Question

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

Short answer

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

Step-by-step solution

Differentiate the function with respect to $b$

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

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 $b=E[X]$.

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: $\frac{d^{2}E[(X-b{)}^2]}{d{b}^2}=2$. Since the second derivative is positive, it indicates that the function $E[(X-b{)}^2]$ is a minimum at $b=E[X]$.