Suppose $X_1, X_2 \ldots X_n$ are i.i.d., and for some real coefficients $a_i$ we have $$ \sum_{i=1}^n a_i X_i \overset D =\mathcal N(0, 1). $$ Is it necessary, that $X_1$ has normal distribution?

My attempt: note that not all $a_i$ are zero. If all $a_i$ were equal, then using characteristic functions we would have $$ \varphi_{X_1} (a_1 t)^n = \exp\left(-\frac {t^2} 2\right)\quad \implies \quad \varphi_{X_1} (t) = \exp\left(-\frac {t^2} {2n a_1^2}\right)$$ so $X_1$ must have normal distribution. But we can't quite use this approach for different $a_i$.


1 Answer 1


This statement is similar to the Cramér's decomposition theorem, which states which states that if the sum of two independent nonconstant random variables is normally distributed, then each of the summands is also normally distributed. It generalizes straightforwardly for the case of $n$ variables.

So, denoting $\widetilde{X_i} = a_i X_i$, we can rewrite the statement as $\sum_{i=1}^n \widetilde{X_i} \overset D =\mathcal N(0, 1)$, where the summands are independent, but not identically distributed unless $a_1 = a_2 = ... = a_n$. From the Cramér's theorem we obtain that $\widetilde{X_i}$ are normally distributed, and so do $X_i$. Also, observe that

$$ \mathrm{Var}\left(\sum_{i=1}^n \widetilde{X_i}\right) = \mathrm{Var}\left(\sum_{i=1}^n a_i {X_i}\right) = \sum_{i=1}^n a_i^2 \cdot \mathrm{Var}X_i = \mathrm{Var}X_1 \cdot \sum_{i=1}^n a_i^2 = \mathrm{Var}\left(\mathcal N(0, 1)\right) = 1. $$

  • 3
    $\begingroup$ I guess another way to say it is to let $Y = \sum_{i=2}^n a_i X_i$ which is independent of $X_1$, and satisfies $a_1 X_1 + Y \sim N(0,1)$. Then by Cramér, $a_1 X_1$ in particular is normally distributed. $\endgroup$ Aug 25, 2022 at 15:04

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