Say we have X~N(10, 100). It seems to hold that X+X~N(20, 200), however, if we multiply X with constant we have to multiply the variance with the square of the constant. Take for example 2, then we have 2X~N(20, 100 * 2^2) and thus 2X~N(20, 400).
Doesn't X+X and 2X denote exactly the same thing? I feel like there's a difference between the two that I do not understand.
Thanks
On
As variables, $X+X$ and $2X$ are the same. But these now represent random variables coming from a distribution. Perhaps it is better to label one of the $X$'s as $X'$ to distinguish it from $X$ (even though they have the same distribution).
In the case of $X+X'$, you are looking at the distribution where you take a random number from $X$ and another from $X'$ and adding them. In the case of $2X$, you are looking at the distribution obtained by taking a single random number from $X$ and doubling it.
So the numbers you are taking in the case of $X+X'$ are 'not' the same. [Every 'once in awhile', yes. But the point is in the sum $x+x'$, $x,x'$ are random. So 'most' of the time, they are different.] It should then make sense that $2X$ must have the larger variance of the two. To get 'large' numbers from $X+X'$, the two numbers $x,x'$ both need to be 'large.' But that's not very likely - both large?! Preposterous! But in the case of $2X$, you only need to draw a single 'larger' number from $X$ to obtain a 'large' number from $2X$. Similar for 'small' (or really more likely very negative) numbers from $X+X'$ and $2X$. Now that we see that the variances for the distribution are different (though this does not explain what they are, only that they must be different) the distributions cannot be exactly the same.
Here is a simpler example. Suppose $A$ and $B$ are each independent uniform random variables between $0$ and $1$. We could have $A = 0.25$ and $B = 0.8$, for example, as an outcome. So their sum is $$A+B = 1.05.$$ But now $2A = 0.5$. The variable $2A$ is uniform on $[0,2]$, because the outcome of $A$ is uniform on $[0,1]$ and you are just scaling up the result by $2$. But $A+B$ is not uniform on $[0,2]$, although its support is on this interval. Intuitively, this is because in order for $A+B$ to be "close to" $2$, both $A$ and $B$ have to be close to $2$. But there are many more ways for $A+B$ to be "close to" $1$, because either $A$ can be large and $B$ can be small, or vice versa.
To see this explicitly, we can compare this to a discrete distribution--rolling two fair dice numbered from $1$ to $6$. How many ways are there to get a sum of $12$? There is only one way: $(6,6)$. But how many ways are there to get a sum of $7$? There are six ways: $$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1).$$
So now that we understand that the distribution of $A+B$ is not the same as $2A$, it is not too difficult to see that their variances will also be different.
To bring our conversation back to the normal distribution, we can see that you have a misapprehension here. $X+X$ is not an appropriate way to describe a random variable that represents the sum of two independent but identically distributed (IID) normal random variables. In other words, if by $X+X$ you mean to say, "draw two realizations from a normal distribution with mean $\mu$ and variance $\sigma^2$," then this is not the correct notation. Instead, you should write $$X_1 + X_2,$$ where $$X_i \sim \operatorname{Normal}(\mu,\sigma^2), \quad i = 1, 2, \ldots.$$ Then the sum of these IID normal random variables is also normal: $$X_1 + X_2 \sim \operatorname{Normal}(2\mu, 2\sigma^2).$$ But the random variable $2X_1$ does not represent drawing two normal random variables. It means drawing one random variable and multiplying it by $2$. And as we explained with our examples above, this is a different distribution than the sum of independent normal random variables; while it is normal, it has a different variance.