/*! 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 12 Let \(Y_{1}\) and \(Y_{2}\) deno... [FREE SOLUTION] | 91影视

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Let \(Y_{1}\) and \(Y_{2}\) denote the proportions of two different types of components in a sample from a mixture of chemicals used as an insecticide. Suppose that \(Y_{1}\) and \(Y_{2}\) have the joint density function given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 2, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1,0 \leq y_{1}+y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ (Notice that \(Y_{1}+Y_{2} \leq 1\) because the random variables denote proportions within the same sample.) Find a. \(P\left(Y_{1} \leq 3 / 4, Y_{2} \leq 3 / 4\right)\). b. \(P\left(Y_{1} \leq 1 / 2, Y_{2} \leq 1 / 2\right)\).

Short Answer

Expert verified
a. 7/8 b. 3/4

Step by step solution

01

Understand the Problem

We need to find the probability of joint events using the given joint density function. The joint density function is defined within a triangular region (0 鈮 y鈧 鈮 1, 0 鈮 y鈧 鈮 1, and 0 鈮 y鈧 + y鈧 鈮 1) with a constant value of 2.
02

Graph the Region

For both parts, graph the inequalities to find the region of integration: - Part a: y鈧 鈮 3/4, y鈧 鈮 3/4, y鈧 + y鈧 鈮 1 forms a smaller triangle within the main triangular region. - Part b: y鈧 鈮 1/2, y鈧 鈮 1/2, y鈧 + y鈧 鈮 1 forms an even smaller region inside the earlier triangle.
03

Express Region as Integral for Part (a)

Determining the bounds, y鈧 varies from 0 to 3/4 and for each y鈧, y鈧 varies from 0 to 1-y鈧 (intersecting lines y鈧 = 3/4 and y鈧 = 1-y鈧). Hence, set up the integral: \[\int_{0}^{3/4} \int_{0}^{1-y_{1}} 2 \, dy_{2} \, dy_{1}\]
04

Compute the Integral for Part (a)

Calculate the internal integral \(\int_{0}^{1-y_{1}} 2 \, dy_{2}\) first:\[ \int_{0}^{1-y_{1}} 2 \, dy_{2} = 2(1-y_{1}) \] Calculate the external integral next:\[ \int_{0}^{3/4} 2(1-y_{1}) \, dy_{1} = 2 \left[ y_{1} - \frac{y_{1}^2}{2} \right]_{0}^{3/4} \]\[ = 2 \left( \frac{3}{4} - \frac{9}{32} \right) = \frac{7}{8} \]
05

Express Region as Integral for Part (b)

Y鈧 varies from 0 to 1/2 and for each y鈧, y鈧 varies from 0 to 1-y鈧 (intersecting lines y鈧 = 1/2 and y鈧 = 1-y鈧). Set up the integral as:\[\int_{0}^{1/2} \int_{0}^{1-y_{1}} 2 \, dy_{2} \, dy_{1}.\]
06

Compute the Integral for Part (b)

Similar to part a:Calculate the internal integral:\[ \int_{0}^{1-y_{1}} 2 \, dy_{2} = 2(1-y_{1}) \]Calculate the external integral:\[ \int_{0}^{1/2} 2(1-y_{1}) \, dy_{1} = 2 \left[ y_{1} - \frac{y_{1}^2}{2} \right]_{0}^{1/2} \]\[ = 2 \left( \frac{1}{2} - \frac{1}{8} \right) = \frac{3}{4} \]
07

Conclusion

a. The probability that Y鈧 鈮 3/4 and Y鈧 鈮 3/4 is 7/8. b. The probability that Y鈧 鈮 1/2 and Y鈧 鈮 1/2 is 3/4.

