_2) = 1 \) within the square in our problem, every point in the range has the same probability density, simplifying many computations.
Calculating and Understanding Probability
Probability computation, especially in continuous random variables, involves finding the likelihood that a variable falls within a specified range. In our exercise, we computed probabilities both marginally and conditionally.

For instance, to find \( P(0.3 < Y_1 < 0.5) \), we integrated the marginal density function of \( Y_1 \) over that interval. This type of integration gives you the area under the curve of the density function within the selected bounds, representing the probability of \( Y_1 \) lying within that range.

Moreover, conditional probabilities, such as \( P(0.3 < Y_1 < 0.5 | Y_2 = y_2) \), involve using the joint density and marginal densities to find the likelihood of an event, given additional known information. The calculation involves integrating the conditional density over the specified range, which in our problem yielded consistent results due to the uniform nature of the density. Such computations highlight vital concepts like independence, where marginal probabilities are equal to conditional ones when there's no dependency between variables.

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Most popular questions from this chapter

Let the discrete random variables \(Y_{1}\) and \(Y_{2}\) have the joint probability function $$p\left(y_{1}, y_{2}\right)=1 / 3, \quad \text { for }\left(y_{1}, y_{2}\right)=(-1,0),(0,1),(1,0)$$ Find \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\). Notice that \(Y_{1}\) and \(Y_{2}\) are dependent. (Why?) This is another example of uncorrelated random variables that are not independent.

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In Section 5.2 , we argued that if \(Y_{1}\) and \(Y_{2}\) have joint cumulative distribution function \(F\left(y_{1}, y_{2}\right)\) then for any \(a

A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. Let \(Y_{1}\) and \(Y_{2}\) denote the numbers of customers who spend more than \(\$ 50\) on groceries at the respective counters. Suppose that \(Y_{1}\) and \(Y_{2}\) are independent binomial random variables, with the probability that a customer at counter I will spend more than \$50 equal to .2 and the probability that a customer at counter II will spend more than \(\$ 50\) equal to .3. Find the a. joint probability distribution for \(Y_{1}\) and \(Y_{2}\) b. probability that not more than one of the three customers will spend more than \(\$ 50 .\)

A learning experiment requires a rat to run a maze (a network of pathways) until it locates one of three possible exits. Exit 1 presents a reward of food, but exits 2 and 3 do not. (If the rat eventually selects exit 1 almost every time, learning may have taken place.) Let \(Y_{i}\) denote the number of times exit \(i\) is chosen in successive runnings. For the following, assume that the rat chooses an exit at random on each run. a. Find the probability that \(n=6\) runs result in \(Y_{1}=3, Y_{2}=1,\) and \(Y_{3}=2\). b. For general \(n\), find \(E\left(Y_{1}\right)\) and \(V\left(Y_{1}\right)\). c. Find \(\operatorname{Cov}\left(Y_{2}, Y_{3}\right)\) for general \(n\). d. To check for the rat's preference between exits 2 and \(3,\) we may look at \(Y_{2}-Y_{3} .\) Find \(E\left(Y_{2}-Y_{3}\right)\) and \(V\left(Y_{2}-Y_{3}\right)\) for general \(n\).

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