/*! 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 83 Suppose a randomly chosen indivi... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose a randomly chosen individual's verbal score \(X\) and quantitative score \(Y\) on a nationally administered aptitude examination have joint pdf $$ \begin{aligned} &f(x, y) \\ &=\left\\{\begin{array}{cl} \frac{2}{5}(2 x+3 y) & 0 \leq x \leq 1, \quad 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ You are asked to provide a prediction \(t\) of the individual's total score \(X+Y\). The error of prediction is the mean squared error \(E\left[(X+Y-t)^{2}\right]\). What value of \(t\) minimizes the error of prediction?

Short Answer

Expert verified
The value of \( t \) that minimizes the error is \( \frac{13}{20} \).

Step by step solution

01

Set Up the Problem

We need to predict the total score \( X+Y \) by finding a value of \( t \) that minimizes the mean squared error \( E[(X+Y-t)^2] \). Given the joint probability density function (pdf) \( f(x,y) = \frac{2}{5}(2x+3y) \) for \( 0 \leq x, y \leq 1 \), let's proceed to calculate the expectation values.
02

Calculate Expected Value of X and Y

To find \( E(X) \) and \( E(Y) \), we need to evaluate \( \int_0^1 \int_0^1 x \cdot f(x,y) \, dx \, dy \) and \( \int_0^1 \int_0^1 y \cdot f(x,y) \, dx \, dy \) respectively.
03

Calculate \( E(X) \)

Calculate \( E(X) = \int_0^1 \int_0^1 x \cdot \frac{2}{5}(2x+3y) \, dx \, dy \). We first integrate with respect to \( x \): \[ E(X) = \int_0^1 \left( \int_0^1 \frac{2}{5}(2x^2 + 3xy) \, dx \right) dy \]This results in:\[ E(X) = \int_0^1 \left( \frac{1}{5}x^2 + \frac{3}{10}xy^2 \right)_0^1 dy \]\[ = \int_0^1 \left( \frac{1}{5} + \frac{3}{10}y \right) dy \] \[ = \left( \frac{1}{5}y + \frac{3}{20}y^2 \right)_0^1 = \frac{1}{5} + \frac{3}{20} = \frac{7}{20} \]
04

Calculate \( E(Y) \)

Calculate \( E(Y) = \int_0^1 \int_0^1 y \cdot \frac{2}{5}(2x+3y) \, dx \, dy \). Integrate with respect to \( y \): \[ E(Y) = \int_0^1 \left( \int_0^1 \frac{2}{5}(2xy + 3y^2) \, dy \right) dx \]This results in:\[ E(Y) = \int_0^1 \left( \frac{2}{5}xy^2 + \frac{1}{5}y^3 \right)_0^1 dx \]\[ = \int_0^1 \left( \frac{1}{5}x + \frac{1}{5} \right) dx \]\[ = \left( \frac{1}{10}x^2 + \frac{1}{5}x \right)_0^1 = \frac{1}{10} + \frac{1}{5} = \frac{3}{10} \]
05

Calculate Expected Total Score E(X+Y)

Now that we have \( E(X) = \frac{7}{20} \) and \( E(Y) = \frac{3}{10} \), the expected total score is:\[ E(X+Y) = E(X) + E(Y) = \frac{7}{20} + \frac{3}{10} = \frac{7}{20} + \frac{6}{20} = \frac{13}{20} \]
06

Minimize Mean Squared Error

The mean squared error is minimized when \( t = E(X+Y) \). Thus, the value of \( t \) that minimizes the error of prediction is \( \frac{13}{20} \).

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Key Concepts

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

Mean Squared Error
In statistical analysis, the mean squared error (MSE) is an important concept used to evaluate the accuracy of a prediction or estimator. It measures the average of the squares of the errors, where error is the amount by which the estimator differs from the actual value. To put it simply, MSE is a way to quantify how close a set of predicted values are to the actual outcomes.
Mathematically, for predictions \[ E[(X+Y-t)^2] \]MSE considers the squared difference between predicted scores \(t\) and actual outcomes \(X+Y\). These differences are squared so that positive and negative errors do not offset each other and ensure that larger errors are penalized more significantly.
The goal is to find the value of \(t\) which minimizes the mean squared error, essentially making your predictions as accurate as possible. To achieve this, you'll often rely on concepts such as expectation and integration from probability and calculus.
Expectation
Expectation, also referred to as the expected value, is a fundamental concept in the field of probability and statistics. It gives us a measure of the central tendency of a random variable, essentially indicating what value we can "expect" to observe on average.
In the context of the exercise, we deal with the expectation of random variables \(X\) and \(Y\), found using the joint probability distribution. The expected values \(E(X)\) and \(E(Y)\) are calculations like:\[E(X) = \int_0^1 \int_0^1 x \, f(x,y) \, dx \, dy\]These integrals quantify the weighted average of possible outcomes, with the weighting determined by the probability density function \(f(x, y)\). By understanding expectation, one can determine the most likely average result of a random process, which is a vital part of predicting future outcomes, such as in finding the optimal \(t\) for mean squared error.
Integration
Integration is a key mathematical tool used in calculus to find areas under curves, which can translate into finding probabilities and expected values in statistics. When dealing with continuous random variables and their probability distributions, integration helps in calculating quantities like expectations.
For example, in order to find \(E(X)\) and \(E(Y)\), double integration is used over the joint probability distribution:\[E(X) = \int_0^1 \left( \int_0^1 \frac{2}{5}(2x^2 + 3xy) \, dx \right) dy\]This technique takes into account all possible values of \(x\) and \(y\), integrating over the entire space to give the expected outcome. Integration allows us to aggregate all infinitesimally tiny probabilities into a meaningful measure, leading to results that aid in minimizing mean squared error.
Probability Distribution
A probability distribution describes how the values of a random variable are spread or distributed. In this exercise, the joint probability density function \(f(x, y)\) provides information about how the scores \(X\) and \(Y\) are likely to occur together.
The joint probability density function is defined as:\[f(x, y) = \frac{2}{5}(2x+3y) \quad \text{for} \quad 0 \leq x, y \leq 1\]This function gives the likelihood of \(X\) and \(Y\) taking on particular values within specified ranges. The joint distribution covers the combined outcome of both variables, and integrating over its domain provides necessary measures like expected values and variances.
Understanding probability distributions is crucial because it sets the stage for calculating expectations and minimizing prediction errors. These distributions mean we can analyze the relationship between variables in random processes, leading to more precise predictions and calculations like minimizing mean squared error.

