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Chapter 1: Problem 91

Let \(X\) be a random variable such that \(E\left[(X-b)^{2}\right]\) exists for all real \(b\). Show that \(E\left[(X-b)^{2}\right]\) is a minimum when \(b=E(X)\).

Short Answer

Expert verified
The value of \(b\) that minimizes \(E[(X-b)^{2}]\) is \(b=E[X]\).

Step by step solution

01

Expand the Expected Value

Start by expanding \(E[(X-b)^{2}]\) into a more workable format. So, \(E[(X-b)^{2}] = E[X^2 - 2bX + b^2]\). Because the expectation operator is linear, this can be further simplified to \(E[X^2] - 2bE[X] + b^2\).
02

Differentiate

Differentiate \(E[X^2] - 2bE[X] + b^2\) with respect to \(b\) to determine where the function is a minimum. The derivative is \(-2E[X] + 2b\).
03

Set the Derivative Equal to Zero

Setting the derivative to zero will yield the value of \(b\) at the minimum of \(E[(X-b)^{2}]\). Thus, \(-2E[X] + 2b = 0\) which simplifies to \(b = E[X]\).
04

Second Derivative Test

Taking the second derivative of \(E[X^2] - 2bE[X] + b^2\), we get \(2\). Since the second derivative is greater than zero, \(b = E[X]\) is indeed a minimum.

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Most popular questions from this chapter

Let the probability set function of the random variable \(X\) be $$P(A)=\int_{A} e^{-x} d x, \quad \text { where } \mathscr{A}=\\{x ; 0

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