# Why does the summation property of Gaussian distributions hold?

According to the summation property of Gaussian distributions: If there are 2 Gaussian distributions as given by:

$$y$$~$$N (\mu, \Sigma)$$ and $$y'$$~$$N (\mu', \Sigma')$$

then $$y~+y'$$~$$N(\mu+ \mu', \Sigma+\Sigma')$$

Consider 2 boxes that contain chits of papers with numbers written on them.

The first box contains chits numbered with either 1, 2 or 3 and written in a way, such that the probability of getting a number by randomly picking a chit from the box is a Gaussian distribution with $$\mu_1=3$$ and $$\Sigma_1=1$$.

Similarly, the numbers on the chits in box 2 can be either 4, 5 or 6, with a Gaussian distribution with $$\mu_2=5$$ and $$\Sigma_2=1$$

Now if I mix up the chits of both these boxes into one large box, then according to the summation rule, the new distribution has $$\mu_3=7$$ and $$\Sigma=2$$

Is this correct? It doesnt seem so because I dont have a chit with the nuber 7 on it, but the resultant distribution is telling me that getting a chit with 7 has the highest probability.

• Gaussian is a continuous distribution, your statement is invalid. A random variable that takes values on 1, 2 and 3 cannot be Gaussian distributed. – Freguglia Nov 7 '18 at 19:04
• Also following your reasoning (ignoring the Gaussian factor), you have the most probable value at 7 for the sum a number taken from the first box and a number from the second box. You're adding one of each box, not mixing up the boxes. – Freguglia Nov 7 '18 at 19:06
• Be careful not to conflate distributions with random variables. The statement in the question is untrue in general. That "the sum of two Gaussian variates is Gaussian" is not always true (many counterexamples are on site already); joint normality is sufficient (but not necessary, since it's possible under other circumstances, for which there are also examples on site already). It is also untrue that you can just sum the covariance matrices.in general; you need the vectors to be uncorrelated. – Glen_b Nov 8 '18 at 0:46

• If the two sets contains only $0$'s :) – gunes Nov 8 '18 at 5:10