# Minimizing Expected Value

## Problem

Suppose $$X$$ is a random variable with $$E[X^2] < \infty$$. What is the constant $$c$$ that minimizes $$E[(X - c)^2]$$?

## Solution

As with most optimization problems, we will begin by taking the derivative of $$E[(X - c)^2]$$ with respect to $$c$$. To do so, we replace the expectation with its definition, letting $$f(x)$$ represent the probability density function of $$X$$.

$\begin{split} \dfrac{d}{dc}E\left[(X - c)^2\right] &= \dfrac{d}{dc} \int_{-\infty}^{\infty} (x-c)^2f(x)dx \\ &= -2\int_{-\infty}^{\infty}(x-c)f(x)dx \\ &= -2\left[ \int_{-\infty}^{\infty}xf(x)dx - \int_{-\infty}^{\infty}cf(x)dx \right] \\ &= -2\left( E[X] - c \right) \\ \end{split}$

Next, we set the derivative equal to $$0$$ and solve for $$c$$:

$-2\left( E[X] - c \right) = 0$ $c = E[X]$

Thus, $$E[(X - c)^2]$$ is minimized by $$c = E[X]$$. We confirm this by noting $$\dfrac{d^2}{dc}E[(X - c)^2] = 2$$, indicating upward concavity.

## Alternative Approach

There is another, cleaner approach to finding the solution. We begin by utilizing the linearity of expectation property:

$E[(X - c)^2] = E[X^2 - 2cX + c^2] = E[X^2] - 2cE[X] +c^2$

Then, we take the derivative of this new formulation and set the result to $$0$$:

$\dfrac{d}{dc} \left[ E[X^2] - 2cE[X] +c^2 \right] = 0$ $-2E[X] + 2c = 0$ $c = E[X]$