Let [math]X[/math] be a random variable with [math]E(X) = \mu[/math] and [math]Var(X) = \sigma^2[/math]. Suppose

[[math]]E[(X-c_0)^2] \leq E[(X-c)^2][[/math]]

for all [math]c[/math]. Determine [math]E[(X-c_0)^2] [/math].

• [math]\frac{\sigma^2}{2}[/math]
• [math]\mu^2[/math]
• [math]\sigma^2[/math]
• [math]\min(\sigma^2,\mu^2)[/math]
• [math]\frac{\sigma^2 + \mu^2}{2}[/math]

Solution: C

The function [math]e(c) = E[(X-c)^2] [/math] is a quadratic in [math]c[/math] therefore its minimum is achieved when its derivative equals zero. Take the derivative and set it to zero:

[[math]] e'(c_0) =-2E[(X-c_0)] = 0 \implies c_0 = E[X]. [[/math]]

Hence the minimizer equals the expected value, and therefore the answer is automatically [math]\sigma^2[/math].