# distribution of sample mean to sample standard deviation

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?

• 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. Commented May 23 at 6:39

$$\def\ed{\stackrel{\text{def}}{=}}$$ 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 $$\ x\le0\ .$$