# If $X,Y$ are independent and $X+Y$ is normal, are $X, Y$ necessarily normal?

The sum of two independent random variables $$X, Y$$ is normal iff $$X$$ and $$Y$$ are normally distributed.

I came across this fact recently, and it totally baffles me. am not convinced it is true. Is there a proof for this somewhere?

UPDATE:

The direction of starting with two independent random variables that are normally distributed ($$X$$, $$Y$$) to proving that $$X+Y$$ is also normally distributed is not confusing. I am more confused on the other direction of the proof. It seems like we could have two random variables that are independent and NOT normally distributed were the sum could be normally distributed. Suppose we have a random variable $$Y$$ that is standard normal, and $$U$$, $$V$$ which are iid with finite mean and variance from a non-normal distribution. If we have another variable $$W = Y -U$$, and then take $$W + V$$ is it really the case that $$W + V$$ would not be normally distributed? I have done a few arbitrary examples that concur with the statement. I am curious what the proof would look like so I know it is true always.

UPDATE #2

I think I found a counter example. Suppose that $$X\sim I(2)$$ that is, $$X$$ is a degenerate random variable, and $$Y \sim N(0,1)$$. Then the MGF of $$Y$$ is $$\exp(t^2/2)$$ and the MGF of $$X$$ is $$\exp(2t)$$, hence the MGF of $$X+Y$$ is $$\exp(2t + t^2/2)$$, which implies that $$X+Y \sim N(2, 1)$$.

• math.stackexchange.com/questions/2090019/…
– user486983
Commented Jan 24, 2020 at 23:10
• Wikipedia gives multiple proofs in Sum of normally distributed random variables. Why does it baffle you? It is a simple consequence of the product of two exponents with sums of squares and first powers being another exponent with a sum of squares and first powers. Commented Jan 24, 2020 at 23:41
• @Conifold The interesting half of the question, that $X+Y$ normal implies $X$ normal and $Y$ normal, is not covered by that wikipedia article. Commented Jan 25, 2020 at 0:49
• See also arxiv.org/pdf/1810.01768.pdf which has what looks like a proof (I've not checked it). Commented Jan 25, 2020 at 0:54
• I think if you're dedicated, you can use the central limit theorem to prove this; if a normal distribution is a sum of a whole bunch of independent variables with specified mean and variance, the sum of two normal distributions must just be the sum of a whole bunch (well, two whole bunches, but we're working in the limit where the bunches are very large) of independent variables with specified mean and variance - oh wait. Of course, this, not the formulas, is why we like normal distributions. Commented Jan 25, 2020 at 4:22

## 1 Answer

Suppose $$X+Y$$ is normal and $$X,Y$$ are independent.

$$X+Y \sim N(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2)$$ since we know its mean and variance.

Thus the MGF of $$X+Y$$ is $$e^{(\mu_X+\mu_Y)t+\frac{1}{2}t^2(\sigma_X^2 + \sigma_Y^2)}$$.

By independence this is equal to $$E[e^{tX}]E[e^{tY}]$$. I can now deduce $$E[e^{tX}]$$ is proportional to $$e^{\mu_X t+\frac{1}{2}t^2 \sigma_X^2 }$$. Hence it is a normal random variate. The same can be said of $$Y$$.

The other direction is quite easy (see comments).

• can you point out the error in this Commented Jan 25, 2020 at 14:06
• @IGDu94 I didn't downvote the answer, but one place that is not clear to me is how do you know that $E[e^{tX}]$ is proportional to $e^{\mu_X t+\frac{1}{2}t^2 \sigma_X^2 }$? If that part could be explained, then the rest looks correct to me. Commented Feb 15, 2021 at 19:48