I think the following two formulas are true:

$$ \mathrm{Var}(aX)=a^2 \mathrm{Var}(X) $$ while a is a constant number $$ \mathrm{Var}(X + Y)=\mathrm{Var}(X)+\mathrm{Var}(Y) $$ if $X$, $Y$ are independent

However, I am not sure what is wrong with the below:

$$\mathrm{Var}(2X) = \mathrm{Var}(X+X) = \mathrm{Var}(X) + \mathrm{Var}(X) $$ which does not equal to $2^2 \mathrm{Var}(X)$, i.e. $4\mathrm{Var}(X)$.

If it is assumed that $X$ is the sample taken from a population, I think we can always assume $X$ to be independent from the other $X$s.

So what is wrong with my confusion?

  • 8
    $\begingroup$ Variance isn't linear -- your first statement shows this (if it were, you'd have $Var(aX) = a Var(X)$. Covariance on the other hand is bilinear. $\endgroup$
    – Batman
    Dec 4 '15 at 12:47
  • $\begingroup$ $Var(X+X)=Var(X)+Var(X) + 2cov(X,X) = 4V(X)$ since $cov(X,X) = V(X)$. $\endgroup$ Mar 16 '20 at 18:42

$\DeclareMathOperator{\Cov}{Cov}$ $\DeclareMathOperator{\Corr}{Corr}$ $\DeclareMathOperator{\Var}{Var}$

The problem with your line of reasoning is

"I think we can always assume $X$ to be independent from the other $X$s."

$X$ is not independent of $X$. The symbol $X$ is being used to refer to the same random variable here. Once you know the value of the first $X$ to appear in your formula, this also fixes the value of the second $X$ to appear. If you want them to refer to distinct (and potentially independent) random variables, you need to denote them with different letters (e.g. $X$ and $Y$) or using subscripts (e.g. $X_1$ and $X_2$); the latter is often (but not always) used to denote variables drawn from the same distribution.

If two variables $X$ and $Y$ are independent then $\Pr(X=a|Y=b)$ is the same as $\Pr(X=a)$: knowing the value of $Y$ does not give us any additional information about the value of $X$. But $\Pr(X=a|X=b)$ is $1$ if $a=b$ and $0$ otherwise: knowing the value of $X$ gives you complete information about the value of $X$. [You can replace the probabilities in this paragraph by cumulative distribution functions, or where appropriate, probability density functions, to essentially the same effect.]

Another way of seeing things is that if two variables are independent then they have zero correlation (though zero correlation does not imply independence!) but $X$ is perfectly correlated with itself, $\Corr(X,X)=1$ so $X$ can't be independent of itself. Note that since the covariance is given by $\Cov(X,Y)=\Corr(X,Y)\sqrt{\Var(X)\Var(Y)}$, then

The more general formula for the variance of a sum of two random variables is

$$\Var(X+Y) = \Var(X) + \Var(Y) + 2 \Cov(X,Y)$$

In particular, $\Cov(X,X) = \Var(X)$, so

$$\Var(X+X) = \Var(X) + \Var(X) + 2\Var(X) = 4\Var(X)$$

which is the same as you would have deduced from applying the rule

$$\Var(aX) = a^2 \Var(X) \implies \Var(2X) = 4\Var(X)$$

If you are interested in linearity, then you might be interested in the bilinearity of covariance. For random variables $W$, $X$, $Y$ and $Z$ (whether dependent or independent) and constants $a$, $b$, $c$ and $d$ we have

$$\Cov(aW + bX, Y) = a \Cov(W,Y) + b \Cov(X,Y)$$

$$\Cov(X, cY + dZ) = c \Cov(X,Y) + d \Cov(X,Z)$$

and overall,

$$\Cov(aW + bX, cY + dZ) = ac \Cov(W,Y) + ad \Cov(W,Z) + bc \Cov(X,Y) + bd \Cov(X,Z)$$

You can then use this to prove the (non-linear) results for variance that you wrote in your post:

$$\Var(aX) = \Cov(aX, aX) = a^2 \Cov(X,X) = a^2 \Var(X)$$

$$ \begin{align} \Var(aX + bY) &= \Cov(aX + bY, aX + bY) \\ &= a^2 \Cov(X,X) + ab \Cov(X,Y) + ba \Cov (X,Y) + b^2 \Cov(Y,Y) \\ \Var(aX + bY) &= a^2 \Var(X) + b^2 \Var(Y) + 2ab \Cov(X,Y) \end{align} $$

The latter gives, as a special case when $a=b=1$,

$$\Var(X+Y) = \Var(X) + \Var(Y) + 2 \Cov(X,Y)$$

When $X$ and $Y$ are uncorrelated (which includes the case where they are independent), then this reduces to $\Var(X+Y) = \Var(X) + \Var(Y)$. So if you want to manipulate variances in a "linear" way (which is often a nice way to work algebraically), then work with the covariances instead, and exploit their bilinearity.

  • 2
    $\begingroup$ Yes! I think you pinpointed at the beginning that the confusion was essentially a notational one. I found it very helpful when one book (very explicitly, some might say laboriously) explained the interpretation of and rules of evaluating a probabilistic statement (so that, e.g., even if you know what you mean by $\Pr (X+X=n)$ where $X \sim \text{Uniform}(1..6)$, it is technically incorrect if you're considering throwing a $n$ in craps (and $X+X=2X$ would never yield an odd roll); the event would be properly expressed using $X_1,X_2$ i.i.d.). $\endgroup$ Dec 4 '15 at 18:53
  • 1
    $\begingroup$ This is in contrast to (and I think my misapprehension might have stemmed from) how 2+PRNG(6)+PRNG(6) often is how you would toss dice as above and/or notation/conventions such as $2 \text{d}6 = \text{d}6 + \text{d}6$ in which different instances are genuinely intended to be independent. $\endgroup$ Dec 4 '15 at 18:53
  • 1
    $\begingroup$ @Vandermonde That's an interesting point. I initially considered mentioning the use of subscripts to distinguish between "different $X$s" but didn't bother - think I might edit it in now. The argument that "you'd never get an odd total score if the sum was $2X$" is very clear and convincing to someone who can't see the need to distinguish: thanks for sharing it. $\endgroup$
    – Silverfish
    Dec 4 '15 at 19:09

Another way of thinking about it is that with random variables $2X \neq X + X$.

$2X$ would mean two times the value of the outcome of $X$, while $X + X$ would mean two trials of $X$. In other words, it's the difference between rolling a die once and doubling the result, vs rolling a die twice.

  • $\begingroup$ +1 This is a perfectly clear and correct answer. Welcome to our site! $\endgroup$
    – whuber
    Jul 12 '19 at 2:51
  • $\begingroup$ Thanks @whuber! $\endgroup$
    – BBrooklyn
    Jul 13 '19 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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