4

I wonder

  1. what families of distributions can satisfy that the sum of their any two random variables still have a distribution in the same family?
  2. what families of distributions can satisfy that the product of their any two random variables still have a distribution in the same family?

My questions arose when I read this reply

  • The sum of normal (Cauchy, Levy) random variables is normal (Cauchy, Levy).

  • The sum of gamma random variables is gamma if the distributions have a common scale parameter.

  • The product of log-normal random variables is log-normal.

I know it is not true that the sum of two normal distributed random variables is still normally distributed. For example, if $X$ is normally distributed, define $Y$ to be $X$ if $|X| > c$ and $Y = −X$ if $|X| < c$, where $c > 0$. Then $X+Y$ is not normally distributed.

I am not sure how to verify if the other claims are right or not.

What can be say about a family of distribution that satisfies the above requirements?

Thanks and regards!

Community
  • 1
Tim
  • 47,382
  • ?? Obviously, the author means sums of independent random variables, end of story. – Did Jul 01 '12 at 19:59
  • 1
    The statement that you have highlighted is incomplete: it is necessary to include the condition that the random variables considered are independent random variables. Otherwise, as your example shows, the sum of normal random variables is not necessarily a normal random variable. But, the sum of independent normal random variables is normal. The sum of jointly normal random variables is normal. In summary, the claims you read are correct if you take them with the grain of salt that all the random variables are independent even though the writer regretfully forgot to say so. – Dilip Sarwate Jul 01 '12 at 20:03
  • @DilipSarwate: Okay. (1) Are all the statements assuming independence? (2) Are there families of distributions that satisfy the requirements without requiring independence? – Tim Jul 01 '12 at 20:15
  • The dependent case you gave is a sum of dependent variables that are not both normal. Y is not normal unless c=0. There is no reason to expect dependent random variables to have their sum in the family. Suppose X has a distribution that is symmetric about 0 then Y=-X has the same distribution as X but X+Y is identically 0. – Michael R. Chernick Jul 01 '12 at 20:32
  • @MichaelChernick: In my example, forgot to say $X$ has mean zero. Now is $Y$ still not normally distributed? – Tim Jul 01 '12 at 20:34
  • 1
    @MichaelChernick Assuming that $X$ is a zero-mean normal random variable, the OP's $X$ and $Y$ both are marginally normal but they are dependent and they are not jointly normal, and so their sum is not normal. It is true that the sum of jointly normal random variables (which of course are marginally normal too) is a normal random variable. Independence is not needed but joint normality is. – Dilip Sarwate Jul 01 '12 at 20:42
  • Yes if c=0 then Y is normal. – Michael R. Chernick Jul 01 '12 at 21:07
  • 2
    @MichaelChernick "Yes if c=0 then Y is normal." True but irrelevant to the question raised by the OP, who asks that if $E[X] = 0$, then is $Y$ normal? Note that $c > 0$ is still part of the statement, and so saying that $Y$ is normal if $c = 0$ does not really answer the question. And incidentally, the answer to the OP's question is Yes, if $E[X] = 0$, then $Y$ as the OP defines it is normal. It is not necessary to insist that $c$ must be $0$. – Dilip Sarwate Jul 02 '12 at 01:21
  • @DilipSarwate Okay I see the distinction. I pointed to a discontinuity in the value of Y as X increases. I guess I should not have followed that by saying "how is this discontinuous distribution normal?" I should have said "How is the distribution with this discontinuity in Y normal?" But now I think I see my error for |X|<c f(y)=f(-x) but this just flips the density from 0 to c with the density from-c to 0 which does not change the density. So the density of Y is the same as for X at every point. – Michael R. Chernick Jul 02 '12 at 15:37
  • @did: Where is Didier's fun story now? – Tim Jul 02 '12 at 16:09

0 Answers0