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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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  • $\begingroup$ If only you could enclose the excel sheet. How are you computing the variance of both dice? $\endgroup$
    – Gautam Shenoy
    Nov 14, 2012 at 16:24
  • $\begingroup$ Michael Hardy's answer, though downvoted, is correct. $\endgroup$
    – Jonathan Christensen
    Nov 14, 2012 at 19:28
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    $\begingroup$ 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. $\endgroup$
    – Dilip Sarwate
    Nov 14, 2012 at 23:34

2 Answers 2

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$\newcommand{\Var}{\operatorname{Var}}$

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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  • $\begingroup$ Thanks for clearing that up for me! When would you use $Var(X+X)$? i.e. what do you mean by "nowhere near independent"? $\endgroup$
    – CodyBugstein
    Nov 14, 2012 at 20:17
  • $\begingroup$ 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. $\endgroup$
    – Michael Hardy
    Nov 15, 2012 at 2:19
  • $\begingroup$ 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 $X+Y$. $\endgroup$
    – Michael Hardy
    Nov 15, 2012 at 2:22
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X and itself aren't independent. X and itself are correlated with rho = 1.

In general, for two events, X (the first die) and Y (the second die), VAR(X+Y) = VAR(X) + VAR(Y) + 2COV(XY).

Since rolling two dice ARE independent (rho = 0), COV(XY) = 0, so VAR(X+Y) = 35/12 + 35/12 + 0.

VAR(X+X) is 4*VAR(X), but this recognizes that X and itself are perfectly correlated and we could just as well calculate this as VAR(X+X) = 35/12 + 35/12 + 2*35/12 since we know that COV(XX) is the same as VAR(X).

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