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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 ?

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  • $\begingroup$ 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, 2016 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, 2016 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, 2016 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, 2016 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, 2016 at 17:36

1 Answer 1

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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.

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