Probability distribution of sum of IID gaussian random variables

Question

I often see people say that if you have 2 IID gaussian RVs, say $X \sim \mathcal{N}(\mu_x, \sigma_x^2)$ and $Y \sim \mathcal{N}(\mu_y, \sigma_y^2)$, then the distribution of their sum is $\mathcal{N}(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2)$.

This is only true when $X$ and $Y$ have the same units right, otherwise you can't even sum them to begin with without standardization?

e.g., if $X$ was some measure of distance in meters and $Y$ was some measure of velocity in $\frac{meters}{second}$, then you can't simply just add their means and variances together. That wouldn't make sense. You'd have to standardize them first so they're both unitless before you can do the above.

Answer 1

If $X$ and $Y$ are independent , $X \sim N(\mu_X,\sigma_X^{2})$ and $Y \sim N(\mu_Y,\sigma_Y^{2})$ then $X+Y \sim N(\mu_X+\mu_Y,\sigma_X^{2}+\sigma_Y^{2})$. [IID would force $\mu_X=\mu_Y, \sigma_X=\sigma_Y$ which is not required here. Independence is enough]. There is no need to standardize the random variables for this.

[There are many ways of proving this and one way is to use characteristic functions. Use the fact that $Ee^{it(X+Y)}=Ee^{itX}Ee^{itY}$ and you would be able to supply a proof].

Answer 2

I think what you are looking for is a Multivariate Gaussian — assign each unit to a different dimension and then you have a random Gaussian vector. If there is no correlation between the two dimensions this reduces to each coming from its own Gaussian.

Answer 3

I think your confusion does not really have anything to do with random variables, but rather simply with interpreting units. If I tell you I have $10$ apples and $3$ oranges, then $10 + 3 = 13$, no matter what, but what $13$ represents is a question that is orthogonal to the mathematical question of what is the sum of $10$ and $3$.

If $X$ is a random number of apples and $Y$ a random number of oranges, then there is nothing that prevents the random number $X + Y$ from existing. The question of what $X+Y$ should represent is orthogonal to the mathematical question of what is the distribution of $X+Y$.