Re: Normal distribution

In article <ee1ei4$3v86@xxxxxxxxxxxxxxxxxxxx>,
Herman Rubin <hrubin@xxxxxxxxxxxxxxxxxxxx> wrote:
In article <1157890187.289066.126980@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
ManOfLight <mladensavov@xxxxxxxxx> wrote:
Hello everybody,

I have the following question which I will formulate briefly. We know
that given a standard normal distribution we have

P(X>x) assymp C(1/x)e^{-x^{2}/2} as x goes to infinity.

We could find similar expressions for a multidimensional standard
normal distribution X for
P(||X||>x)=P(X is outside the ball with radius x).

My question is what can be said for
P(X is outside an ellipse ) assymp ?
as the ellipse gets bigger and bigger or
equivalently
P(c_{1}x_{1}^{2}+...+ c_{k}x_{k}^{2}>=t) assympt ?
as t goes to infinity.

Assuming the covariance matrix is the identity,
the asymptotic rate for the logarithm is given
by the largest circle contained in the ellipse
with center at the origin. The coefficient
(the analog of C/x in the univariate case) is
harder to get, but it will be like a power of
the radius, and there will be an asymptotic
expansion with further terms.

I think it's the smallest circle not contained in
the ellipse, not the largest.
Consider the 2d case, X and Y with covariance matrix I.
In polar coordinates (R, Theta), R has density
f_R(r) = r exp(-r^2/2), and Theta is uniform on
[0, 2 pi). The ellipse (x/a)^2 + (y/b)^2 = t
becomes r^2 (cos^2(theta)/a^2 + sin^2(theta)/b^2) = t
so Prob((X/a)^2 + (Y/b)^2 >= t)
= 1/(2 pi) int_0^(2 pi) int_{sqrt(t/g(theta))}^infty
r exp(-r^2/2) dr dtheta
= 1/(2 pi) int_0^{2 pi} exp(-t/(2 g(theta))) dtheta

where g(theta) = cos^2(theta)/a^2 + sin^2(theta)/b^2.
If a < b, the maximum of g is at theta=0 and theta=Pi,
with 1/g(theta) = a^2 - (a^4/b^2 - a^2) theta^2 + O(theta^4)
so, if I haven't made a mistake,

Prob((X/a)^2 + (Y/b)^2 >= t)
~ 1/pi int_{-infty}^infty exp(-t (a^2 - (a^4/b^2 - a^2) theta^2)/2)
dtheta
= sqrt(2/(pi t (b^2 - a^2))) b/a exp(-t a^2/2)

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

Relevant Pages

  • Re: Normal distribution
    ... that given a standard normal distribution we have ... P(X is outside an ellipse) assymp? ... the radius, and there will be an asymptotic ... Herman Rubin, Department of Statistics, Purdue University ...
    (sci.math)
  • Normal distribution
    ... that given a standard normal distribution we have ... P(X is outside an ellipse) assymp? ... as the ellipse gets bigger and bigger or ...
    (sci.math)