I think the following two formulas are true:

Var(aX)=a2Var(X) while a is a constant number

Var(X+Y)=Var(X)+Var(Y) if X, Y are independentHowever, I am not sure what is wrong with the below:

Var(2X)=Var(X+X)=Var(X)+Var(X) which does not equal to 22Var(X), i.e. 4Var(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 Xs.

So what is wrong with my confusion?

**Answer**

The problem with your line of reasoning is

“I think we can always assume X to be independent from the other Xs.”

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. X1 and X2); 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 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

\Cov(X,X)=1\sqrt{\Var(X)^2}=\Var(X)

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.

**Attribution***Source : Link , Question Author : lanselibai , Answer Author : Community*