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Chapter 4: Problem 35

If \(Y\) is a continuous random variable such that \(E\left[(Y-a)^{2}\right]<\infty\) for all \(a,\) show that \(E\left[(Y-a)^{2}\right]\) is minimized when \(a=E(Y) .[\) Hint: \(\left.E\left[(Y-a)^{2}\right]=E\left(\\{[Y-E(Y)]+[E(Y)-a]\\}^{2}\right) .\right]\)

Short Answer

Expert verified
The expression is minimized when \(a = E(Y)\).

Step by step solution

01

Express the variance formula

Using the hint, we rewrite the expression for mean squared error: \[ E[(Y-a)^2] = E\left(\{[Y-E(Y)]+[E(Y)-a]\}^{2}\right) \] This expands to: \[ E[(Y-a)^2] = E\left( (Y - E(Y))^2 + 2(Y-E(Y))(E(Y)-a) + (E(Y)-a)^2 \right) \]
02

Simplify using linearity of expectation

Due to the linearity of expectation, we can distribute the expectation operator: \[ E[(Y-a)^2] = E[(Y-E(Y))^2] + 2E\left((Y-E(Y))(E(Y)-a)\right) + E[(E(Y)-a)^2] \] Now we examine each term separately.
03

Evaluate the terms

1. **First term:** \[ E[(Y-E(Y))^2] = \text{Var}(Y) \] which is a constant.2. **Second term:** \[ 2(E(Y)-a)E[(Y-E(Y))] = 2(E(Y)-a)\cdot0 = 0 \] because \(E[Y-E(Y)] = 0\).3. **Third term:** \[ E[(E(Y)-a)^2] = (E(Y)-a)^2 \] as it does not depend on \(Y\) and is treated as a constant.
04

Finalize the expression and find the minimum

Putting it all together:\[ E[(Y-a)^2] = \text{Var}(Y) + 0 + (E(Y)-a)^2 \]Since \(\text{Var}(Y)\) is a constant, the expression is minimized when \( (E(Y)-a)^2 = 0 \), i.e., \( a = E(Y) \).
05

Conclusion

Thus, the value of \(a\) that minimizes \(E[(Y-a)^2]\) is indeed \(a = E(Y)\), as the proof confirms.

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Key Concepts

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

Mean Squared Error
Mean squared error (MSE) is a fundamental concept that often shows up in statistics and data analysis. MSE measures how much the values of a variable deviate from a chosen target or expected value, represented by \( a \).
To understand MSE, consider the formula \( E[(Y-a)^2] \). Here, \( Y \) is a continuous random variable, and \( a \) is a constant we are trying to adjust for minimum error. Essentially, MSE is about finding the average of the squared differences between actual values \( Y \) and the constant \( a \).
The squaring part serves two purposes:
  • It transforms all deviations to non-negative values, eliminating issues with negative differences.
  • It places greater emphasis on larger deviations, making it a more sensitive measure of error.
By minimizing the MSE, we aim to choose an \( a \) that makes the actual data as close as possible to the expected value.
Linearity of Expectation
Linearity of expectation is a crucial rule in probability and statistics useful in simplifying complex expectation expressions.
This principle states that the expected value of a sum of random variables is the same as the sum of their individual expected values. Formally, it can be written as: \( E[X + Y] = E[X] + E[Y] \), regardless of whether \( X \) and \( Y \) are independent.
In the solution, linearity of expectation helps break down the expression:
  • \( E[(Y-a)^2] = E[(Y-E(Y))^2] + 2E[(Y-E(Y))(E(Y)-a)] + E[(E(Y)-a)^2] \).
  • Each component can be evaluated separately using this property.
The beauty of linearity is its universal applicability, making it an essential tool for both theoretical analysis and practical problem-solving.
Variance
Variance is another statistical measure, closely related to mean squared error, that quantifies the spread or variability of a set of data points.
The variance of a random variable \( Y \), denoted as \( \text{Var}(Y) \), is calculated using the formula \( E[(Y - E(Y))^2] \). This aligns directly with the first term in our expression \( E[(Y-a)^2] \).
Variance tells us how much the values of \( Y \) are expected to spread around their mean \( E(Y) \). When \( E[(Y-a)^2] \) is rewritten using variance, it highlights that:
  • The portion related to variance \( \text{Var}(Y) \) remains constant because it does not depend on \( a \).
  • The other terms are influenced solely by the choice of \( a \).
Understanding variance's role helps in recognizing that minimizing variance leads to selecting \( a \) as close as possible to the mean.
Minimization Problem
The minimization problem in this context involves determining the value of \( a \) that will minimize \( E[(Y-a)^2] \). This type of problem is central in optimization and statistical estimation tasks.
By setting the derivative of \( E[(Y-a)^2] \) with respect to \( a \) to zero, we identify that the minimum is achieved when \( a = E(Y) \). This is because:
  • The expression \( (E(Y)-a)^2 \) achieves its lowest value, zero, when \( a = E(Y) \).
  • Consequently, \( E[(Y-a)^2] = \text{Var}(Y) + (E(Y)-a)^2 \) is minimized.
Minimization ensures the expected differences between \( Y \) and \( a \) are as small as possible, optimizing the representation of \( Y \) with respect to \( a \). This strategy is vital for effective data modeling and accurate risk assessment.

