Prove that the SD of an asymmetric function is larger than the SD of a symmetric one

#1
How can I prove that the standard deviation of an asymmetric function (mean>median>mode) is larger than the standard deviation of a symmetric one (mean=median=mode) of the same variable?

I was asked if the statement "the standard deviation of an asymmetric function (mean>median>mode) is larger than the standard deviation of a symmetric one (mean=median=mode) of the same variable?" true or false and asked to prove the answer.
 
Last edited:

Dason

Ambassador to the humans
#2
Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

Is this homework?
 

Dason

Ambassador to the humans
#4
Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

You edited the question! That makes a big difference.

What do you think the answer is?
 
#5
Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

I think that the SD of an asymmetric function is larger than the SD of a symmetric one but can not prove it.
 

Dason

Ambassador to the humans
#6
Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

Let's say you have a A is an asymmetric function by your definition and S is a symmetric one. You claim A(X) > S(X).

What can you tell me about another function f such that f(X) = A(X)/c for some constant c. Is it asymmetric? What will the standard deviation of this be?
 

rogojel

TS Contributor
#7
Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

Hi,
this can not generally be true, can it? I mean for any assymetrical function one could construct a symmetrical one with a larger SD - is there some constraint on the functions that was not mentioned?

regards
rogojel
 

Dason

Ambassador to the humans
#8
Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

Hi,
this can not generally be true, can it?
I don't believe it is. Then again I've never heard the term asymmetric function as defined here before. But you point out the big reason I don't think it's true - there doesn't seem to be any constraint on the functions and as I was hinting to before you can arbitrarily increase or decrease the standard deviation just by multiplying the result by a constant.
 

rogojel

TS Contributor
#9
Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

Hi,
I completely missed the hint. I thought that it shows how one can prove the statement and wondered what I am missing.

regards
rogojel