Part 3: Independent Normal Distributions

In statistics, a normally distributed random variable with mean m and standard deviation s has a Gaussian density, which is function of the form

p1( x) =        1 s 2p   e -( x-m)2/( 2s2)

(3)

It follows that the joint density for two independent, normally distributed events is a function of two variables of the form

p( x,y) = p1( x) p2( y) =  1 2ps1s2 e -( x-m1) 2/( 2s12)     e -( x-m2) 2/( 2s22)

For simplicity, we will consider here only independent, normally distributed events with mean m = 0 in both and standard deviations s = s₁ = s₂. In such cases, the joint density function is

p( x,y) =  1 2ps2 e -( x2+y2)/( 2s2)

However, if either of  m₁ or m₂ are not zero, then substitutions of the form u = x-m will reduce the joint density into the form considered here.

EXAMPLE 5    Let ( X,Y) be the coordinates of the final resting place of a ball which is released from a position on the z-axis toward the xy-plane, and suppose the two coordinates are independently normally distributed with a mean of 0 and a standard deviation of 3 feet.

What is the probability that the ball's final resting place will be no more than 5 feet from the origin?

       

Solution: Since s = 3, the joint density function is

p( x,y) =  1 18p e -( x2+y2) /18

and we want to know the probability that ( X,Y) will be in a circle with radius 5 centered at the origin. Since such a circle corresponds to r = 0 to r = 5 for q in [ 0,2p] , the probability is

P[ X 2+Y 2  £  25]  óó õõ   R    1 18p   e -(x2+y2) /18    dA

Converting to polar coordinates then yields

P[ X 2  +  Y 2  £  25] =  1 18p ó õ 2p 0  ó õ 5 0  e -r2/18      rdrdq

and if we now let u = r², du = 2rdr, then u( 0) = 0 and u( 5) = 25 implies that

P[ X 2  +  Y 2  £  25] =  1 36p ó õ 2p 0  ó õ 25 0  e-u/18dudq =  1 36p ó õ 2p 0  ( 18-18e-25/18) dq = 1-e-25/18 = 0.750648

Thus, there is about a 75% chance that the ball's final resting place will be no more than 5 feet from the origin.