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Suppose ${X_i}$ follows the standard normal distributon N(0,1), what is the distribution of $\frac{\sum{x_i}}{\sqrt{\sum{x_i^2}}}$

Xi'an
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Ethannn
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    Welcome to CV. If this question relates to a class exercise, please see https://stats.stackexchange.com/tags/self-study/info and add the tag to modify the question accordingly. – Pitouille Nov 19 '21 at 08:38
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    Assuming $X_i$'s are independent, the vector $\left(\frac{X_i}{\sqrt{\sum X_i^2}}\right)_{1\le i\le n}$ is uniformly distributed on the surface of a unit sphere. So you are looking for the distribution of sum of the components of this vector. Where do you need this? – StubbornAtom Nov 19 '21 at 10:38
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    That this is a uniform distribution on the sphere is shown at https://stats.stackexchange.com/questions/7977. The distribution of the components is derived at https://stats.stackexchange.com/questions/85916. These two threads together therefore (fully) answer the question. – whuber Nov 19 '21 at 14:14

2 Answers2

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Hint: you can rewrite $\sum x_i^2 = \bar{x}^2 + \sum (x_i -\bar{x})^2$.

Note that the terms $n\bar{x}^2$ and $\sum (x_i -\bar{x})^2$ are independent chi-squared distributed variables.

Then you can rewrite $Y = \frac{\sum{x_i}}{\sqrt{\sum{x_i^2}}}$ as

$$\begin{array}{} Y^2 &= &\frac{\left(\sum{x_i}\right)^2}{{\sum{x_i^2}}}\\ &=& \frac{n^2\bar{x}^2}{\bar{x}^2+{\sum{(x_i-\bar{x})^2}}}\\ &=& \frac{n}{1/n+{\left(\sum{(x_i-\bar{x})^2}\right)/(n\bar{x}^2)}}\\ \end{array}$$

And this term in the denominator ${\left(\sum{(x_i-\bar{x})^2}\right)/(n\bar{x}^2)}$ follows an F-distribution

So we have that $n/Y^2-1/n$ follows a F-distribution.

Possibly this can be simplified further (edit: and as Henry noted in the comments you can 'turn it into' a beta distribution. Or... $Y^2$ follows a beta distribution.)

Sextus Empiricus
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  • $Y$ seems to be a symmetric Beta distribution relocated and rescaled to have mean $0$ and variance $1$ (might that be called a *standardised symmetric Beta distribution*?) – Henry Nov 19 '21 at 12:00
  • @Henry you are right you can simplify it to a beta distribution. The beta distribution and F-distribution [are related](https://en.m.wikipedia.org/wiki/F-distribution#Properties_and_related_distributions). My answer was mostly to give the hint to rewrite $\sum x_i^2 = \bar{x}^2 + \sum (x_i -\bar{x})^2$. – Sextus Empiricus Nov 19 '21 at 12:11
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Some empirical assertions if there are $n$ terms and $Y_n = \frac{\sum_1^n{X_i}}{\sqrt{\sum_1^n{X_i^2}}}$ with the $X_i$ iid $N(0,1)$:

  • $Y_n$ has expectation $0$ and variance $1$

  • $Y_n$ has support on $[-\sqrt{n},\sqrt{n}]$ (unless $n=1$, in which case the support is just $\{-1,1\}$)

  • The density for $Y_n$ on its support is proportional to $(n-y^2)^{(n-3)/2}$

  • In particular if $n=3$ this is a uniform distribution; if $n=4$ it is a Wigner semicircle distribution; if $n=2$ it is a version of the arcsine distribution

  • $\frac{Y_n}{\sqrt{n}}+\frac12$ has a Beta distribution with parameters $\alpha=\frac{n-1}{2}$ and $\beta=\frac{n-1}{2}$; perhaps $Y_n$ could be called a standardised symmetric Beta distribution

  • As $n$ increases, the distribution of $Y_n$ converges to a standard normal distribution

  • If $Z_{i,n} = \frac{{X_i}}{\sqrt{\sum_1^n{X_i^2}}}$ with the $X_i$ iid $N(0,1)$ then $Y_n =\sum_1^n Z_{i,n}$ and, as StubbornAtom commented, the vector $\mathbf Z_{n}$ is uniformly distributed on the surface of a unit hypersphere. $Z_{i,n}$ has the same distribution as $\frac{Y_n}{\sqrt{n}}$, so with mean $0$ and variance $\frac1n$ supported on $[-1,1]$

Henry
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  • This density $(n-y^2)^{(n-3)/2}$ is not of a beta distribution, but the density of a variable that is the square root of a variable that follows a beta distribution. https://en.m.wikipedia.org/wiki/Kumaraswamy_distribution#Relation_to_the_Beta_distribution – Sextus Empiricus Nov 19 '21 at 12:21
  • @SextusEmpiricus it does have a relocated rescaled beta distribution: if you let $w=\frac{y}{2\sqrt{n}} + \frac12$ you would get $(n-y^2)^{(n-3)/2}=(4n)^{(n-3)/2} w^{(n-3)/2}(1-w) ^{(n-3)/2}$ which is proportional to the density of a beta distribution – Henry Nov 19 '21 at 12:28
  • Many thanks to @Henry. I'm working on standardizing portfolio weighting and this is confusing me. Thanks and I'm new to this community. – Ethannn Nov 22 '21 at 07:27