6

What would be $\operatorname{Var}(X^2)$, if $\operatorname{Var}(X)=\sigma^2$?

Michael Hardy
  • 7,094
  • 1
  • 20
  • 38
MYaseen208
  • 2,379
  • 7
  • 32
  • 46
  • 1
    $Var[X] \stackrel{d}{=} \mathbb{E}[X^2] - (\mathbb{E}[X])^2 \to Var[X^2] = \mathbb{E}[X^4] - (\mathbb{E}[X^2])^2 $ or perhaps even more confusingly $Var[X^2] = \mathbb{E}[X^4] - (Var[X^2] + (\mathbb{E}[X])^2)$ – user2168 May 05 '11 at 04:18
  • 3
    In general there is no readily available formula. You can use delta method to get the approximation: here is [the relevant question](http://stats.stackexchange.com/questions/5782/variance-of-a-function-of-one-random-variable/5790#5790). – mpiktas May 05 '11 at 05:59

4 Answers4

13

As a simple example of the responses of @user2168 and @mpiktas: The variance of the set of values 1,2,3 is 0.67, while the variance of its square is 10.89. On the other hand, the variance of 2,3,4 is also 0.67, but the variance of the squares is 24.22.

These are just variances for finite sets of data, but the idea extends to distributions.

Nick Sabbe
  • 12,119
  • 2
  • 35
  • 43
  • 11
    Another similar example: if +1 and -1 are equally likely then the variance is 1, but the variance of the squares is 0; if 0 and 2 are equally likely then the variance is again 1 but the variance of the squares is 4. – Henry May 05 '11 at 07:43
4

Error propagation via Taylor's rule (aka "delta" method) --

$$\operatorname{Var}(X^2) \approx 4\operatorname{\mathbb{E}}(X)^2 \operatorname{Var}(X)$$

Michael Hardy
  • 7,094
  • 1
  • 20
  • 38
by Taylor
  • 64
  • 1
  • 2
    Do you mean $4 E(X)^2 Var(X)$ ? – Elvis Feb 02 '12 at 15:34
  • 2
    Hmmm... I wonder what happens in the delta method when $E(x)=0$, as in the standard Normal distribution ;-). – whuber Feb 02 '12 at 15:38
  • 1
    @whuber - you can add another term to the Taylor expansion underlying the delta method; if you do this, you get $\text{Var}(X^2) \approx \mu_4(X)$, the fourth central moment of $X$. – jbowman Feb 03 '12 at 01:59
  • @Elvis - you're right; I've fixed the formatting to make it clearer. – jbowman Feb 03 '12 at 02:03
  • @Elvis, no, I meant X^2, but you make a good point. This applies only to heteroskedastic errors, so.. not very useful to the actual question that was asked! – by Taylor Feb 03 '12 at 14:26
  • 1
    ?! $\mathrm{Var} \left(X^2\right) = 4 X^2 \mathrm{Var} (X)$ doesn’t make any sense, on the left side you have a constant and on the right a random variable. – Elvis Feb 04 '12 at 10:23
1

It's easy to see that the relationship between then is not constant by taking $X'=X+c$. Shifting a distribution by a constant doesn't affect the variance, but $Var((X+c)^2)$ can be made arbitrarily large. $Var(X^2)$ is a fourth-order statistic (i.e. is a combination of moments of order four and smaller), and cannot be written in terms of lower order statistics such as variance and mean.

Acccumulation
  • 3,688
  • 5
  • 11
-1

Error propagation via Taylor's rule (aka "delta" method) --

$$\operatorname{Var}(X^2) \approx 4\operatorname{\mathbb{E}}(X)^2 \operatorname{Var}(X) - \operatorname{Var}(X)^2 $$

Sorry i have expanded the taylor's rule in one extra order, because to just approximate the $\operatorname{Var}(X)$ linearly caused some problem with my algorithm, thought it would help other people to realize it's not linear...

Michael Hardy
  • 7,094
  • 1
  • 20
  • 38
yupbank
  • 141
  • 4
  • -1 You propose a solution that is *negative* whenever $E(X)$ is sufficiently small, which should make you uncomfortable, and you also suggest, incorrectly, that there is an "exact solution." There is no universal solution without making an explicit assumption about $E(X^4).$ – whuber Dec 18 '18 at 17:47
  • sorry i ran into this answer https://stats.stackexchange.com/questions/5782/variance-of-a-function-of-one-random-variable/5790#5790 which have \begin{align} Var(f(X))=[f'(EX)]^2Var(X)-\frac{[f''(EX)]^2}{4}Var^2(X)+\tilde{T}_3 \end{align} . i was using the \begin{align} \text{Var}(X^2) \approx 4\mathbb{E}(X)^2\text{Var}(X) \end{align} only and run into problems . and you are right since $ \tilde{T}_3 $ contains more than f'''(EX)* something, but ( f'''(EX)+something)*something – yupbank Dec 18 '18 at 20:43