# A student asks: Why does variance do THAT?

A student asks:

The mark scheme says $Var(2 - 3X) = 9 Var(X)$. Where on earth does that come from?

Great question, which I’m going to answer in two ways. Firstly, there’s the instinctive reasoning; secondly, there’s the maths behind it, just to make sure.

### Instinctively

Well, instinctively, you’d think it was $2 - 3 Var(X)$, obviously, but it’s not. You need better instincts than that!

In this example, you’ve taken your random variable X, you’ve stretched it by a scale factor of -3 and moved it across two. Both of those have an effect on the mean: $E(2-3X)$ *does* behave as you’d instinctively think, and give you $2 - 3E(X)$.

However, the variance is a measure of spread - in fact, it’s the square of the standard deviation. Moving the variable sideways doesn’t change the spread. Multiplying it by -3 does, though: it multiplies the *standard deviation* by three ((The standard deviation is always positive, so we ignore the minus sign.)) That means the variance is multiplied by 9, as the mark scheme says.

### Mathematically

Let’s take a look at the definition of the variance:

$Var(X) = \frac 1n (x - \bar x)^2$ (other equivalent definitions are available; in most cases, I prefer “mean of the squares minus the square of the mean”).

Now, replace $x$ with $(2-3x)$ and $\bar x$ with $(2 - 3\bar x)$. You’ve got:

$Var(2-3X) = \frac 1n \left( (2-3x) - (2-3\bar x) \right)^2$

Tidying that bracket up:

$Var(2-3X) = \frac 1n \left ( -3x + 3 \bar x \right) = \frac 1n \left( -3(x - \bar x \right)^2$ $ = \frac 1n (-3)^2 (x-\bar x )^2 = 9 Var(X)$

You can do the same thing more generally with $a + bX$ if you like, but it works out in exactly the same way.