Mean Absolute Deviation and Standard Deviation average, standard, absolute, positive, 2011

agulani

For a normal distribution , if the mean and the median values are zero , and the average of positive values and the absolute value of average of negative values is 1% , what is the standard deviation?



Edited 1 time(s). Last edit at Wednesday, June 15, 2011 at 03:38PM by janakisri.

ryanlb

This is only true for E(X) = 0. The formula for E(x) = m is much more complicated.

x ~ normal(0, var(x))
1% = E(|x|) = ((2*var(x))/pi)^(1/2)

so

stdev(x) = (pi/2)^.5 = 1.2533...%

tango_gs

Thanks MathMan , you're really good

bleach

What? Really?
So for the set {1, -1, 1, -1, 1, -1 ... } the standard deviation is 1.2533?

JonnyKay

Is {1, -1, 1, -1, 1, -1 ... } normally distributed?

seemorr Wrote:
-------------------------------------------------------
> What? Really?
> So for the set {1, -1, 1, -1, 1, -1 ... } the
> standard deviation is 1.2533?

Viceroy

MathMan's derivation is correct. If you use

E(|x|)=(2/(sqrt(2pi)*sigma)*intergral from 0 to infinity of x*exp(-x^2/2sigma^2)dx, then change variable (for example, y = x^2/2sigma^2), you will get his formula.

Since the question was about normal distribution with zero mean, the answer is correct. If distributional assumption is different, the answer would be different as well. If you'd like to come up with a range of values for a wide range of symmetric distributions, you can try to minimize and maximize sqrt((1/n)*sum(x_k^2)) given 1/n*(sum(x_k^2))=1% and all x_k>=0. That might a fun problem to think about.

ba736

Oh I guess {1, -1, 1, -1 ...} isn't normally distributed ...



Edited 1 time(s). Last edit at Tuesday, June 21, 2011 at 04:37PM by seemorr.

lucasg85

Not true that "{1, -1, 1, -1, 1, -1 ...} is clearly normally distributed"

If your pattern contains only 1's and -1s , how can it be normally distributed? Every normal distribution chart I've seen is shaped like a bell ( hence bell shaped curve).

The pattern of alternating 1's and -1s results in a frequency diagram with two vertical , equal height lines , one at +1 and one at -1

lunarfollies

Variance bound is [E(|x|)^2 – E(x)^2, inf).

Regarding the variance bounds for this type of problem, the following is true regardless of symmetry.

Var(|x|) = E(|x|^2) – E(|x|)^2 = E(x^2) – E(|x|)^2
Var(x) = E(x^2) – E(x)^2

So, Var(|x|) + E(|x|)^2 = Var(x) + E(x)^2

In this case, E(|x|) = 1 and E(x) = 0. So,

Var(|x|) + 1 = Var(x)

Since we know Var(|x|) >= 0 then the minimum value of Var(x) is 0 + 1 = 1. Seemorr already produced a distribution where Var(|x|) = 0 is possible. (Generally the minimum is E(|x|)^2 – E(x)^2, note that E(|x|)^2 >= E(x)^2)

So what is the max….well, assume a case where the distribution of |x| is lognormal. It should be clear that when we set E(|x|) = 1 (or any constant), Var(|x|) is unbounded above.

This is far from mathematical proof but it should also be obvious from above that combinations of seemorr’s distribution and a lognormal distribution for |x| should be able to produce a set where E(|x|) = 1, E(x) = 0, and Var(x) is equal to whatever we want.