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If \(Y_{1}\) is the total time between a customer's arrival in the store and leaving the service window and if \(Y_{2}\) is the time spent in line before reaching the window, the joint density of these variables, according to Exercise 5.15 , is $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll}e^{-y_{1}}, & 0 \leq y_{2} \leq y_{1} \leq \infty \\\0, & \text { elsewhere }\end{array}\right.$$ Are \(Y_{1}\) and \(Y_{2}\) independent?

Short Answer

Expert verified
No, \(Y_1\) and \(Y_2\) are not independent; their joint density can't be expressed as a product of their marginals.

Step by step solution

01

Understanding Independence in Probability

In probability and statistics, two random variables \( Y_1 \) and \( Y_2 \) are independent if the joint probability density function (pdf) is equal to the product of their individual (marginal) pdfs. This means: \[ f(y_1, y_2) = f_{Y_1}(y_1) \cdot f_{Y_2}(y_2) \] for all values of \( y_1 \) and \( y_2 \).
02

Identifying Joint Density Function

The joint density given is \( f(y_1, y_2) = e^{-y_1} \) for \( 0 \leq y_2 \leq y_1 < \infty \) and zero elsewhere. Our task is to see if this can be expressed as the product of two separate functions: one involving only \( y_1 \) and one only \( y_2 \).
03

Finding Marginal Densities

To see if \( Y_1 \) and \( Y_2 \) are independent, we first need to find the marginal densities. The marginal density of \( Y_1 \) is \( f_{Y_1}(y_1) = \int_{0}^{y_1} e^{-y_1} \ dy_2 = y_1 e^{-y_1} \) because the integral is taken over \( y_2 \) from 0 to \( y_1 \). The marginal density of \( Y_2 \) is \( f_{Y_2}(y_2) = \int_{y_2}^{\infty} e^{-y_1} \ dy_1 = e^{-y_2} \) due to the region of integration being from \( y_2 \) to \( \infty \).
04

Checking Independence

With the marginal densities \( f_{Y_1}(y_1) = y_1 e^{-y_1} \) and \( f_{Y_2}(y_2) = e^{-y_2} \), for \( Y_1 \) and \( Y_2 \) to be independent, \( f(y_1, y_2) = e^{-y_1} \) should equal \( f_{Y_1}(y_1) \cdot f_{Y_2}(y_2) = (y_1 e^{-y_1}) \cdot (e^{-y_2}) = y_1 e^{-(y_1 + y_2)} \). Clearly, these functions are not the same because the terms do not match. \( f(y_1, y_2) = e^{-y_1} \) does not decompose into the product of these functions. Therefore, \( Y_1 \) and \( Y_2 \) are not independent.

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Key Concepts

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

Joint Density Function
When studying probability, encountering a joint density function is common when dealing with continuous random variables. It provides a complete picture of how two random variables behave together. Let's consider two random variables, say \(Y_1\) and \(Y_2\), which might represent different aspects of a situation, like total time in a store for \(Y_1\) and waiting time as \(Y_2\).

For these variables, their joint density function \(f(y_1, y_2)\) describes how probabilities are distributed over the possible values of these variables. Specifically, it gives us the probability of \(Y_1\) and \(Y_2\) occurring simultaneously. For instance, the expression \(f(y_1, y_2) = e^{-y_1}\) reveals how the time variables \(Y_1\) and \(Y_2\) relate to each other. In most cases, the function is positive where the probabilities are likely, and zero elsewhere.

Understanding the joint density function is essential for determining how the variables are interrelated. By examining it, students can ascertain whether the interplay between \(Y_1\) and \(Y_2\) affects their individual behaviors.
Marginal Density
To delve deeper into the probability landscape, one can explore the concept of marginal density. Marginal density is essentially extracting a single random variable's probability behavior from a joint density by integrating out the other variable.

To find the marginal density of \(Y_1\), we integrate out \(Y_2\) from the joint density function. Mathematically, for \(Y_1\) this would be expressed as \(f_{Y_1}(y_1) = \int_0^{y_1} e^{-y_1} \, dy_2 = y_1 e^{-y_1}\). This result shows us how likely \(Y_1\) is, irrespective of \(Y_2\)'s values.

Similarly, the marginal density of \(Y_2\) is found by integrating \(Y_1\) out from the joint density function. This is done as \(f_{Y_2}(y_2) = \int_{y_2}^{\infty} e^{-y_1} \, dy_1 = e^{-y_2}\). This computation represents the independent probabilities associated with \(Y_2\).

Marginal densities provide insights into how one variable behaves without any influence from the other and are crucial for conducting further checks for any dependencies.
Random Variables Independence
Random variables independence is a cornerstone concept in probability, crucial for understanding complex systems. Two random variables \(Y_1\) and \(Y_2\) are considered independent if their joint distribution can be expressed as the product of their marginal distributions.

Symbolically, this independence is tested by checking if the equation \(f(y_1, y_2) = f_{Y_1}(y_1) \cdot f_{Y_2}(y_2)\) holds. If this equality is true for all possible values of \(y_1\) and \(y_2\), the variables do not influence each other, behaving independently.

In our example, the functions were defined as \(f(y_1, y_2) = e^{-y_1}\), \(f_{Y_1}(y_1) = y_1 e^{-y_1}\), and \(f_{Y_2}(y_2) = e^{-y_2}\). Upon checking, these functions do not satisfy the independence equation because the joint density \(e^{-y_1}\) does not equal the product of the marginal densities \((y_1 e^{-y_1}) \cdot (e^{-y_2}) = y_1 e^{-(y_1 + y_2)}\).

Therefore, \(Y_1\) and \(Y_2\) are not independent, indicating a relationship exists between the time spent waiting and the total time inside the store. Understanding independence helps us model and predict outcomes in real-world scenarios more effectively.

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Most popular questions from this chapter

We considered two individuals who each tossed a coin until the first head appears. Let \(Y_{1}\) and \(Y_{2}\) denote the number of times that persons \(A\) and \(B\) toss the coin, respectively. If heads occurs with probability \(p\) and tails occurs with probability \(q=1-p,\) it is reasonable to conclude that \(Y_{1}\) and \(Y_{2}\) are independent and that each has a geometric distribution with parameter p. Consider \(Y_{1}-Y_{2}\), the difference in the number of tosses required by the two individuals. a. Find \(E\left(Y_{1}\right), E\left(Y_{2}\right),\) and \(E\left(Y_{1}-Y_{2}\right)\) b. Find \(E\left(Y_{1}^{2}\right), E\left(Y_{2}^{2}\right),\) and \(E\left(Y_{1} Y_{2}\right)\) (recall that \(Y_{1}\) and \(Y_{2}\) are independent). c. Find \(E\left(Y_{1}-Y_{2}\right)^{2}\) and \(V\left(Y_{1}-Y_{2}\right)\) d. Give an interval that will contain \(Y_{1}-Y_{2}\) with probability at least \(8 / 9\)

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. \(P\left(Y_{1} \leq 1 / 2, Y_{2} \leq 1 / 2\right)\).

Suppose that the probability that a head appears when a coin is tossed is \(p\) and the probability that a tail occurs is \(q=1-p .\) Person A tosses the coin until the first head appears and stops. Person B does likewise. The results obtained by persons \(A\) and \(B\) are assumed to be independent. What is the probability that \(A\) and \(B\) stop on exactly the same number toss?

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

If \(c\) is any constant and \(Y\) is a random variable such that \(E(Y)\) exists, show that \(\operatorname{Cov}(c, Y)=0\).

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