# Thread: Linear combination of normal random variables

1. ## Linear combination of normal random variables

"Fact: Any linear combination of independent normal random variables has a normal distribution."

Is the condition "independent" here absolutely necessary? If we remove the word "independent", would the linear combination still be normally distributed? Why or why not?

Thank you!

2. yes you need indep...

Let X be any normal rv.
Then Y=X-X is not a normal.
P(Y=0)=1 is not normal.

3. Say if we have X~Normal(0,1), Y~Normal(0,1)

E(X+Y) = E(X)+E(Y) = 0
V(X+Y) = V(X)+V(Y)+2Cov(X,Y) = 2+2Cov(X,Y)

Can we now say that X+Y~Normal(0, 2+2Cov(X,Y) ) ?

Thanks!

4. Originally Posted by kingwinner
Say if we have X~Normal(0,1), Y~Normal(0,1)

E(X+Y) = E(X)+E(Y) = 0
V(X+Y) = V(X)+V(Y)+2Cov(X,Y) = 2+2Cov(X,Y)

Can we now say that X+Y~Normal(0, 2+2Cov(X,Y) ) ?

Thanks!
No!

(This counterexample is wrong, see Laurents post below) Counterexample:Lets say your r.v pair X,Y has a joint char function $\phi(s_1,s_2) = \mathbb{E}(e^{s_1X+s_2Y}) = \text{exp}\left({\frac{s_1 ^2 + s_2 ^2}2 + s_1s_2}\right)$. Clearly they are marginally gaussian as you require and the sum distribution is not gaussian.

However if your question is if "independence" is necessary,then its not. Counterexample: What if X = 3Z+Y and where Z and Y are independent normal?

5. Unfortunately, I haven't learnt joint characteristic function in my class yet (I've learnt moment generating functions, however), so I don't have the required background to understand your example. I am sorry about that.

What if X = 3Z+Y and where Z and Y are independent normal?
Since they are indepednent, X must be normally distributed as well. Now my question is, if the random variables are NOT indepednent (see my example below), would a linear combination of those random variables also be normally distributed?

In short, my question is:
Suppose that X~Normal(0,1), Y~Normal(0,1), where X and Y are NOT independent
Can we say that X+Y~Normal(0, 2+2Cov(X,Y) ) for sure?

Thanks!

6. Originally Posted by kingwinner
Since they are indepednent, X must be normally distributed as well.
You didnt get my point. I wanted to say X+Y is normal distributed even though X and Y are not independent.

Now my question is, if the random variables are NOT indepednent (see my example below), would a linear combination of those random variables also be normally distributed?

In short, my question is:
Suppose that X~Normal(0,1), Y~Normal(0,1), where X and Y are NOT independent
Can we say that X+Y~Normal(0, 2+2Cov(X,Y) ) for sure?

Thanks!
My previous post is trying to tell you that there is no general answer. Some times the sum could be normal, sometimes not.

7. Originally Posted by Isomorphism
No!

Counterexample:Lets say your r.v pair X,Y has a joint char function $\phi(s_1,s_2) = \mathbb{E}(e^{s_1X+s_2Y}) = \text{exp}\left({\frac{s_1 ^2 + s_2 ^2}2 + s_1s_2}\right)$. Clearly they are marginally gaussian as you require and the sum distribution is not gaussian.
To be precise, what you're dealing with is rather a moment generating function (kind of Laplace transform) than a characteristic function (kind of Fourier transform).
By the way, it is not easy to prove that a function is a moment generating function for some probability distribution (it involves Bochner's theorem, which is uneasy to check in general), so that you should prove first that your function is indeed a m.g.f.. Actually it is a m.g.function, because this is that of a Gaussian vector...

Marginals are indeed standard Gaussian r.v., but the m.g.f. of the sum is $\mathbb{E}[e^{s(X+Y)}]=\exp(2s^2)$ (take $s_1=s_2=s$), which is the m.g.f. of a centered Gaussian with variance 4... Hence this is no counterexample.

