Chapter 4: Problem 50
\( \quad\) Let \(X\) have finite variance, and set \(v(x)=\mathbf{E}(X-x)^{2}\). Show that \(\mathbf{E} v(X)=2 \operatorname{var} X\).
Short Answer
Expert verified
The result holds: \( \mathbf{E}[v(X)] = 2\operatorname{var}(X) \).
Step by step solution
01
Expand the Expression
The function given is \( v(x) = \mathbf{E}(X-x)^2 \), which represents the expected value of the squared difference between \(X\) and \(x\). We can expand \((X-x)^2\) using the binomial expansion: \((X-x)^2 = (X^2 - 2Xx + x^2)\).
02
Compute the Expected Value of the Expression
Substitute the expanded form into the expectation: \[ \mathbf{E}[v(X)] = \mathbf{E}[\mathbf{E}(X^2 - 2Xx + x^2)] = \mathbf{E}[\mathbf{E}(X^2) - 2\mathbf{E}(X)x + \mathbf{E}(x^2)] \] Using the linearity of expectation, this simplifies to: \[ \mathbf{E}(X^2) - 2x\mathbf{E}(X) + x^2 \].
03
Simplify with Properties of Variance
Recognize that \( \mathbf{E}(X^2) = \operatorname{var}(X) + \mathbf{E}(X)^2 \). Thus, substituting in our equation we have: \[ \mathbf{E}[v(X)] = \operatorname{var}(X) + \mathbf{E}(X)^2 - 2x\mathbf{E}(X) + x^2 \] Further, variance is defined as \( \operatorname{var}(X) = \mathbf{E}(X^2) - (\mathbf{E}(X))^2 \).
04
Use Definition of Variance
Substitute back \( \operatorname{var}(X) = \mathbf{E}(X^2) - (\mathbf{E}(X))^2 \) and rewrite: \[ \mathbf{E}[v(X)] = \operatorname{var}(X) + \mathbf{E}(X)^2 - (\mathbf{E}(X))^2 \] This simplifies to: \[ 2\operatorname{var}(X) \].
05
Conclude the Verification
We have shown that \( \mathbf{E}[v(X)] = 2\operatorname{var}(X) \) through correct expansion, usage of variance properties, and expectation properties. Therefore, affirmative: the problem statement is valid and proven.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Expected Value
The expected value is a fundamental concept in probability and statistics. It represents the average or mean value that you would expect from a random variable if the experiment were repeated many times. For a random variable \(X\), the expected value is denoted by \(\mathbf{E}(X)\). It is calculated by summing the products of all possible outcomes and their respective probabilities:
- \[ \mathbf{E}(X) = \sum_{i} x_i \, P(x_i) \]
Binomial Expansion
The binomial expansion is a powerful algebraic method for expanding expressions that are squared or raised to a power. In the context of the exercise, it applies when you need to expand \((X-x)^2\). The binomial theorem outlines the expansion of expressions like this:
- \[ (X-x)^2 = X^2 - 2Xx + x^2 \]
Linearity of Expectation
Linearity of expectation is a property that greatly simplifies the calculation of expected values, particularly in expressions with sums. This property states that the expected value of a sum of random variables equals the sum of their expected values, regardless of whether these variables are dependent or independent:
- \[ \mathbf{E}(aX + bY) = a\mathbf{E}(X) + b\mathbf{E}(Y) \]
Properties of Variance
Variance is a measure of the spread or dispersion of a random variable's possible values. It tells us how much the values deviate from the mean. The variance of a random variable \(X\) with expected value \(\mathbf{E}(X)\) is given by:
- \[ \operatorname{var}(X) = \mathbf{E}(X^2) - (\mathbf{E}(X))^2 \]
- \(\operatorname{var}(aX) = a^2 \operatorname{var}(X)\)