# Sum of two Normals

I know when $$X\sim \mathcal{N}\left(\mu_X,\sigma^2_x\right)$$ and $$Y\sim \mathcal{N}\left(\mu_Y,\sigma^2_y\right)$$, then $$Z = X+Y \sim \mathcal{N}\left(\mu_X + \mu_Y,\sigma^2_x+\sigma^2_y\right)$$.

I also know the proof for this. However, I'm not sure what's wrong with this "proof" that would suggest elsewise:

$$\begin{split} Z &= X + Y = \sigma_XV + \sigma_yV +\mu_x + \mu_y \\ &= (\sigma_X + \sigma_y)V + (\mu_x + \mu_y) \\ &\sim \mathcal{N}\left(\mu_X + \mu_Y,(\sigma_x+\sigma_y)^2\right). \end{split}$$

Where $$V$$ is the standard normal.

• $$\sigma_X V+\sigma_Y V\ne(\sigma_X+\sigma_Y)V$$Probability distributions do not add together in this way as proven by the erroneous result. Otherwise a Binomial distribution would just be a Bernoulli distribution multiplied by a constant. – Peter Foreman Nov 10 '19 at 19:19
• Ahhhh it would be more like $Z = X + Y = \sigma_XV_1 + \sigma_yV_2 +\mu_x + \mu_y$, which then don't just 'add up'? – analysis1 Nov 10 '19 at 19:27

You seem to be assuming that $$X = \sigma_X V + \mu_X$$ and $$Y = \sigma_Y V + \mu_Y$$ where you are able to treat $$V$$ just like a random variable.
But if $$V$$ is a random variable, then your assumption implies that $$V = \frac1{\sigma_X} (X - \mu_X),$$ which we can plug into the equation for $$Y$$ to find that
$$Y = \sigma_Y \left(\frac1{\sigma_X} (X - \mu_X)\right) + \mu_Y = \frac{\sigma_Y}{\sigma_X} X - \frac{\sigma_Y \mu_X}{\sigma_X} + \mu_Y.$$
In other words, $$Y$$ is just $$X$$ scaled and translated; the two variables are perfectly correlated. And indeed when you add perfectly correlated normal variables, the variances combine in the way your "proof" says.
The usual formula depends on an assumption you did not state, namely that $$X$$ and $$Y$$ are independent. If you remove that assumption then there are many possible results of $$X+Y$$, not just the two in the question.