Matheagle gave a working counterexample. Since Dirac measures can be seen as "limit cases" of Gaussian distribution when the variance goes to 0, I tend to prefer the following one: Let $X,\varepsilon$ be independent r.v. where $X$ is a standard Gaussian r.v., and $\varepsilon$ has a distribution given by $P(\varepsilon=+1)=P(\varepsilon=-1)=1/2$. Then let $Y=\varepsilon X$, so that $Y$ is a standard Gaussian, while $X+Y=(1+\varepsilon)X$ is 0 with probability 1/2, hence it is not Gaussian, and it is not degenerate.

8. Originally Posted by Laurent
To be precise, what you're dealing with is rather a moment generating function (kind of Laplace transform) than a characteristic function (kind of Fourier transform).

By the way, it is not easy to prove that a function is a moment generating function for some probability distribution (it involves Bochner's theorem, which is uneasy to check in general), so that you should prove first that your function is indeed a m.g.f.. Actually it is a m.g.function, because this is that of a Gaussian vector...

Marginals are indeed standard Gaussian r.v., but the m.g.f. of the sum is $\mathbb{E}[e^{s(X+Y)}]=\exp(2s^2)$ (take $s_1=s_2=s$), which is the m.g.f. of a centered Gaussian with variance 4... Hence this is no counterexample.
Whoops!

Perhaps I should have gone with my standard counter $\phi(s_1,s_2) = \mathbb{E}(e^{s_1X+s_2Y}) = (1 + s_1s_2)\text{exp}\left({\frac{s_1 ^2 + s_2 ^2}2}\right)
$

In my basic random process classes these examples were considered good enough. So does this new example confirm to the rigorous approach?

Your counterexample is nice

9. Originally Posted by Isomorphism
Whoops!

Perhaps I should have gone with my standard counter $\phi(s_1,s_2) = \mathbb{E}(e^{s_1X+s_2Y}) = (1 + s_1s_2)\text{exp}\left({\frac{s_1 ^2 + s_2 ^2}2}\right)
$
Where do you get this example from? If my computations are correct, this still doesn't work, because what you gave seems to be the m.g.f. of the "density" $f(x,y)=\frac{1}{2\pi}(1+xy)\exp\left(-\frac{x^2+y^2}{2}\right)$ on $\mathbb{R}^2$, which is negative sometimes... This illustrates what I said about Bochner's theorem in my last post: not any function, even if it equals 1 at 0 and is "smooth", is a m.g.f.. Being a m.g.f. is a strong condition that can't be checked at first sight.

10. Again, if Y=-X, then W=X+Y is not normal. P(W=0)=1.
The negative of a N(0,1) is a N(0,1), so both X and Y are st normals, but their sum is not a normal, not even continuous.

11. However if your question is if "independence" is necessary,then its not. Counterexample: What if X = 3Z+Y and where Z and Y are independent normal?
You didnt get my point. I wanted to say X+Y is normal distributed even though X and Y are not independent.
Then X=3Z-Y would be a better example

However, he talked about independence for "any linear combination" of normal distribution. Not for X+Y in particular.
With Gaussian vectors, we were told that $X_1,\dots,X_n$ form a Gaussian vector (meaning that any linear combination of its components follows a normal distribution) if (and only if ?) they're independent.

12. He's going to keep asking this question until the COWS (MOOO) come home.
He knows the answer by now.

13. How about this?
"ANY linear combination of uncorrelated normal random variables has a normal distribution."

Is this a correct statement? Why or why not?

Thanks!

14. Originally Posted by kingwinner
How about this?
"ANY linear combination of uncorrelated normal random variables has a normal distribution."

Is this a correct statement? Why or why not?

Thanks!
Did you check on our counterexamples first? In mine (post #7), the r.v. $X$ and $Y$ are uncorrelated.

15. I guess the cows aren't home yet.

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