if $X$ and $Y$ are i.i.d., and if $X+Y$ and $X-Y$ are independent, are $X$ and $Y$ normally distributed?

Question

Just recently come across Normal Distribution, and the following statement seems to be quite true, but is it? Can someone provide some general proof sketch if so please:

For X and Y identically and independently distributed with mean 0 and variance 1.

Suppose X+Y and X-Y are independent

Does it imply X and Y follow normal distribution?

See this answer on stats.SE for a generalization of this idea. — Dilip Sarwate, May 06 '14 at 01:23
http://math.stackexchange.com/questions/780236/tricky-probability-problem/780256#780256 — Arash, May 06 '14 at 01:51

score 2 · Answer 1 · answered May 06 '14 at 01:10

2

Yes, those assumptions imply that $X$ and $Y$ are normal, but the theorem is difficult. (In fact, you do not need to assume they are identically distributed; it is enough that they are independent.) It was proven by Sergei Bernstein in 1941. You can find a proof, and much more about characterizations of the normal distribution, in this book. (Bernstein's theorem is proved in Section 5.1.)

answered May 06 '14 at 01:10

symplectomorphic

19,197

Thank you so much for the information about this proof but the book link is not available. Could you put the book name here? Thanks again! – user431550 Oct 25 '20 at 20:12

if $X$ and $Y$ are i.i.d., and if $X+Y$ and $X-Y$ are independent, are $X$ and $Y$ normally distributed?

1 Answers1

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