Standard deviation of {1, -1, i, -i}

Question

(just in case: $i^2 = -1$)

mean: $$m = \frac{1 - 1 + i - i}{4} = 0$$

standard deviation: $$\sigma = \sqrt{\frac{1^2 + (-1)^2 + i^2 + (-i)^2}{4}} = \sqrt{\frac{1 + 1 - 1 - 1}{4}} = 0$$

It feels wrong, but I don't understand how or why. Would anyone tell me why it's wrong?

Answer 1

Because you are using complex numbers in a context where only real numbers make sense.

Answer 2

It's wrong because you need to use the magnitude of the each element not its square. This is because complex numbers ($a + bi$) are two dimensional vectors where their length is given by $\sqrt{a^2+b^2}$. This magnitude can also be found by multiplying by the complex conjugate (ie: $|a+bi|^2 = (a+bi)(a-bi)$)

This now will give the correct answer to

$$\sigma = \sqrt{\frac{|1|^2+|-1|^2+|i|^2+|-i|^2}{4}} = \sqrt{\frac{(1)(1)+(-1)(-1)+(i)(-i)+(-i)(i)}{4}} = 1 $$

Answer 3

I would convert them as follows:

$1 → 1 + 0i$

$-1 → -1 + 0i$

$i → 0 + 1i$

$-i → 0 + -1i$

Then take the mean and standard deviation of the real and imaginary components separately, e.g. mean and standard deviation of $\{1, -1, 0, 0\}$. For both, mean is $0.000$ and standard deviation is $(⅔)^½ ≈ 0.816$.

Real: $(0.000 ± 0.816)$

Imaginary: $(0.000 ± 0.816)i$

Or $(0.000 ± 0.816) + (0.000 ± 0.816)i$, which you could plot as an ellipse centered at $(0, 0)$ with both radii at $0.816$ representing one standard deviation.

Using this representation and by systemically handling the components, you can define addition, multiplication, negation $-E$, and inverse $E⁻¹$ operations on ellipses. Whether this has any use is debatable!