Exercise on Expected Value and Variance

Question

Let $A$ be a discrete random variable with expected value $a$ and variance $b^2$

$$a:=\Bbb E(A)~~\,\\b^2:=\Bbb{Var}(A)$$

The function F is defined as $$F(x):=\Bbb E[(A-x)^2]$$ Show that $$b^2=(1/2)\cdot \Bbb E(F(A))$$

Hint: Express the function $F(x)$ through $a$, $b^2$ and $A$.

I tried a few things but nothing really worked. Thanks in advance!

Answer 1

We have the constants $a$ and $b$ defined as: $a:=\Bbb E(A)$ and $b^2:=\Bbb {Var}(A)$.   It will be useful to note that this means: $\Bbb E(A^2)=b^2+a^2$, by the definition of variance.

$F(x)$, being defined as $\Bbb E((A-x)^2)$, is a function of free variable $x$, with no free occurrence of random variable $A$, since the linear operation (expectation) binds that random variable.

$$\begin{align}F(x)&=\Bbb E(A^2-2Ax+x^2)\\&=\text{a function of free variable $x$, and constants $a,b$}\end{align}$$

$F(A)$ is that function with all free occurrences of $x$ replaced by $A$.   This replacement must occur after the evaluation of the expectation to avoid collision with the bound occurrence. In short:$$\begin{align}F(A) &= \text{a function of random variable $A$, and constants $a,b$}\end{align}$$

The random variable $A$ occurs free in this function, so we take its expectation as we normally do.

$$\begin{align}\Bbb E(F(A))&=\\&~~\vdots\\&=2b^2\end{align}$$

And so it is done.

Answer 2

You are told $\mathbb{E}[A] = a$, $Var[A] = b^2$, from which we obtain $\mathbb{E}[A^2] = a^2+b^2$

Assumption: To clarify notations, will assume x to be an instance of Random Variable X. For clarity, I will mark this as $F(X=x)$ instead of $F(x)$. Further, since we are not told in $F(x)$, what the expectation is over, we will presume this to be over A. In this notation, we will write $$F(X=x) := \mathbb{E}_{A}[(A-x)^2]$$

Then we can write: \begin{align*} F(X=x)&=\mathbb{E}_{A}[(A-x)^2]\\ &= \mathbb{E}_{A}[A^2] -2 \mathbb{E}_{A}[Ax]+\mathbb{E}_{A}[x^2]\\&= \mathbb{E}_{A}[A^2] -2x \mathbb{E}_{A}[A]+x^2\\& = a^2+b^2-2ax+x^2 \end{align*} Then, \begin{align*} \mathbb{E}_X[F(X=x)]&=\mathbb{E}_X[a^2+b^2-2ax+x^2]\\ &=a^2+b^2-2a\mathbb{E}_X[X=x]+\mathbb{E}_X[(X=x)^2]\\ \mathbb{E}_A[F(A)]&=a^2+b^2-2a\mathbb{E}_A[A]+\mathbb{E}_A[A^2]\\ &=a^2+b^2-2a\cdot a+(a^2+b^2)\\ &=2b^2\\ \end{align*} which proves $b^2=\frac{\mathbb{E}_A[F(A)]}{2}$ as required

Observation: The question is rather unclear in the first part on how X and A are related. For instance, if one uses $\mathbb{E}_{A|X=x}(\cdot)$ instead of $\mathbb{E}_{A}(\cdot)$ for $F(X=x)$, then, one ends up with an answer of 0 for the given question. Thus one has to assume that x is a free variable (as @Graham Kemp points out in comments)