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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?

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    Your question could be much improved by adding a bit of context. However, both of the first two equations are correct under ["typical" assumptions](http://en.wikipedia.org/wiki/Brownian_motion#Mathematics). This should immediately hint as to why the last is "incorrect" (i.e., again, under the "typical" assumptions).2011-05-13

3 Answers 3

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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)

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    @hhh : there you have. YOu can also express it using the correlation coefficient, of course.2011-05-13
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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}}].$

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    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.2011-05-13
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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

or

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