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The formula for the variance of the sum of two independent random variables is given $$ \Var (X +X) = \Var(2X) = 2^2\Var(X)$$

How then, does this happen:

Rolling one dice, results in a variance of $\frac{35}{12}$. Rolling two dice, should give a variance of $2^2\Var(\text{one die}) = 4 \times \frac{35}{12} \approx 11.67$. Instead, my Excel spreadsheet sample (and other sources) are giving me 5.83, which can be seen is equal to only $2 \times \Var(X)$.

What am I doing wrong?

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If only you could enclose the excel sheet. How are you computing the variance of both dice? – Gautam Shenoy Nov 14 '12 at 16:24
Michael Hardy's answer, though downvoted, is correct. – Jonathan Christensen Nov 14 '12 at 19:28
Symbols should stand for the same thing wherever they appear in a formula. While it is perfectly acceptable to use $X$ to denote the result of first roll of the die, it is not appropriate to use $X$ to also denote the result of the second roll of the die, unless you are considering a weird die that always shows the same number on two successive rolls. That is, for an ordinary die, $X+X$ is not the sum of the results of the two successive rolls, and the variance of the sum is not $4$var$(X)$. Instead, the variance is var$(X) + $var$(Y) = 2$var$(X)$ as Michael Hardy points out. – Dilip Sarwate Nov 14 '12 at 23:34
up vote 2 down vote accepted


The formula you give is not for two independent random variables. It's for random variables that are as far from independent as you can get. If $X,Y$ are independent, then you have $\Var(X+Y)=\Var(X)+\Var(Y)$. If, in addition, $X$ and $Y$ both have the same distribution, then this is equal to $2\Var(X)$. It is also the case that, as you say, $\Var(X+X)=4\Var(X)$. But that involves random variables that are nowhere near independent.

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Thanks for clearing that up for me! When would you use $Var(X+X)$? i.e. what do you mean by "nowhere near independent"? – CodyBugstein Nov 14 '12 at 20:17
I would write $\operatorname{Var}(X+X)$ only when that is what I meant. Suppose $X=\left.\begin{cases} 0 & \text{with probability }1/3, \\ 1 & \text{with probability }1/2, \\ 2 & \text{with probability }1/6. \end{cases}\right\}$ Then $X+X=\left.\begin{cases} 0 & \text{with probability }1/3, \\ 2 & \text{with probability }1/2, \\ 4 & \text{with probability }1/6. \end{cases}\right\}$ On the other hand, suppose $Y$ is _independent of $X$ and has that same distribution. Then $X+Y$ could be $0$, $1$, $2$, $3$, or $4$, each with some probability that follows from the above. – Michael Hardy Nov 15 '12 at 2:19
Specifically, $X+Y=\left.\begin{cases} 0 & \text{with probability }1/9, \\ 1 & \text{with probability }1/3, \\ 2 & \text{with probability }13/36 ,\\ 3 & \text{with probability }1/6, \\ 4 & \text{with probability }1/36. \end{cases}\right\}$ So the distribution of $X+X$ is quite different from the distribution of $X+Y$. – Michael Hardy Nov 15 '12 at 2:22

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