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Most popular questions from this chapter

. An ecologist selects a point inside a circular sampling region according to a uniform distribution. Let \(X=\) the \(x\) coordinate of the point selected and \(Y=\) the \(y\) coordinate of the point selected. If the circle is centered at \((0,0)\) and has radius \(R\), then the joint pdf of \(X\) and \(Y\) is $$ f(x, y)=\left\\{\begin{array}{cl} \frac{1}{\pi R^{2}} & x^{2}+y^{2} \leq R^{2} \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the probability that the selected point is within \(R / 2\) of the center of the circular region? [Hint: Draw a picture of the region of positive density \(D\). Because \(f(x, y)\) is constant on \(D\), computing a probability reduces to computing an area.] b. What is the probability that both \(X\) and \(Y\) differ from 0 by at most \(R / 2\) ? c. Answer part (b) for \(R / \sqrt{2}\) replacing \(R / 2\) d. What is the marginal pdf of \(X ?\) Of \(Y ?\) Are \(X\) and \(Y\) independent?

Let \(X_{1}\) and \(X_{2}\) be quantitative and verbal scores on one aptitude exam, and let \(Y_{1}\) and \(Y_{2}\) be corresponding scores on another exam. If \(\operatorname{Cov}\left(X_{1}, Y_{1}\right)=5, \operatorname{Cov}\left(X_{1}, Y_{2}\right)=1, \operatorname{Cov}\left(X_{2}, Y_{1}\right)=2\), and \(\operatorname{Cov}\left(X_{2}, Y_{2}\right)=8\), what is the covariance between the two total scores \(X_{1}+X_{2}\) and \(Y_{1}+Y_{2}\) ?

Let \(X_{1}\) and \(X_{2}\) be independent, each having a standard normal distribution. The pair \(\left(X_{1}, X_{2}\right)\) corresponds to a point in a two-dimensional coordinate system. Consider now changing to polar coordinates via the transformation, $$ \begin{gathered} Y_{1}=X_{1}^{2}+X_{2}^{2} \\ Y_{2}=\left\\{\begin{array}{cl} \arctan \left(\frac{X_{2}}{X_{1}}\right) & X_{1}>0, X_{2} \geq 0 \\ \arctan \left(\frac{X_{2}}{X_{1}}\right)+2 \pi & X_{1}>0, X_{2}<0 \\ \arctan \left(\frac{X_{2}}{X_{1}}\right)+\pi & X_{1}<0 \\ 0 & X_{1}=0 \end{array}\right. \end{gathered} $$ from which \(X_{1}=\sqrt{Y_{1}} \cos \left(Y_{2}\right), X_{2}=\sqrt{Y_{1}} \sin \left(Y_{2}\right)\). Obtain the joint pdf of the new variables and then the marginal distribution of each one. [Note: It would be nice if we could simply let \(Y_{2}=\arctan\) \(\left(X_{2} / X_{1}\right)\), but in order to insure invertibility of the arctan function, it is defined to take on values only between \(-\pi / 2\) and \(\pi / 2\). Our specification of \(Y_{2}\) allows it to assume any value between 0 and \(2 \pi\).]

A store is expecting \(n\) deliveries between the hours of noon and 1 p.m. Suppose the arrival time of each delivery truck is uniformly distributed on this one-hour interval and that the times are independent of each other. What are the expected values of the ordered arrival times?

Show that if \(X, Y\), and \(Z\) are rv's and \(a\) and \(b\) are constants, then \(\operatorname{Cov}(a X+b Y, Z)=a \operatorname{Cov}(X, Z)+\) \(b \operatorname{Cov}(Y, Z)\)

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