Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My instructor insists that $$\mathrm{Var}(X_{2\text{months}}) = 2 \mathrm{Var}(X_{1\text{month}}) $$ with independent random variables $X_{1\text{month}}$ and $X_{2\text{months}}$, but generally with a random variable $Y$ $$\mathrm{Var}(2Y) = 2^2\mathrm{Var}(Y).$$ So $$\mathrm{Var}(X_{2\text{months}}) = \mathrm{Var}(2X_{1\text{month}}) = 2^2 \mathrm{Var}(X_{1\text{month}}).$$ Which one is right and why?

share|improve this question
    
Your question could be much improved by adding a bit of context. However, both of the first two equations are correct under "typical" assumptions. This should immediately hint as to why the last is "incorrect" (i.e., again, under the "typical" assumptions). –  cardinal May 13 '11 at 12:58

3 Answers 3

up vote 6 down vote accepted

This is a typical confusion. For any independent random variables $A$ and $B$ the following two equations hold:

$Var(k A) = k^2 Var(A) \Rightarrow Var(2 A) = 4 Var(A)$

$Var(A + B) = Var(A) + Var(B)$

The problem arises when one tries to use the second to find $Var(2 A)$, by replacing $B$ for $A$:

$Var(2 A) = Var(A + A) = Var(A) + Var(A) = 2 Var(A)$

which contradicts the first equation. But this is wrong, because the second equation (variance of sum equals sum of variances) does not necessary hold if the variables are not independent -and certainly A is not independent of A...

Added: To get the complete picture, the general formula is easy to get:

$Var(A + B) = Var(A) + Var(B) + 2 Cov(A, B)$

This holds for any $A,B$, and it reduces to $Var(A + B) = Var(A) + Var(B)$ if $A$ and $B$ are independent (uncorrelated is enough). In the other extreme case (extreme correlation), when $B=A$, $Cov(A,B)=Var(A)$ and $Var(A+A)=4 Var(A)$. (You could also check the other extreme case: $B = -A$, correlation coeffient = -1)

share|improve this answer
    
I definitely wouldn't say your second equation applies only to independent random variables. –  cardinal May 13 '11 at 14:20
    
True. I edited it. –  leonbloy May 13 '11 at 14:22
    
what about if the two random variable correlate to some extent? Where do you have the correlation coefficient? –  hhh May 13 '11 at 14:27
    
@hhh : there you have. YOu can also express it using the correlation coefficient, of course. –  leonbloy May 13 '11 at 14:36

The assumption made here is that $X_{2\text{months}}$ is equal to the sum of two copies of $X_{1\text{month}}$, say $X_{\text{July}}$ and $X_{\text{August}}$. Then $$ \mathrm{Var}[X_{2\text{months}}] = \mathrm{Var}[X_{\text{July}} + X_{\text{August}}] = \mathrm{Var}[X_{\text{July}}] + \mathrm{Var}[X_{\text{August}}] = 2\mathrm{Var}[X_{1\text{month}}].$$

share|improve this answer
    
Plus $X_\text{July}$ and $X_\text{August}$ are independent. In the second part of the original question, $2X_\text{1 month} = X_\text{1 month} + X_\text{1 month}$ is not the sum of two independant variables. –  Sebastien May 13 '11 at 13:40

From wikipedia, I found a proper answer that solved the problem.

Let the sum of random variables be $Y = \sum_{i=1}^{N} X_{i}$.

$Var(Y) = \sum_{i=1}^{N} Var(X_{i}) + 2 \sum_{i<j} Cov(X_{i}, X_{j})$

or

$Var(Y) = \sum_{i=1}^{N} \sum_{j=1}^{N} Cov(X_{i} , X_{j})$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.