# Expansion of variance in terms of expected value

From page 1 (page 13 in the .pdf file), equations 1.3 and 1.4, in this book I find out that

$var[r] = E[(r - E[r])^2] = E[r^2] - (E[r])^2$

Is this true in general and how do I show that it is true?

I have started by expanding

$E[(r - E[r])^2] = E[(r - E[r])^2] = E[r^2 + (E[r])^2 - 2 \cdot E[r] \cdot r]$

How do I continue?

• Expectation is linear. Let $\mu=E[r]$. Then $E[r^2-2\mu r+\mu^2]=E[r^2]-2\mu E[r]+\mu^2=E[r^2]-2\mu^2+\mu^2$. – André Nicolas Apr 29 '15 at 16:28

You may proceed as follows:

$$E[r^2+(E[r])^2-2E[r]r] = E[r^2]+E[E[r]^2]-E[2E[r]r] = E[r^2]+E[r]^2-2E[r]E[r] = E[r]^2-E[r]^2$$ Note: $E[r]$ is a constant. So it can come out of $E$ as you see for example in $E[2E[r]r] = 2E[r]E[r]$.

This question is a bit old, but I would still like to clarify in details how do you actually obtain $E[r^2]−(E[r])^2$ starting from:

$$E[(r - E[r])^2] = E[(r - E[r])^2] = E[r^2 + (E[r])^2 - 2 \cdot E[r] \cdot r]$$

because it could not be obvious for a newbie.

So far, we are here:

$$E[r^2 + (E[r])^2 - 2 \cdot E[r] \cdot r]$$

If we assert that $r$ is our set of $n$ elements:

$$r = \{ r_1, r_2, r_3, r_4, ... r_n \}$$

and that $E[r]$ is the mean $\mu$, then we know that:

$$E[r] = \mu = \frac{1}{n} \sum\limits_{i=1}^n r_i$$

So we can go on and rewrite the formula as a summation:

$$E[r^2 + (E[r])^2 - 2 \cdot E[r] \cdot r] = \\ \frac{1}{n} \biggl(\sum\limits_{i=1}^n r^2_i + E[r]^2 - 2 \cdot E[r] \cdot r_i \biggr)$$

And leverage the properties of summations:

$$\frac{1}{n} \biggl(\sum\limits_{i=1}^n r^2_i + E[r]^2 - 2 \cdot E[r] \cdot r_i \biggr) = \\ \\ \frac{1}{n} \biggl(\sum\limits_{i=1}^n r^2_i + \sum\limits_{i=1}^n E[r]^2 - \sum\limits_{i=1}^n 2 \cdot E[r] \cdot r_i \biggr) = \\ \\ \frac{\sum\limits_{i=1}^n r^2_i}{n} + \frac{\sum\limits_{i=1}^n E[r]^2}{n} - \frac{\sum\limits_{i=1}^n 2 \cdot E[r] \cdot r_i}{n}$$

Now, take a look at the last term:

$$\frac{\sum\limits_{i=1}^n 2 \cdot E[r] \cdot r_i}{n}$$

Again, if we use the summations' properties:

$$\frac{\sum\limits_{i=1}^n 2 \cdot E[r] \cdot r_i}{n} = \\ 2 \cdot E[r] \cdot \frac{\sum\limits_{i=1}^n r_i}{n}$$

We can effectively see that this multiplication has the mean in one of its terms!

$$2 \cdot E[r] \cdot \frac{\sum\limits_{i=1}^n r_i}{n} = 2 \cdot E[r] \cdot \frac{1}{n} \sum\limits_{i=1}^n r_i = \\ 2 \cdot E[r] \cdot \mu = 2 \cdot E[r] \cdot E[r]$$

So, let's go back to the whole formula we are expanding:

$$\frac{\sum\limits_{i=1}^n r^2_i}{n} + \frac{\sum\limits_{i=1}^n E[r]^2}{n} - \frac{\sum\limits_{i=1}^n 2 \cdot E[r] \cdot r_i}{n} = \\ \frac{\sum\limits_{i=1}^n r^2_i}{n} + \frac{\sum\limits_{i=1}^n E[r]^2}{n} - 2 \cdot E[r] \cdot E[r]$$

We proceed similarly with the other two terms:

1) $$\frac{\sum\limits_{i=1}^n r^2_i}{n} = \\ \frac{1}{n}\sum\limits_{i=1}^n r^2_i = E[r^2]$$

2) $$\frac{\sum\limits_{i=1}^n E[r]^2}{n} = \\ \frac{1}{n}\sum\limits_{i=1}^n E[r]^2 = E[E[r]^2] = \frac{1}{n} \cdot \biggl(n \cdot E[r]^2\biggr) = E[r]^2$$

And finally we get:

$$E[r^2] + E[r]^2 - 2 \cdot E[r] \cdot E[r] = E[r^2] + E[r]^2 - 2E[r]^2 = E[r^2] - E[r]^2$$

Which is what the OP was asking.

Hope this helps someone.