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Chapter 1: Problem 8

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
By expanding and taking the derivative of the expected squared difference, it can be shown that the expectation is minimized when \(b\) is equal to the expected value of the random variable.

Step by step solution

01

Expand the squared term

Start by expanding the term \((X-b)^{2}\), which gives us \(X^{2} - 2bX + b^{2}\). Then take the expectation of the result.
02

Rewrite the expected value

Using the properties of expectations, the expected value of the expanded term can be written as \(E\left[(X-b)^{2}\right] = E[X^{2}] - 2bE[X] + b^{2}\).
03

Take the derivative with respect to \(b\)

Now take the derivative of this expression with respect to \(b\). This leads to an equation \(0 = -2E[X] + 2b\). Solve it to find the value of \(b\) which minimizes the expression.
04

Solve for \(b\)

Solve the equation for \(b\) to get \(b = E[X]\). This shows that \(E\left[(X-b)^{2}\right]\) is indeed a minimum when \(b=E[X]\).

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