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Most popular questions from this chapter

A random variable \(Y\) is said to have a log-normal distribution if \(X=\ln (Y)\) has a normal distribution. (The symbol In denotes natural logarithm.) In this case \(Y\) must be non negative. The shape of the log-normal probability density function is similar to that of the gamma distribution, with a long tail to the right. The equation of the log-normal density function is given by $$ f(y)=\left\\{\begin{array}{ll} \left.\frac{1}{\sigma y \sqrt{2 \pi}} e^{-(\ln (y)-\mu)^{2} / 2 \sigma^{2}}\right), & y>0 \\ 0, & \text { elsewhere } \end{array}\right. $$ Because \(\ln (y)\) is a monotonic function of \(y\) $$ P(Y \leq y)=P[\ln (Y) \leq \ln (y)]=P[X \leq \ln (y)] $$ where \(X\) has a normal distribution with mean \(\mu\) and variance \(\sigma^{2} .\) Thus, probabilities for random variables with a log-normal distribution can be found by transforming them into probabilities that can be computed using the ordinary normal distribution. If \(Y\) has a log-normal distribution with \(\mu=4\) and \(\sigma^{2}=1,\) find a. \(P(Y \leq 4)\) b. \(P(Y>8)\)

The length of time required to complete a college achievement test is found to be normally distributed with mean 70 minutes and standard deviation 12 minutes. When should the test be terminated if we wish to allow sufficient time for \(90 \%\) of the students to complete the test?

The length of time required by students to complete a one-hour exam is a random variable with a density function given by $$f(y)=\left\\{\begin{array}{ll}c y^{2}+y, & 0 \leq y \leq 1, \\\0, & \text { elsewhere. } \end{array}\right.$$ a. Find \(c\). b. Find \(F(y)\). c. Graph \(f(y)\) and \(F(y)\). d. Use \(F(y)\) in part (b) to find \(F(-1), F(0),\) and \(F(1)\). e. Find the probability that a randomly selected student will finish in less than half an hour. f. Given that a particular student needs at least 15 minutes to complete the exam, find the probability that she will require at least 30 minutes to finish.

A circle of radius \(r\) has area \(A=\pi r^{2} .\) If a random circle has a radius that is uniformly distributed on the interval \((0,1),\) what are the mean and variance of the area of the circle?

The life length \(Y\) of a component used in a complex electronic system is known to have an exponential density with a mean of 100 hours. The component is replaced at failure or at age 200 hours, whichever comes first. a. Find the distribution function for \(X\), the length of time the component is in use. b. Find \(E(X)\)
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