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Noting this question, it is known that the sum of squares of independent identically distributed gaussian random variables is a Noncentral chi-squared distribution. That is great and all, but what about:

$$ V = \frac{X_1+X_2+\dots+X_k}{\sqrt{\frac{X_1^2+X_2^2+\dots+X_k^2}{k}}} $$

Is this also a known distribution with well known properties?

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  • $\begingroup$ Obscure. The sum $X_1+\cdots+X_n$ is normal, not chi square, not non central chi square. Do you mean: let $X_1,\ldots,X_n$ iid with distribution $N(m,1)$, what is the distribution of $(X_1+\cdots+X_n)/c_n\sqrt{X^2_1+\cdots+X^2_n}?$ edit: I see you have corrected your statement. $\endgroup$ Commented May 23 at 6:39

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$\def\ed{\stackrel{\text{def}}{=}}$ Your expression $\ \frac{X_1+X_2+\dots+X_k}{\sqrt{\frac{X_1^2+X_2^2+\dots+X_k^2}{k}}}\ $ is not the ratio of the sample mean to the sample standard deviation, as implied by the title of the question, but the ratio of the sample sum to the sample root mean square. The sample mean is $$ \overline{X}\ed\frac{X_1+X_2+\dots+X_k}{k}\ . $$ I believe the term "sample standard deviation" is ambiguous, since I have seen it used to refer to both the quantity $$ \overline{S}\ed\sqrt{\frac{(X_1-\overline{X})^2+(X_2-\overline{X})^2+\dots+(X_k-\overline{X})^2}{k-1}}\ , $$ which I believe to be the more common usage, probably because the expression under the square root is an unbiassed estimate of the variance of the $\ X_i\ $, and also to refer to the quantity $$ \left(\sqrt{\frac{k-1}{k}}\right)\overline{S}=\sqrt{\frac{(X_1-\overline{X})^2+(X_2-\overline{X})^2+\dots+(X_k-\overline{X})^2}{k}}\ . $$ Thus the ratio of the sample mean to the sample standard deviation is $$ \frac{X_1+X_2+\dots+X_k}{c\sqrt{(X_1-\overline{X})^2+(X_2-\overline{X})^2+\dots+(X_k-\overline{X})^2}}\ ,\tag{1}\label{e1} $$ where $\ c=\frac{k}{\sqrt{k-1}}\ $ if the first definition of sample standard deviation is used, or $\ c=\sqrt{k}\ $ if the second is used. In either case it's not all that difficult to obtain an expression for the cumulative distribution function for the random variable \eqref{e1}, although the details are somewhat technical.

