1
$\begingroup$

Let $X_1, X_2, ..., X_k$ be an independent and identically distributed random variables. Assume $\mathbb{E}(X_i^2) < \infty$ for $1 \leq i \leq k$ and $$\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}} \ \mbox{and} \ X_1$$ has the same distribution function. Then $$X_1 \ \mbox{is normal with mean} \ 0.$$

I notice that the statement is similar to CLT version with iid, but CLT always works with an infinite sequence of random variables.

Here is what I try :

Since $$\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}} =^{distribution} X_1$$ $$\phi_{\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}}}(t) = \phi_{X_1}$$ where $\phi_X(t) = \mathbb{E}(e^{itX})$ the characteristic function.

By independent and identically distributed, $$(\phi_{\frac{X_1}{\sqrt{k}}}(t))^k=\phi_{\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}}}(t) = \phi_{X_1}(t).$$

That is where I do not know how to go on. Any hints please ?

$\endgroup$
  • $\begingroup$ Rewrite is as $\phi(t)=(\phi(t/\sqrt k))^k$ and Taylor $\phi$ expand to the second order around $0$. $\endgroup$ – A.S. Mar 25 '16 at 2:22
  • $\begingroup$ @A.S. You mean I should do Taylor serie expansion for $\phi(t)$ ? But is the characteristic always has derivative ? I mean characteristic is always continuous, but is it also differentiable ? Could you guide a bit more ? $\endgroup$ – Both Htob Mar 25 '16 at 2:30
  • $\begingroup$ Derivatives of $\phi_X$ at $0$ are closely related to moments of $X$: $\phi^{(k)}_X(0)=i^kE(X^k)$. Since you know the second moment is finite, you know that $\phi$ has a second derivative at $0$. Taylor expand, include and error term $R_3=o(t^2)$ and match powers of $t$ things on both sides. $\endgroup$ – A.S. Mar 25 '16 at 2:57
  • $\begingroup$ @A.S. I do the taylor expansion both sides as you suggests, and get a polynomial. But I cannot figure out why this things will lead me to conclude that $X \sim N(0, \sigma^2)$. What characterization of normal I should use ? Is it $\phi_Z(t) = e^{-t^2\sigma^2/2}$ ? I will get to this result by finding roots of polynomial ? $\endgroup$ – Both Htob Mar 26 '16 at 12:56
  • $\begingroup$ I didn't fully think it through. You need to Taylor expand cumulant generating function $\log \phi_X$ - not characteristic function. You get: $i\mu t-\sigma^2 t^2/2+R_3(t)=k(i\mu t/\sqrt k-\sigma^2 t^2/2k+R_3(t/\sqrt k))$, hence (assuming $k\ge 2$) $\mu=0$ and $R_3(t/\sqrt k)=R_3(t)/k$. Since $R_3=o(t^2)$, $R_3=0$. $\endgroup$ – A.S. Mar 26 '16 at 17:36
1
$\begingroup$

Since the question is tagged "central limit theorem", let me give a way using this result.

We assume $k\geqslant 2$. Consider $\left(Y_i\right)_{i\geqslant 1}$ a sequence of i.i.d. random variables distributed as $X$. By induction on $l$, it is possible to deduce that for each $l\geqslant 1$, $$\frac 1{\sqrt{k^l}}\sum_{i=1}^{k^l}Y_i\mbox{ as the same distribution as }X.$$ Now, since $X$ is necessarily centered and has a finite second moment, we conclude by the central limit theorem that $X$ is a centered (possibly degenerated) Gaussian random variable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.