1 Mathematical considerations

From the mathematical point of view, the necessary condition for the isotropy is the stochastic independency of the sky distribution of the bursts on their observed physical properties. This means that, if f(b,l, x₁,...,x_n) ${\rm d}F {\rm d}x_1...{\rm d}x_2$ is the probability of finding an object in the ${\rm d}F=\cos b\; {\rm d}l\;{\rm d}b$ infinitesimal solid angle and in the $(x_1, x_1+{\rm d}x_1, ..., x_n,x_n+{\rm d}x_n)$ interval, one must have

$$f(l,b,x_1,...,x_n) = \omega(l,b) g(x_1,...,x_n).$$ (1)

Here $0\leq l \leq 360^\circ ,\; -90^\circ \leq b \leq 90^\circ$ give the celestial positions in Galactical coordinates, x_n ($n \geq 1$) measure the physical properties (peak fluxes, fluences, durations, etc.) of GRBs and g is their probability density. One may assume the fulfilment of this equation for GRBs.

In the case of spatial isotropy, assuming that the detection probability does not depend on the celestial direction, one has: $\omega(l,b)=1/(4\pi)$. In general case one may decompose the function $\omega(b,l)$ into the spherical harmonics. One obtains:

$$\omega(b,l) = (4\pi)^{-1/2} \omega_0 -$$

$$(3/(4\pi))^{1/2} (\omega_{1,-1} \cos b \sin l - \omega_{1,1} \cos b \cos l + \omega_{1,0} \sin b) +$$

$$(15/(16\pi))^{1/2}( \omega_{2,-2} \cos^2 b \sin 2l + \omega_{2,2} \cos^2 b \cos 2l -$$

$$\omega_{2,-1} \sin 2b \sin l - \omega_{2,1} \sin 2b \cos l) +$$

$$(5/(16\pi))^{1/2}\omega_{2,0} (3 \sin^2 b - 1) + {\rm higher\; order\; harm.}$$ (2)

The first term on the right-hand side is the monopole term, the following three ones are the dipole terms, the following five ones are the quadrupole terms. Since $\omega$ is constant for isotropic distribution, on the right-hand side any terms, except for $\omega_0$,should be identically zeros. To test this hypothesis one has to compute the values of the corresponding spherical harmonic at the celestial positions of the observed GRBs and apply, e.g., the Student test in order to see that the mean of the computed values significantly differs from zero.

A further trivial consequence of the isotropy is the expected equal number of bursts in celestial regions of equal areas. For example, one may divide the celestial sphere into two equal areas, e.g., taking those regions in which the sign of a given harmonic is either positive or negative, respectively. Then one may compare the number of GRBs in these regions by the standard binomial (Bernoulli) test. The details (together with the relevant references) of this test and also of the test based on the spherical harmonics are discussed in [Balázs et al. 1998].