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Please help me to find the limiting distribution (as $n \rightarrow \infty$) of the following: $$ U_n = \frac{X_1 + X_2 + \ldots + X_n}{X_1^3 + X_2^3 + \ldots X_n^3},$$ where $X_i$ are iid $N(0,1)$.

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    $\begingroup$ Have you tried looking at transformations of random variables? For instance, one might try characteristic functions, Laplace-Stieltjes transforms, etcetera. $\endgroup$ – Stijn May 13 '13 at 11:28
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    $\begingroup$ Hint: The numerator and denominator are asymptotically bivariate normal. You may compute their moments directly: their means are obviously zero, the variance of the numerator is $n$, the variance of the denominator is $15n$, and the covariance is $3n$. (Thus the correlation is $3/\sqrt{15} \approx 0.775$.) To find the limiting distribution, express any zero-mean bivariate normal $(U,V)$ in the form $(A, \beta A + B)$ for independent zero-mean normals $A$ and $B$ and constant $\beta$, then note that the ratio $V/U = \beta + B/A$ is a shifted scaled Cauchy distribution. $\endgroup$ – whuber Feb 7 '17 at 17:07
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If the formulation was $$ U_n = \frac{X_1 + X_2 + \ldots + X_n}{Y_1^3 + Y_2^3 + \ldots Y_n^3}$$ where $X_i\sim N(0,1)$ and $Y_i \sim N(0,1)$ are independent, it would be just a classic textbook exercise. You use the fact that $$F_n \stackrel{d}{\to} F,\quad G_n \stackrel{d}{\to}G \Rightarrow \frac{F_n}{G_n} \stackrel{d}{\to} \frac{F}{G}$$ and we can conclude that $U$ asymptotes to scaled Cauchy distribution.

But in your formulation, we can't apply the theorem due to dependence. My Monte-Carlo suggests that the limit distribution of $U_n$ is non-degenerate and it has no first moment and is not symmetric. I'd be interested in whether there is a explicit solution to this problem. I feel like the solution can only be written in terms of Wiener process.

[EDIT] Following whuber's hint, note that

$$(\frac{1}{\sqrt{n}}\sum{X_i},\frac{1}{\sqrt{n}}\sum{X_i^3})\stackrel{d}{\to}(Z_1,Z_2)$$ where $$(Z_1,Z_2) \sim N(0,\pmatrix{1 &3 \\3 &15})$$ by noting that $E[X_1^4]=3$ and $E[X_1^6]=15$. (moments of standard normal, $(n-1)!!$ for even $n$) Then by continuous mapping theorem, we have $$U_n \stackrel{d}{\to} \frac{Z_1}{Z_2}$$ Noting that we can write $Z_1 = \frac{1}{5}Z_2+\sqrt{\frac{2}{5}}Z_3$ where $Z_3\sim N(0,1)$ and independent of $Z_2$, we conclude that $$U_n \stackrel{d}{\to} \frac{1}{5}+\sqrt{\frac{2}{5}}\frac{Z_3}{Z_2} \equiv \frac{1}{5}+\sqrt{\frac{2}{75}}\Gamma$$ where $\Gamma \sim Cauchy$

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Some comments, not a full solution. This is to long for a comment, but really only a comment. Some properties of the solution. Since the $X_i$ are iid standard normal, which is a symmetric (about zero) distribution, $X_i^3$ will also have symmetric distributions, and sums of (independent) symmetric rv's will be symmetric. So this is a ratio with the numerator and denominator both symmetric, so will be symmetric. The denominator will have a continuous density which is positive at zero, so we will expect the ratio to lack expectation (It is a general result that if $Z$ is a random variable with continuous density positive at zero, then the $1/X$ will lack expectation. See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?). But here, there is dependence between numerator and denominator which complicates the matter ... (Clearly needs more thought here).

The interesting paper https://projecteuclid.org/download/pdf_1/euclid.aop/1176991795 shows that $x_i^3$ above, the cube of standard normal variables, has an indeterminate distribution "in the hamburger sense", that is, it is not determined by its moments! So the comment above about using transforms, might indicate a difficult way to proceed!

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