Suppose that $X$ and $Y$ are i.i.d. random variables such that $\mathscr{E}(X)=\mathscr{E}(Y)=0$ and $\mathscr{E}(X^2)=\mathscr{E}(Y^2)=1$. If $\frac{X+Y}{\sqrt{2}}$ have the same distribution as $X$, then $X$, $Y$ have the standard normal distribution.
$\bullet$ Proof.
Let $f_X(t)$ be the ch.f. of $X$. Consider the ch.f. of $\frac{X+Y}{\sqrt{2}}$, $$\mathscr{E}\left(\exp{\left\{i\frac{X+Y}{\sqrt{2}}t\right\}}\right)\overset{i.i.d.}{=}\left[\mathscr{E}\left(\exp{\left\{i\frac{X}{\sqrt{2}}t\right\}}\right)\right]^2=\left[f_X\left(\frac{t}{\sqrt{2}}\right)\right]^2\equiv f_X(t).$$Which implies that $$f_X(t)=\left[f_X\left(\frac{t}{\sqrt{2}}\right)\right]^2=\left[f_X\left(\frac{t}{2}\right)\right]^4=\cdots=\left[f_X\left(\frac{t}{2^n}\right)\right]^{4^n},$$and thus $$\begin{array}{rl}f_X(t) &=\left[f_X\left(\frac{t}{2^n}\right)\right]^{4^n}\\
& = \left[f_X(0)+\frac{it}{2^n}f'_X(0)+\frac{1}{2}\left(\frac{it}{2^n}\right)^2f''_X(0)+o\left(\frac{t}{2^n}\right)^2\right]^{4^n}\\
&=\left[1-\frac{t^2}{2}\frac{1}{4^n}+o\left(\frac{t}{2^n}\right)^2\right]^{4^n}\\
&\rightarrow\exp{\left\{-\frac{t^2}{2}\right\}}\mbox{ as }n\rightarrow\infty,\end{array}$$which is the ch.f. of the standard normal distribution. Since $X$, $Y$ are i.i.d. random variables, we have that $X$, $Y$ both have the standard normal distribution.
$\Box$
沒有留言:
張貼留言