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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
The minimum of function \(E\left[(X-b)^{2}\right]\) is achieved when \(b = E[X]\).

Step by step solution

01

Differentiate the function with respect to \(b\)

The first step would be to differentiate the function \(E\left[(X-b)^{2}\right]\) with respect to \(b\). Let's expand it first and then differentiate: \(E\left[(X-b)^{2}\right] = E[X^{2} - 2bX + b^{2}] = E[X^{2}] - 2bE[X] + b^{2}\). The derivative with respect to \(b\) yields: \(\frac{dE\left[(X-b)^{2}\right]}{db} = -2E[X] + 2b\).
02

Find the value of \(b\) at which the function is minimum

To find the minimum of the function, set the derivative to zero and solve for \(b\): \(-2E[X] + 2b = 0\). This results in \(b = E[X]\).
03

Check the second derivative to confirm the minimum

To confirm that this is indeed the minimum, we calculate the second derivative and verify if it is positive: \(\frac{d^{2}E\left[(X-b)^{2}\right]}{db^{2}} = 2\). Since the second derivative is positive, it indicates that the function \(E\left[(X-b)^{2}\right]\) is a minimum at \(b = E[X]\).

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