# Do the mean and standard deviation of the sum of independent normal random variables $X$ and $Y$ equal $\mu_X+\mu_Y$ and $sd_X+sd_Y$?

I was following with Chapter 4 of famous 'Statistical Inference 2nd ed' textbook (Casella, Berger) and discovered that if X is a randomly normally distributed variable, and so is Y, and both of them are independent, therefore the distribution of the sum of equal length random vectors X + Y has the parameters mu_XY = mu_X + mu_Y and sd_XY = sd_X + sd_Y

Just for check that statement I performed a simulation in R with a following code:

> rnd_vec_X <- rnorm(10000, mean = 10, sd = 3)
> rnd_vec_Y <- rnorm(10000, mean = 20, sd = 7)
> rnd_vec_XY_sum <- rnd_vec_X + rnd_vec_Y
> mean(rnd_vec_XY_sum)
 30.05607
> sd(rnd_vec_XY_sum)
 7.564628


Indeed, the mean of the sum of there independent random variables is ~30 (20 + 10), as predicted by theory. But why the standard deviation is so far from 10 (7 + 3)? I performed nearly twenty simulations with different seeds and different sample sizes... No luck. I guess something is wrong with my code or with my understanding of the topic. But what?

## 1 Answer

Under independence (in fact, already under lack of correlation), the variance of the sum is the sum of the variances, $$\sigma^2_{X+Y}=\sigma^2_{X}+\sigma^2_{Y}$$ Incidentally, this result does not rely on normality.

It is, therefore, not true that $$\sqrt{\sigma^2_{X+Y}}=\sqrt{\sigma^2_{X}}+\sqrt{\sigma^2_{Y}},$$ i.e., the sum of the standard deviations is not the standard deviation of the sum - simply because the square root of a sum is not the sum of the square roots.

In your example, $X$ and $Y$ have variances 9 and 49, respectively, so that the population standard deviation is

> sqrt(3^2+7^2)
 7.615773


which agrees with your simulation.