Thread: Marginal Density

Joint density g(y1,y2,y3) = e^(-y3)*y2*y3^2
0<y1<1, 0<y2<1, 0<y3<+inf
marginal density g(y1),g(y2),g(y3) is supposed to be indepent so g(y1)*g(y2)*g(y3) = g(y1,y2,y3)
However, I keep getting wrong anwer and I do not know why?
Maybe there is some trick for setting up the boundary for integration?
I have g(y1) = 1, g(y2) = 1, g(y3) = (e^(-y3)*y3^2)/2.
What wrong with my calculation?

Originally Posted by ninano1205

Joint density g(y1,y2,y3) = e^(-y3)*y2*y3^2
0<y1<1, 0<y2<1, 0<y3<+inf
marginal density g(y1),g(y2),g(y3) is supposed to be indepent so g(y1)*g(y2)*g(y3) = g(y1,y2,y3)
However, I keep getting wrong anwer and I do not know why?
Maybe there is some trick for setting up the boundary for integration?
I have g(y1) = 1, g(y2) = 1, g(y3) = (e^(-y3)*y3^2)/2.
What wrong with my calculation?

The integrations look routine to me (all integral terminals are constants and the integrands are standard) so I can't see where your trouble might be. Please show the details of your calculations.

Since the joint density factors on a cube the three variables are independent.
THERE is NO need to integrate.
I teach out of Walpole Meyers and Wackerly and they seem to leave out that fact.
The densities are obvious by inspection.
Y_1 is uniform on (0,1)
The density of Y_2 is 2y_2 on (0,1)
And Y_3 is a gamma(3,1) and the constant that goes with
the (y_3)^2e^{-y_3} is 1/2.

YOUR density of Y_2 is wrong.