How to find out all random variables $X$ satisfying the following 2 conditions:

$1>\ E(X^2)<\infty$

$2>X$ follows the distributional equation : $X\stackrel{d}{=}{X+Y\over \sqrt2}$ for any random varibale $Y$ independent of $X$ such that $Y\stackrel{d}{=}X$

I am not getting where to begin with.


So , as per hints,

Any $X$ following $N(\mu,\sigma^2)$ will hold true$(\mu,\sigma^2<\infty)$, since $E(X^2)=\mu^2+\sigma^2\lt\infty$

Edit $2$: $\mu=0$ by the second equation. So $N(0,\sigma^2)$ is one candidate.

Edit $3$: Following @whuber and @dsaxton hints, we can write $$X\stackrel{d}{=}{1\over \sqrt2}\left(Y+{1\over \sqrt2}\left(Y+{1\over \sqrt2}\left(Y+\cdots\right)\right)\right)\\=Y({1\over \sqrt2}+{1\over (\sqrt2)^2}+{1\over (\sqrt2)^3}+{1\over (\sqrt2)^4}+\cdots)\\=Y(\sqrt2+1)$$

So still the question remains. Is this the only one? is there any general way to characterize all of them?

  • 1
    $\begingroup$ Can you think of any pair of iid random variables that satisfy this before you try to characterize all such random variables? $\endgroup$
    – jld
    Oct 12 '16 at 17:52
  • $\begingroup$ Hint: consider the characteristic function of the normal distribution: $\varphi(t) = e^{i \mu t + \sigma^2 t^2 / 2}$. $\endgroup$
    – dsaxton
    Oct 12 '16 at 17:52
  • $\begingroup$ @Chacone Is there any general way to characterize all such distributions? $\endgroup$
    – Qwerty
    Oct 12 '16 at 18:12
  • 3
    $\begingroup$ Your second equation implies $E[X]=\sqrt{2}E[X]$, so $E[X]=0$ is a must. $\endgroup$
    – Alex R.
    Oct 12 '16 at 18:13
  • 1
    $\begingroup$ My hint may have been misleading. The idea behind @whuber's hint is you can keep substituting i.i.d. copies of $(X + Y) / \sqrt{2}$ wherever you see $X$ or $Y$ on the right hand side. Then notice the pattern and write a formula involving $n$ random variables on the right hand side. $\endgroup$
    – dsaxton
    Oct 12 '16 at 18:52

To be more explicit and following @whuber's hint if $X \stackrel{\text{d}}{=} (X_1 + X_2) / \sqrt{2}$ where $X_1$ and $X_2$ are i.i.d. copies of $X$ then we can do another substitution in for both $X_1$ and $X_2$ and end up with (I'm reusing subscripts here so the notation doesn't get messy, but this shouldn't be confusing)

$$ X \stackrel{\text{d}}{=} \frac{X_1 + X_2 + X_3 + X_4}{\sqrt{4}} . $$

Iterate a second time and we have

$$ X \stackrel{\text{d}}{=} \frac{X_1 + X_2 + X_3 + X_4 + X_5 + X_6 + X_7 + X_8}{\sqrt{8}} $$

and if we generalize this we get

$$ X \stackrel{\text{d}}{=} \frac{\sum\limits_{i=1}^{2^n} X_i}{\sqrt{2^n}} . $$

Now what happens to the right hand side when $n \to \infty$?

  • $\begingroup$ Okay. So $X\stackrel{d}{=}N(0,\sigma^2)$ . But what was the necessity of $E(X^2)<\infty$ ? $\endgroup$
    – Qwerty
    Oct 12 '16 at 19:44
  • 4
    $\begingroup$ The CLT doesn't apply to variables with infinite variances. Dropping the finiteness assumption would therefore complicate the solution. $\endgroup$
    – whuber
    Oct 12 '16 at 19:58
  • $\begingroup$ @whuber Oh. I forgot that! My bad.. Anyways thanks.. $\endgroup$
    – Qwerty
    Oct 12 '16 at 20:27

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