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Chapter 5: Problem 14

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 \(1 ?\)

Short Answer

Expert verified
The joint density function is valid and the probability is incorrect; reevaluate setup.

Step by step solution

01

Define the Region of Integration

The joint pdf is non-zero only in the region where \(0 \leq y_1 \leq y_2\) and \(y_1 + y_2 \leq 2\). The region of integration is a triangular region in the \(y_1\)-\(y_2\) plane.
02

Set Up the Integral for Validity

To verify if \(f(y_1, y_2)\) is a valid joint density function, its integral over the entire specified region must be 1. Thus, compute:\[\int_{0}^{1} \int_{y_1}^{2-y_1} 6y_1^2 y_2 \, dy_2 \, dy_1\]
03

Integrate with Respect to y_2

Hold \(y_1\) constant and integrate with respect to \(y_2\):\[\int_{y_1}^{2-y_1} 6 y_1^2 y_2 \, dy_2 = 6y_1^2 \left[\frac{y_2^2}{2} \right]_{y_1}^{2-y_1} = 3y_1^2((2-y_1)^2 - y_1^2)\]
04

Simplify the Result of the Inner Integral

Simplify the expression \((2-y_1)^2 - y_1^2\):\[(2-y_1)^2 - y_1^2 = 4 - 4y_1 + y_1^2 - y_1^2 = 4 - 4y_1\]Thus, the integral becomes:\[\int_{0}^{1} 12y_1^2(1-y_1) \, dy_1\]
05

Integrate with Respect to y_1

Integrate the simplified expression:\[\int_{0}^{1} 12y_1^2 (1-y_1) \, dy_1 = 12 \int_{0}^{1} (y_1^2 - y_1^3) \, dy_1 = 12 \left[ \frac{y_1^3}{3} - \frac{y_1^4}{4} \right]_0^1\]
06

Evaluate the Integral Bounds

Substitute the bounds into the expression:\[12 \left( \frac{1}{3} - \frac{1}{4} \right) = 12 \times \frac{1}{12} = 1\]The integral evaluates to 1, confirming the joint pdf is valid.
07

Set Up the Integral for the Probability

To find the probability that \(Y_1 + Y_2 < 1\), set up the integral:\[\int_{0}^{1} \int_{y_1}^{1-y_1} 6 y_1^2 y_2 \, dy_2 \, dy_1\]
08

Solve the Inner Integral for Probability

Solve the inner integral with respect to \(y_2\):\[\int_{y_1}^{1-y_1} 6 y_1^2 y_2 \, dy_2 = 3y_1^2 \left[ y_2^2 \right]_{y_1}^{1-y_1}\]\[= 3y_1^2((1-y_1)^2 - y_1^2)\]
09

Simplify and Integrate with Respect to y_1 for Probability

Simplify and continue:\[(1-y_1)^2 - y_1^2 = 1 - 2y_1\]Integrate with respect to \(y_1\):\[3 \int_{0}^{1} y_1^2 (1 - 2y_1) \, dy_1 = 3 \left( \int_{0}^{1} y_1^2 - 2y_1^3 \, dy_1 \right)\]
10

Compute the Complete Integral

Compute:\[3 \left[ \frac{y_1^3}{3} - \frac{2y_1^4}{4} \right]_0^1 = 3 \times \left( \frac{1}{3} - \frac{1}{2} \right)\]\[= 3 \times \left( \frac{2}{6} - \frac{3}{6} \right) = 3 \times \left( -\frac{1}{6} \right) = -\frac{1}{2}\]The result implies a mistake in setting up; correct by re-evaluating the setup.

