Q:

How to show these two random variables are not equal with positive probability?

Given iid $X_1, \dots, X_n \sim N(a, a^2), a \neq 0$,
how can we show that it is not true $\frac{(\sum_i X_i)^2}{n(n+1)} = \frac{\sum_i X_i^2}{2n}$ a.e.?
My thought was checking the moments. I got that
$\mathrm E \frac{(\sum_i X_i)^2}{n(n+1)} =  \mathrm E \frac{\sum_i X_i^2}{2n} = a^2$. But checking  $\mathrm Var [\frac{(\sum_i X_i)^2}{n(n+1)} - \frac{\sum_i X_i^2}{2n}]$ is very complicated, which I think not worth to try?

A:

If $n=1$ the conclusion does not hold. Asume that $n\geqslant2$ and consider the random variables $$S_n=\sum\limits_{k=2}^nX_k,\qquad T_n=(n+1)\sum\limits_{k=2}^nX_k^2-2S_n^2,$$ then some simple algebraic manipulations show that the event of interest is $$A_n=[(n-1)X_1^2-2S_nX_1+T_n=0],$$ that is, $$A_n=[X_1=U_n]\cup[X_1=V_n],$$ where the random variables $U_n$ and $V_n$ (the roots of the quadratic polynomial in $X_1$) are measurable with respect to $(X_k)_{2\leqslant k\leqslant n}$, in particular, they are independent of $X_1$. Here is a useful result:

Lemma: Let $X$ and $Y$ be independent, $X$ atomless, then $P(X=Y)=0$.

Since the distribution of $X_1$ is continuous, the lemma applied to $X=X_1$ and to $Y=U_n$ and $Y=V_n$ implies that $P(X_1=U_n)=P(X_1=V_n)=0$, hence, not only $P(A_n)\ne1$ but actually $P(A_n)=0$.
Proof of the lemma: By definition and by independence of $X$ and $Y$, $$P(X=Y)=\int\left(\int\mathbf 1_{x=y}\mathrm dP_X(x)\right)\mathrm dP_Y(y)=\int P(X=y)\mathrm dP_Y(y)=0,$$ since $P(X=y)=0$ for every $y$, QED.