If $\ \mu,\sigma\ $ are the mean and standard deviation, respectively, of $\ X_i\ ,$ and $\ x>0\ ,$ then \begin{align} &P\left(\frac{\sum_\limits{i=1}^kX_i}{c\sqrt{\sum_\limits{i=1}^k(X_i-\overline{X})^2}}\le x\right)\\ =&P\left(\sum_\limits{i=1}^kX_i\le0\right)\\ &+P\left(\sum_\limits{i=1}^kX_i>0, \left(\sum_\limits{i=1}^kX_i\right)^2\le (cx)^2\sum_\limits{i=1}^k(X_i-\overline{X})^2\right)\\ =&\mathcal{N}\big(k\mu,k\sigma^2;0\big)\\ &+P\left(\sum_\limits{i=1}^kX_i>0, 0\le (cx)^2\sum_\limits{i=1}^kX_i^2-\left(1+\frac{(cx)^2}{k}\right)\left(\sum_\limits{i=1}^kX_i\right)^2\right)\\ =&\mathcal{N}\big(k\mu,k\sigma^2;0\big)\\ +&P\left(\mathbf{1}_k^TX>0,0\le X^T\left((cx)^2I_{k\times k}-\left(1+\frac{(cx)^2}{k}\right)\mathbf{1}_k\mathbf{1}_k^T\right)X\right)\ . \end{align} Let $\ U\ $ be the unitary matrix whose first column is $\ \frac{1}{\sqrt{k}}\mathbf{1}_k\ ,$ and whose second to $\ k^\text{th}\ $ columns form an orthonormal basis of the subspace of $\ \mathbb{R}^k\ $ orthogonal to $\ \mathbf{1}_k\ .$ Then $$ U^T\left((cx)^2I_{k\times k}-\left(1+\frac{(cx)^2}{k}\right)\mathbf{1}_k\mathbf{1}_k^T\right)U=\pmatrix{-k&0&0&\dots&0\\ 0&(cx)^2&0&\dots&0\\ 0&0&(cx)^2&\dots&0\\ \vdots&\vdots&&\ddots&\vdots\\ 0&0&\dots&\dots&(cx)^2}\ , $$ and if $\ Y\ed U^TX\ ,$ then \begin{align} X^T\left((cx)^2I_{k\times k}-\left(1+\frac{(cx)^2}{k}\right)\mathbf{1}_k\mathbf{1}_k^T\right)X&=X^TUU^T\left((cx)^2I_{k\times k}-\left(1+\frac{(cx)^2}{k}\right)\mathbf{1}_k\mathbf{1}_k^T\right)UU^TX\\ &=Y^T\pmatrix{-k&0&0&\dots&0\\ 0&(cx)^2&0&\dots&0\\ 0&0&(cx)^2&\dots&0\\ \vdots&\vdots&&\ddots&\vdots\\ 0&0&\dots&\dots&(cx)^2}Y\\ &=(cx)^2\sum_{i=2}^kY_i^2-kY_1^2 \end{align} and $\ Y_1=\frac{1}{\sqrt{k}}\sum_\limits{i=1}^kX_i\ .$ Therefore, \begin{align} P\left(\frac{\sum_\limits{i=1}^kX_i}{c\sqrt{\sum_\limits{i=1}^k(X_i-\overline{X})^2}}\le x\right)&=\mathcal{N}\big(k\mu,k\sigma^2;0\big)+P\left(Y_1>0,0\le(cx)^2\sum_{i=2}^kY_i^2-kY_1^2\right)\ . \end{align} Since $\ X\ $ is multivariate normal with mean $\ \mu\mathbf{1}_k\ $ and covariance matrix $\ \sigma^2I_{k\times k}\ ,$ then $\ Y=U^TX\ $ is also multivariate normal with mean $$ \mu U^T\mathbf{1}_k=\mu\pmatrix{\sqrt{k}\\0\\0\\\vdots\\0} $$ and covariance matrix $$ \sigma^2UI_{k\times k}U^T=\sigma^2I_{k\times k}\ . $$ Thus, $\ Y_1,Y_2,Y_ 3,\dots,Y_k\ $ are independent normal variates, all with standard deviation $\ \sigma\ ,$ and all having zero mean with the exception of $\ Y_1\ ,$ which has mean $\ \mu\sqrt{k}\ $. Therefore $\ Z\ed\sum_\limits{i=2}^k\frac{Y_i^2}{\sigma^2}\ $ follows a chi-squared distribution with $\ k-1\ $ degrees of freedom, and \begin{align} P\left(Y_1>0,0\le(cx)^2\sum_{i=2}^kY_i^2-kY_1^2\right)&=P\left(0<\sqrt{k}Y_1\le cx\sigma\sqrt{Z}\right)\\ &=\int_0^\infty P\left(\left.0<\sqrt{k}Y_1\le cx\sigma\sqrt{z}\,\right|Z=z\right)d\chi_{k-1}^2(z)\\ &=\frac{1}{\sigma2^\frac{k}{2}\Gamma\left(\frac{k-1}{2}\right)\sqrt{\pi k}}\int_0^\infty z^\frac{k-3}{2}e^\frac{-z}{2}\int_0^{cx\sigma\sqrt{z}}e^\frac{-(y-k\mu)^2}{2k\sigma^2}dydz\ . \end{align} Finally, therefore \begin{align} P&\left(\frac{\sum_\limits{i=1}^kX_i}{c\sqrt{\sum_\limits{i=1}^k(X_i-\overline{X})^2}}\le x\right)\\ &=\frac{1}{\sigma2^\frac{k}{2}\Gamma\left(\frac{k-1}{2}\right)\sqrt{\pi k}}\int_0^\infty z^\frac{k-3}{2}e^\frac{-z}{2}\int_{-\infty}^{cx\sigma\sqrt{z}}e^\frac{-(y-k\mu)^2}{2k\sigma^2}dydz\ , \end{align} because $\ \mathcal{N}\big(k\mu,k\sigma^2;0\big)=\frac{1}{\sigma\sqrt{2\pi k}}{\displaystyle\int_{-\infty}^0}e^\frac{-(y-k\mu)^2}{2k\sigma^2}dy\ $ and $\ {\displaystyle\int_0^\infty}d\chi_{k-1}^2(z)=1\ .$ A similar calculation shows that the same formula holds also for $\ x\le0\ .$

  • A comprehensive explanation of how to derive the distributions of sampe statistics like this is given in Chapter 11 of Kendall and Stuart's The Advanced Theory of Statistics, although this particular example doesn't appear to be among those treated explicitly.
  • This statistic is a fairly natural one for which one might want to know the exact distribution, so I'm fairly sure that its derivation must appear somewhere in the statistical or mathematical literature, although, to the best of my memory, I've never seen one.
  • On the other hand, since the distribution depends explicitly on the true mean and standard deviation of the samples, which are very often not known, the statistic would be of limited use for statistical testing purposes. I'm therefore guessing that the distribution and its properties are not particularly well known.
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