All random variables following a distributional equation

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.

EDIT :

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?

• Can you think of any pair of iid random variables that satisfy this before you try to characterize all such random variables?
– jld
Oct 12 '16 at 17:52
• Hint: consider the characteristic function of the normal distribution: $\varphi(t) = e^{i \mu t + \sigma^2 t^2 / 2}$. Oct 12 '16 at 17:52
• @Chacone Is there any general way to characterize all such distributions? Oct 12 '16 at 18:12
• Your second equation implies $E[X]=\sqrt{2}E[X]$, so $E[X]=0$ is a must. Oct 12 '16 at 18:13
• 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. 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 $n \to \infty$?

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