# Thread: joint probability density function, using the Jacobian

1. ## joint probability density function, using the Jacobian

if X1 and X2 are independent random variables from the standard normal distribution i have worked out the joint probability density function of X1,X2 to be
(1/2π) exp[-1(x12 + x22 ) /2 ] (is this correct?)

then if Y1=X1 and Y2=X2/X1 the joint probability density function of Y1,Y2 is
(1/2π) exp[-1y12 (1+y22)/2) ] |J|

i worked out J to be y1

is that all correct and what does the joint probability density function of Y1, Y2 then become?
thanks

2. *where it says x12 and y22 etc this is x1 squared and y2 squared etc, dont no why it came out like that

3. good job.

Although your notations are really hard to understand, your method looks correct.. You just need to plug in the value of |J| and you have the joint pdf

4. i will try and write it out a bit fuller and clearer;

X1 and X2 are independent random variables from a standard normal distribution
so the joint p.d.f of X1, X2 = p.d.f X1 x p.d.f X2
= ( 1/(2pie)^1/2) ) exp[ -1(x1^2)/2 ] ( 1/(2pie)^1/2) ) exp[ -1(x2^2)/2 ]
= (1/2pie) exp[ -1(x1^2 + x2^2)/2 ]

If Y1 = X1 and Y2 = X2/X1 then the joint p.d.f of Y1, Y2 is
= (1/2pie) exp[ -1(y1^2 + (y1y2)^2)/2 ] |J|
= (1/2pie) exp[ -1(y1^2(1+y2^2))/2 ] |J|

where J = y1, so joint p.d.f of Y1, Y2 is
= (y1/2pie) exp[ -1(y1^2(1+y2^2))/2 ]

is this all correct?

5. Do not forget the SUPPORT of the distribution.

and,

Pie: the one you eat.

Pi: $\displaystyle \pi \;=\; 3.14159$