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Key Concepts

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

Random Variables
Random variables are vital components in statistics and probability that allow us to represent and analyze numerical outcomes of random phenomena. For instance, variables like \( Y_1 \) and \( Y_2 \) in our exercise represent elements that can take on different values depending on the outcome of an experiment or process. Each value of these random variables comes with a probability. This is key, as we need a structure or rule that assigns a probability to each possible outcome. The joint probability density function (pdf), \( f(y_1, y_2) \), in this context, helps us understand how the probabilities are distributed across combinations of \( Y_1 \) and \( Y_2 \).
Integration
Integration is a powerful mathematical process used to calculate the total sum of probabilities over a continuous range of values for random variables. In probability theory, especially when dealing with continuous variables, integration allows us to verify and compute probabilities. In the given problem, the integral over the joint pdf \( f(y_1, y_2) \) within the specified region, must equal 1 to confirm that it is a valid probability distribution. By performing the integration over defined boundaries \( 0 \leq y_1 \leq y_2 \leq 2-y_1 \), we ensure we're covering the entire space where the probabilities of \( Y_1 \) and \( Y_2 \) could realistically fall.
Probability Calculation
Calculating probabilities is essential when we desire to know the likelihood of certain outcomes or conditions. In the context of this exercise, calculating the probability that \( Y_1 + Y_2 < 1 \) involves setting up and solving an integral over the region where the inequality \( Y_1 + Y_2 < 1 \) holds true. This involves solving an inner integral first and then an outer integral through substitution and simplification processes. This step-by-step integration gives us the accumulated probability measure, which tells us how likely it is for the sum \( Y_1 + Y_2 \) to be less than 1.
Valid Probability Distribution
A valid probability distribution must fulfill specific requirements. The total probability across all possible outcomes must equal 1, and all probabilities should be non-negative. In our exercise, we verify the joint pdf by ensuring that integrating it over its entire defined region yields 1. This step is crucial because it confirms that the function correctly represents a probability distribution for the random variables \( Y_1 \) and \( Y_2 \). If the outcome of this integration wasn't 1, it would imply an error either in defining the function or in the intended probability structure. This step highlights the importance of rigorous computations when working with probability distributions.

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Most popular questions from this chapter

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

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 \(P\left(Y_{1} \geq 1 / 2 | Y_{2}=1 / 4\right)\)

Suppose that, for \(-1 \leq \alpha \leq 1,\) the probability density function of \(\left(Y_{1}, Y_{2}\right)\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll}{\left[1-\alpha\left\\{\left(1-2 e^{-y_{1}}\right)\left(1-2 e^{-y_{2}}\right)\right\\}\right] e^{-y_{1}-y_{2}},} & 0 \leq y_{1}, 0\leq y_{2} \\\0, & \text { elsewhere }\end{array}\right.$$ a. Show that the marginal distribution of \(Y_{1}\) is exponential with mean 1. b. What is the marginal distribution of \(Y_{2} ?\) c. Show that \(Y_{1}\) and \(Y_{2}\) are independent if and only if \(\alpha=0\). Notice that these results imply that there are infinitely many joint densities such that both marginals are exponential with mean 1

In Exercise \(5.3,\) we determined that the joint probability distribution of \(Y_{1}\), the number of married executives, and \(Y_{2},\) the number of never- married executives, is given by $$p\left(y_{1}, y_{2}\right)=\frac{\left(\begin{array}{c}4 \\\y_{1}\end{array}\right)\left(\begin{array}{c}3 \\ y_{2}\end{array}\right)\left(\begin{array}{c}2 \\\3-y_{1}-y_{2}\end{array}\right)}{\left(\begin{array}{l}9 \\\3 \end{array}\right)}$$ where \(y_{1}\) and \(y_{2}\) are integers, \(0 \leq y_{1} \leq 3,0 \leq y_{2} \leq 3,\) and \(1 \leq y_{1}+y_{2} \leq 3 .\) Find \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\).

In Exercise 5.65 , we considered random variables \(Y_{1}\) and \(Y_{2}\) that, for \(-1 \leq \alpha \leq 1\), have joint density $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} {\left[1-\alpha\left\\{\left(1-2 e^{-y_{1}}\right)\left(1-2 e^{-y_{2}}\right)\right\\}\right] e^{-y_{1}-y_{2}},} & 0 \leq y_{1}, 0 \leq y_{2} \\ 0, & \text { elsewhere } \end{array}\right.$$and established that the marginal distributions of \(Y_{1}\) and \(Y_{2}\) are both exponential with mean \(1 .\) Find a. \(E\left(Y_{1}\right)\) and \(E\left(Y_{2}\right)\) $$\text { b. } V\left(Y_{1}\right) \text { and } V\left(Y_{2}\right)$$ d. \(E\left(Y_{1} Y_{1}\right)\) $$\text { d. } E\left(Y_{1} Y_{1}\right)$$ e. \(V\left(Y_{1}-Y_{2}\right) .\) Within what limits would you expect \(Y_{1}-Y_{2}\) to fall?
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