View Full Version : Joint Density Function Question

02-21-2010, 03:48 AM
So (fy1,fy2) = e^-(y1 + y2) ) , for y1>0, y2>0
and 0 elsewhere.

What is P(Y1 + Y2 < 3)

The answer is .8009 I just can't work out how to get it. I can do other joint probability questions so perhaps my calculus is bad. Help would be greatly appreciated. Thank you.

02-21-2010, 04:01 AM
Oh and the way I approached it was to find P(Y1< 3 - Y2) where we work out joint distribution of the fuction of y1 from 0 to 3 - Y2 and y2 from 0 to 3.
Like I say I can do these questions normall, so I guess there is a problem with my calculus, unless there is something special about this particular distribution?

02-21-2010, 07:05 AM
You are in the right track.

First integrate with y1 then y2
Joint distribution of the function y2 from 0 to 3 and y1 from 0 to 3 - Y2
When you integrate with Y1... you will get e^-y2 *( 1-e^(y2 - 3) ) and you need to integrate this with y2

Other way you can solve this is ...

Let http://upload.wikimedia.org/math/5/0/4/504f8859b07e7a07cf3eaec4a89fc067.png be exponentially distributed and independent and http://upload.wikimedia.org/math/8/b/a/8baf08055b3be5c954c6fa8f5e729d22.png. Then http://upload.wikimedia.org/math/8/8/4/8849fd5230033e1b36fd44f801a4c2fa.png, i.e. Y has a Gamma distribution (http://en.wikipedia.org/wiki/Gamma_distribution).

But first approach is the direct.

02-21-2010, 07:27 AM
Ok I got that part right, I think it was when i was trying to integrate with y2 that I went astray. Should i use the integration by parts method here?

02-21-2010, 07:52 AM
Not required.

e^-y2 *( 1-e^(y2 - 3) ) = e^-y2 - e^-y2 * e^(y2 - 3) =e^-y2 - e^(-y2+y2 - 3)
=e^-y2 - e^(- 3) ( and this is is a simple function in y2)
Finally you get the answer 1- e^-3 -3*e^-3