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Suppose that X is a random variable (say, a normal random variable with mean a and variance b). Then is the sum $X + X$ equal to $2X$?

I am asking this because I know that $2X$ has mean $2a$ and variance $4b$. If we just apply $var(X + X) = var(X) + var(X) = 2b$, we get a different answer because $var$ cannot be applied this way to dependent random variables?

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  • $\begingroup$ Yes; variance is not a linear operator, so we cannot do this. $\mathsf E(X+Y) = \mathsf E(X)+\mathsf E(Y)$ but $\mathsf{Var}(X+Y)=\mathsf{Var}(X)+\mathsf{Var}(Y)+2\mathsf{Cov}(X,Y)$ $\endgroup$ – Graham Kemp Jan 20 '16 at 2:55
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    $\begingroup$ And since $\mathsf{Cov}(X,X)=\mathsf{Var}(X)$ then $\mathsf {Var}(2X)=4 \mathsf{Var}(X)$. $\endgroup$ – Graham Kemp Jan 20 '16 at 3:04
  • $\begingroup$ Only when the variables are uncorrelated is the variance of the sum the sum of the variances. $\mathsf {Var}(X+Y)=\mathsf {Var}(X)+\mathsf {Var}(Y)$ if and only if $\mathsf {Cov}(X,Y)=0$. $\endgroup$ – Graham Kemp Jan 20 '16 at 3:29
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It is because you missed the covariance in between. $$ Var[X + X] = Var[X] + 2Cov[X, X] + Var[X] = b + 2b + b = 4b $$

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  • $\begingroup$ It don't shows that $X+X=2X$. It shows that, in case of $X+X=2X$, then $Var(X)=4Var(X)$. $\endgroup$ – sinbadh Jan 20 '16 at 2:52
  • $\begingroup$ @sinbadh Do you mean that $X + X$ is not always equal to $2X$? $\endgroup$ – Student Jan 20 '16 at 2:56
  • $\begingroup$ Yes and No. But... do you remember definition of random variable? A random variable is a function $X$ with domain a probability space $(\Omega,\mathcal{F},P)$ and codomain other measurable space $(X,\sigma)$ such that for every $A\in \sigma$ we have $X^{-1}(A)\in\mathcal{F}$. Thus, the symbol $X+X$ not necesarily has sense $\endgroup$ – sinbadh Jan 20 '16 at 3:00
  • $\begingroup$ I am not quite following the argument here. Do you mean the function $2X$ may not be well defined when the image is out of the codomain? Maybe give an example for this kind of measure space? Thanks. $\endgroup$ – BGM Jan 20 '16 at 3:07
  • $\begingroup$ Yes, I'm saying it. $\endgroup$ – sinbadh Jan 20 '16 at 3:17
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With random variables, it is not true that $X + X = 2X$. More formally, if $X$ and $Y$ are independent random variables, $X + Y$ and $2X$ don't have the same distribution.

For example let $X$ and $Y$ be the outcomes of two die rolls. Then $X + Y$ is the sum of the numbers on the two dice and $2X$ is twice the number on the first die. These don't have the same distribution - for example,$X + Y$ can be odd, and $2X$ is always even.

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  • $\begingroup$ The reason I was asking this question is because I want to show that for Brownian processes $B_t$ and $B_s$, $t > s$, the sum of them is a gaussian random variable with variance $t + 3s$. The reason is that $B_t + B_s = B_t - B_s + B_s + B_s = (B_t - B_s) + 2B_s$. Since $(B_t - B_s)$ is independent of $B_s$ and hence of $2B_s$, the variance of the expression is just the sum of the variance of the two independent gaussian random variables. The first is $t-s$. The second is $4s$. Hence the answer is $t + 3s$. Hence, I asked if $B_s + B_s = 2B_s$. $\endgroup$ – Student Jan 20 '16 at 3:06
  • $\begingroup$ The main point here is stating that in general if $X, Y$ are i.i.d. random variables, then $X + Y$ and $2X$ has different distribution. I am not sure why this is supporting the argument that $X + X = 2X$ is not true. $\endgroup$ – BGM Jan 20 '16 at 3:10
  • $\begingroup$ Yes, $B_s+B_s=2B_s$ and $\mathsf {Var}(2B_s)=4\mathsf{Var}(B_s)$ $\endgroup$ – Graham Kemp Jan 20 '16 at 3:10
  • $\begingroup$ $\begin{align}\mathsf {Var}(B_t+B_s) & = \mathsf {Var}((B_t-B_s)+2B_s) \\ & = \mathsf {Var}(B_t-B_s)+\mathsf{Var}(2B_s) +\mathsf {Cov}(B_t-B_s, 2B_s) \\ & = (t-s)+4s +0 & \text{because $(B_t-B_s)\perp B_s$} \\ & = t+3s\end{align}$ $\endgroup$ – Graham Kemp Jan 20 '16 at 3:16
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It is simpler. Formally, don't forget that $X$ is no more than a function wich satisfies certains properties: it has a domain, a codomain (which is a vector space in most of the cases), a correspondence rule and an extra propertie (measurability, but it is not important for your question).

If $\Omega$ and $V$ is your codomain, where for all $u,v\in V$ we have $u+v$ is well defined (for example $X:\Omega\to\mathbb{R}$ for obtain a way to sum $X(w)+X(w)$), then $X+X:\Omega\to\mathbb{R}$ is, by definition, $(X+X)(w)=X(w)+X(w)$ (remember that $X(w)$ is in certain form a number. Then $X(w)+X(w)$ is a number).

Thus, all depend on whether the codomain has a sum or not, AND, if it is the case, how do you define the sum.

For example, take $U=\{A,B\}$ (letters $A$ and $B$) and $\sigma=2^U$. Now, set $\Omega=[0,1]$ with $\mathcal{F}=\mathcal{B}$, and $P:\mathcal{F}\to\mathbb{R}$ given by $P(A)=\lambda(A)$ (Lebesgue measure).

Let $X:\Omega\to U$ given by $X(w)=A$. Thus $X$ is a random variable and $(X+X)(w)=X(w)+X(w)=A+A$. But what is the value of $A+A$? $A$ is only a letter, and, at least I don't know how sum letters.

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