/*! 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 88 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 $$ f(x, y)=\left\\{\begin{array}{cl} \frac{2}{5}(2 x+3 y) & 0 \leq x \leq 1,0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ 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 mean squared error is the expected value \(E[X+Y]\).

Step by step solution

01

Understand the Problem

We need to predict the total score \(X+Y\) such that the mean squared error \(E[(X+Y-t)^{2}]\) is minimized. Our goal is to find a value of \(t\) that minimizes this function.
02

Formulate the Mean Squared Error

The prediction error is given by the mean squared error function: \[ E\left[(X+Y-t)^{2}\right] = \int\int (x+y-t)^2 f(x,y)\, dx\, dy \] where \(f(x,y)\) is the joint pdf.
03

Optimal Value of t

To minimize the mean squared error \( E\left[(X+Y-t)^{2}\right] \), \(t\) should be the expected value of \(X+Y\), which is \( E[X+Y] \). This is a standard result from probability theory.
04

Find the Expected Values

The expected value \(E[X]\) is given by: \[ E[X] = \int_0^1 \int_0^1 x \cdot f(x,y) \, dy \, dx \] The expected value \(E[Y]\) is given by: \[ E[Y] = \int_0^1 \int_0^1 y \cdot f(x,y) \, dy \, dx \] We need to calculate these integrals using the given joint pdf.
05

Compute E[X]

Substitute \(f(x,y) = \frac{2}{5}(2x+3y)\) into the integral for \(E[X]\): \[ E[X] = \int_0^1 \int_0^1 x \cdot \frac{2}{5}(2x+3y) \, dy \, dx \] Evaluate the inner integral with respect to \(y\), and then evaluate the outer integral with respect to \(x\).
06

Compute E[Y]

Similarly, find \(E[Y]\) using: \[ E[Y] = \int_0^1 \int_0^1 y \cdot \frac{2}{5}(2x+3y) \, dy \, dx \] Evaluate the inner integral with respect to \(y\), followed by the outer integral with respect to \(x\).
07

Calculate E[X+Y]

Now compute \(E[X+Y] = E[X] + E[Y]\), as \(X\) and \(Y\) are defined over the same range with the given joint distribution.
08

Conclusion

Set \(t = E[X+Y]\), since this value minimizes the mean squared error \(E\left[(X+Y-t)^{2}\right]\).

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Key Concepts

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

Joint Probability Density Function
A joint probability density function (joint pdf) describes the likelihood of two random variables, such as verbal score \(X\) and quantitative score \(Y\), occurring together.
It gives us a way to simultaneously study their behavior and dependence on each other. In our scenario, the joint pdf is defined as: \[ f(x, y)=\left\{\begin{array}{cl} \frac{2}{5}(2 x+3 y) & 0 \leq x \leq 1, 0 \leq y \leq 1 \ 0 & \text{ otherwise } \end{array}\right. \] This function is valid where both \(x\) and \(y\) are between 0 and 1.
Outside this region, the joint pdf is zero, meaning those combinations are impossible.
  • The joint pdf must integrate to 1 over all possible \(x\) and \(y\) values.
  • It helps us find marginal pdfs by integrating out one variable.
  • It supports calculating expectations and variances for joint distributions.
In practice, having a joint pdf allows us to explore the extent to which variables like \(X\) and \(Y\) contribute to outcomes such as the total score \(X + Y\).
Expected Value
The expected value of a random variable is like a weighted average where probabilities weigh the outcomes.
It's an important concept because it provides a single summary metric for overall behavior of random variables.
For our problem, it is crucial for predicting the total score \(X + Y\). To find the expected value \(E[X + Y]\), we calculate each component's expected value first. - \(E[X]\) involves the integral \[\int_0^1 \int_0^1 x \cdot f(x, y) \, dy \, dx\]- Similarly, \(E[Y]\) can be found with \[\int_0^1 \int_0^1 y \cdot f(x, y) \, dy \, dx\]The combined value \(E[X + Y] = E[X] + E[Y]\) is then found by adding these results together.
Once calculated, \(E[X + Y]\) is the value of \(t\) that minimizes the mean squared prediction error, as it represents the most likely combined verbal and quantitative score given our joint pdf.
Prediction Error
Prediction error occurs when the predicted value deviates from the actual observed value.
In statistics, we often use the Mean Squared Error (MSE) to quantify this deviation. The formula for MSE in this context is: \[ E\left[(X+Y-t)^{2}\right] = \int\int (x+y-t)^2 f(x,y)\, dx\, dy \] This expression measures how much the prediction \(t\) differs by squaring the error, making large differences more significant. We sum over all possible \((x, y)\) since each has a probability from our joint pdf.
By minimizing this difference, we ensure our prediction \(t\) is as close to \(X + Y\) as possible, in average terms, across all realizations.
Minimization
Minimization refers to the process of finding the smallest possible value of a function.
In statistics, it often deals with reducing errors or variances to improve predictions. For mean squared error \(E\left[(X+Y-t)^{2}\right]\), minimization is important because it helps in identifying the best prediction \(t\) to represent \(X + Y\). To minimize MSE:
  • Calculate the derivative of the MSE with respect to \(t\).
  • Set this derivative to zero and solve for \(t\).
  • Verify that this \(t\) is indeed a minimum point by checking the second derivative.
In our case, solving this gives \(t = E[X + Y]\), meaning the expected value is the optimal prediction. This step ensures our prediction closely aligns with the observed values, reducing average prediction error across all possible outcomes.

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Most popular questions from this chapter

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