Thread: Guassian distribution and polar co-ordinates

Hi
Not sure about this problem

Let X and Y be independent random variable each with a standard gaussian distribution. Let 0<M calculate P ((X^2)+(Y^2)<=(M^2)) using polar co-ordinates??????

my solution is as follows

let
x=rcos(p)
y=rsin(p)
then want to find P ((X^2)+(Y^2)<=(M^2)) = P(r<=m)

Joint distribution of X,Y is

FXY(x,y) = (1/(2*Pi))Exp[-0.5*((x^2)+(y^2))]

and so

FXY(r,p) = (1/(2*Pi))Exp[-0.5*(r^2)]

then P(r<=m) = to the double intergral of FXY(r,p) w.r.t r and p between the limits [0,M] and [0,2*pi]

this however isn't very nice

is this right or have i gone wrong some where . . . .

thankssss

Originally Posted by silk23

Hi
Not sure about this problem

Let X and Y be independent random variable each with a standard gaussian distribution. Let 0<M calculate P ((X^2)+(Y^2)<=(M^2)) using polar co-ordinates??????

my solution is as follows

let
x=rcos(p)
y=rsin(p)
then want to find P ((X^2)+(Y^2)<=(M^2)) = P(r<=m)

Joint distribution of X,Y is

FXY(x,y) = (1/(2*Pi))Exp[-0.5*((x^2)+(y^2))]

and so

FXY(r,p) = (1/(2*Pi))Exp[-0.5*(r^2)]

then P(r<=m) = to the double intergral of FXY(r,p) w.r.t r and p between the limits [0,M] and [0,2*pi]

this however isn't very nice

is this right or have i gone wrong some where . . . .

thankssss

YOU left out the Jacobian, r, which makes this a very simple integral.