If $X$ and $Y$ are iid with mean $0$ and variance $1$, and $T= \frac{X+Y}{\sqrt{ 2 }}$ has the same distribution, prove it is standard normal

Asked Apr 02 '19 at 22:11

Active Apr 04 '19 at 12:57

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Let $X$ and $Y$ be two i.i.d. random variables with mean $E[X]=E[Y]=0$ and variance $Var(X)=Var(Y)=1$.

Let $T=\left( \frac{X+Y}{\sqrt{ 2 }} \right)$ and suppose that the distribution of $T$ is the same as the distribution of $X$ and of $Y$.

Prove that $T$ necessarily follows a standard normal distribution.

It is not clear to me how to prove that $T \sim N(0,1)$ in absence of any other information.

edited Apr 04 '19 at 12:57

asked Apr 02 '19 at 22:11

Maria

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You tagged this with "characteristic functions". Perhaps you should try that – Henry Apr 02 '19 at 22:25
The characteristic function of $X$ obeys a remarkable equation, which you can then exploit. – kimchi lover Apr 02 '19 at 22:26
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The title is very misleading. – Kavi Rama Murthy Apr 02 '19 at 23:45
Possible duplicate of $X$ and $Y$ i.i.d., $X+Y$ and $X-Y$ independent, $\mathbb{E}(X)=0 $and $\mathbb{E}(X^2)=1$. Show $X \sim N(0,1)$ – Angina Seng Apr 03 '19 at 03:20
No is duplicate. i haven't information about X+Y or X-Y, only X,Y iid random variables with mean equals 0 and variance equals 1. no more information – Maria Apr 03 '19 at 05:57
It is not a duplicate but there is a similar approach. You can show the characteristic function satisfies $\varphi(t) = \left(\varphi\left(t/\sqrt{2}\right)\right)^2 = \left(\varphi\left(t/2\right)\right)^4 = \left(\varphi\left(t/2^n\right)\right)^{4^n}$. Then consider the approach in the answer https://math.stackexchange.com/a/562891/6460 – Henry Apr 03 '19 at 07:29

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