¿Qué es tan interesante acerca de este resultado es lo mucho que se parece a la distribución de un coeficiente de correlación. Hay una razón.
Supongamos $(X,Y)$ es normal bivariante con la correlación cero y varianza común $\sigma^2$ para ambas variables. Dibujar un alcoholímetro de la muestra $(x_1,y_1), \ldots, (x_n,y_n)$. Es bien conocido, y fácilmente establecido geométricamente (como Fisher hizo hace un siglo) que la distribución de la muestra coeficiente de correlación
$$r = \frac{\sum_{i=1}^n(x_i - \bar x)(y_i - \bar y)}{(n-1) S_x S_y}$$
is
$$f(r) = \frac{1}{B\left(\frac{1}{2}, \frac{n}{2}-1\right)}\left(1-r^2\right)^{n/2-2},\ -1 \le r \le 1.$$
(Here, as usual, $\bar x$ and $\bar y$ are sample means and $S_x$ and $S_y$ are the square roots of the unbiased variance estimators.) $B$ is the Beta function, for which
$$\frac{1}{B\left(\frac{1}{2}, \frac{n}{2}-1\right)} = \frac{\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{n}{2}-1\right)} = \frac{\Gamma\left(\frac{n-1}{2}\right)}{\sqrt{\pi}\Gamma\left(\frac{n}{2}-1\right)} . \tag{1}$$
To compute $r$, we may exploit its invariance under rotations in $\mathbb{R}^n$ around the line generated by $(1,1,\ldots, 1)$, along with the invariance of the distribution of the sample under the same rotations, and choose $y_i/S_y$ to be any unit vector whose components sum to zero. One such vector is proportional to $v = (n-1, -1, \ldots, -1)$. Its standard deviation is
$$S_v = \sqrt{\frac{1}{n-1}\left((n-1)^2 + (-1)^2 + \cdots + (-1)^2\right)} = \sqrt{n}.$$
Consequently, $r$ must have the same distribution as
$$\frac{\sum_{i=1}^n(x_i - \bar x)(v_i - \bar v)}{(n-1) S_x S_v}
= \frac{(n-1)x_1 - x_2-\cdots-x_n}{(n-1) S_x \sqrt{n}}
= \frac{n(x_1 - \bar x)}{(n-1) S_x \sqrt{n}} = \frac{\sqrt{n}}{n-1}Z.$$
Therefore all we need to is rescale $r$ to find the distribution of $Z$:
$$f_Z(z) = \big|\frac{\sqrt{n}}{n-1}\big| f\left(\frac{\sqrt{n}}{n-1}z\right) = \frac{1}{B\left(\frac{1}{2}, \frac{n}{2}-1\right)} \frac{\sqrt{n}}{n-1}\left(1- \frac{n}{(n-1)^2}z^2\right)^{n/2-2}$$
for $|z| \le \frac{n-1}{\sqrt{n}}$. Formula (1) shows this is identical to that of the question.
Not entirely convinced? Here is the result of simulating this situation 100,000 times (with $n=4$, where the distribution is uniform).
![Figure]()
The first histogram plots the correlation coefficients of $(x_i,y_i),i=1,\ldots,4$ while the second histogram plots the correlation coefficients of $(x_i,v_i),i=1,\ldots,4)$ for a randomly chosen vector $v_i$ que permanece fijo durante todas las iteraciones. Ambos son uniformes. El QQ-plot de la derecha confirma estas distribuciones son esencialmente idénticas.
Aquí está la R código que genera la trama.
n <- 4
n.sim <- 1e5
set.seed(17)
par(mfrow=c(1,3))
#
# Simulate spherical bivariate normal samples of size n each.
#
x <- matrix(rnorm(n.sim*n), n)
y <- matrix(rnorm(n.sim*n), n)
#
# Look at the distribution of the correlation of `x` and `y`.
#
sim <- sapply(1:n.sim, function(i) cor(x[,i], y[,i]))
hist(sim)
#
# Specify *any* fixed vector in place of `y`.
#
v <- c(n-1, rep(-1, n-1)) # The case in question
v <- rnorm(n) # Can use anything you want
#
# Look at the distribution of the correlation of `x` with `v`.
#
sim2 <- sapply(1:n.sim, function(i) cor(x[,i], v))
hist(sim2)
#
# Compare the two distributions.
#
qqplot(sim, sim2, main="QQ Plot")
Referencia
R. A. Fisher, de la Frecuencia de distribución de los valores del coeficiente de correlación en las muestras de un indefinidamente grande de la población. Biometrika, 10, 507. Consulte La Sección 3. (Citado en Kendall Avanzado de la Teoría de las Estadísticas, 5ª Ed., sección 16.24.)