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Key Concepts

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

Joint Density Function
Joint density functions are crucial in probability and statistics, especially when dealing with multiple random variables simultaneously. These functions help us understand how two (or more) related random variables behave together. In this exercise, we have the random variables \(Y_1\) and \(Y_2\), which represent proportions within a chemical mixture. The given joint density function is \[ f\left(y_{1}, y_{2}\right)=\begin{cases} 2, & 0 \leq y_{1} \leq 1, 0 \leq y_{2} \leq 1, 0 \leq y_{1}+y_{2} \leq 1 \ 0, & \text{elsewhere} \end{cases}\]This function is defined over a triangular region on a coordinate plane, meaning that the values of both \(Y_1\) and \(Y_2\) lie between 0 and 1, and their sum does not exceed 1. This structure ensures that both variables are considered as proportions (parts of a whole). It is essential to accurately project this triangular region when solving problems related to joint probabilities.
Integral Calculus
Integral calculus plays a significant role when dealing with joint density functions because it allows us to calculate probabilities over specific regions. For example, probabilities of events where joint variables have certain constraints can be determined by integrating the joint density over the desired region. Here, the regions of integration are determined by inequalities like \(Y_1 \leq 3/4\) and \(Y_2 \leq 3/4\). The integration process involves setting up double integrals, which look like this for Part (a):\[\int_{0}^{3/4} \int_{0}^{1-y_{1}} 2 \, dy_{2} \, dy_{1}\]First, we integrate with respect to \(y_2\) while keeping \(y_1\) constant. Then, for each slice defined by a particular \(y_1\), we integrate across the entire valid range of \(y_2\). This method is applied iteratively across the \(y_1\) limits. Calculating this integral accurately gives the probability for the specified conditions of \(Y_1\) and \(Y_2\).
Probability Borders
The concept of probability borders refers to the boundaries that define the region over which we're calculating probability. In joint probability distributions, these borders need considerable attention to ensure they're correctly determined. The inequalities that define these borders hold much importance, such as \(Y_1 \leq 1/2\) and \(Y_2 \leq 1/2\) for Part (b). The boundary condition, \(Y_1 + Y_2 \leq 1\), emphasizes the dependence between \(Y_1\) and \(Y_2\) as it clarifies the amount both can sum up to. Understanding and identifying the probability borders help delineate the precise area within which the integration occurs. Graphing these borders makes it visually apparent, aiding in setting up the integrals accurately and ensuring that we only calculate probabilities over the correctly defined region.

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Most popular questions from this chapter

Suppose that the random variables \(Y_{1}\) and \(Y_{2}\) have joint probability density function \(f\left(y_{1}, y_{2}\right)\) given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 6 y_{1}^{2} y_{2}, & 0 \leq y_{1} \leq y_{2}, y_{1}+y_{2} \leq 2 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Verify that this is a valid joint density function. b. What is the probability that \(Y_{1}+Y_{2}\) is less than \(1 ?\)

If \(Y_{1}\) and \(Y_{2}\) are independent random variables, each having a normal distribution with mean 0 and variance 1, find the moment-generating function of \(U=Y_{1} Y_{2} .\) Use this moment-generating function to find \(E(U)\) and \(V(U)\). Check the result by evaluating \(E(U)\) and \(V(U)\) directly from the density functions for \(Y_{1}\) and \(Y_{2}\)

In Exercise \(5.16, Y_{1}\) and \(Y_{2}\) denoted the proportions of time that employees I and II actually spent working on their assigned tasks during a workday. The joint density of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} y_{1}+y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ Employee I has a higher productivity rating than employee II and a measure of the total productivity of the pair of employees is \(30 Y_{1}+25 Y_{2}\). Find the expected value of this measure of productivity.

Let \(Y_{1}\) and \(Y_{2}\) have a joint density function given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 3 y_{1}, & 0 \leq y_{2} \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the marginal density functions of \(Y_{1}\) and \(Y_{2}\) b. Find \(P\left(Y_{1} \leq 3 / 4 | Y_{2} \leq 1 / 2\right)\) c. Find the conditional density function of \(Y_{1}\) given \(Y_{2}=y_{2}\) d. Find \(P\left(Y_{1} \leq 3 / 4 | Y_{2}=1 / 2\right)\)

Let \(Y_{1}\) and \(Y_{2}\) be jointly distributed random variables with finite variances. a. Show that \(\left[E\left(Y_{1} Y_{2}\right)\right]^{2} \leq E\left(Y_{1}^{2}\right) E\left(Y_{2}^{2}\right) .\) [Hint: Observe that \(E\left[\left(t Y_{1}-Y_{2}\right)^{2}\right] \geq 0\) for any real number t or, equivalently, $$t^{2} E\left(Y_{1}^{2}\right)-2 t E\left(Y_{1} Y_{2}\right)+E\left(Y_{2}^{2}\right) \geq 0$$ This is a quadratic expression of the form \(A t^{2}+B t+C\); and because it is nonnegative, we must have \(B^{2}-4 A C \leq 0 .\) The preceding inequality follows directly.] b. Let \(\rho\) denote the correlation coefficient of \(Y_{1}\) and \(Y_{2} .\) Using the inequality in part (a), show that \(\rho^{2} \leq 1\)

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