Question

Let Y1 and Y2 have the joint probability distribution function given by f(y1,y2)={4y1y2 0<=y1<=y2<=2 and 0 else. what is the joint distribution function?

Answer 1

did you write the distribution down correctly?

Answer 2

yes.

Answer 3

when you integrate over what you have you get 8 (not 1)

Answer 4

\[\int\limits_{0}^{2}\int\limits_{0}^{y_2}4y_1y_2\,dy_1dy_2=8\]

Answer 5

okay, i must have calculated k wrong. when i calculated k I got 4

Answer 6

ah

Answer 7

Can you help me correctly calculate k?

Answer 8

divide both sides by 8

Answer 9

\[\int\limits_{0}^{2}\int\limits_{0}^{y_2}4y_1y_2\,dy_1dy_2=8\]

\[\int\limits_{0}^{2}\int\limits_{0}^{y_2}\frac{1}{2}y_1y_2\,dy_1dy_2=1\]

Answer 10

im sorry, the restrictions should be 0 to 1 not 0 to 2

Answer 11

ah...ok

Answer 12

still not 4

Answer 13

is it then correct? or did I still incorrectly calculate k?

Answer 14

\[\int\limits_{0}^{1}\int\limits_{0}^{y_2}8y_1y_2\,dy_1dy_2=1\]

Answer 15

can you show me how you got 8?

Answer 16

\[\int\limits_{0}^{1}\int\limits_{0}^{y_2}ky_1y_2\,dy_1dy_2\]

\[=\int\limits_{0}^{1}\left[k\frac{y_1^2}{2}y_2\right]_{0}^{y_2}dy_2\]
\[=\int\limits_{0}^{1}k\frac{y_2^2}{2}y_2dy_2\]

\[=\int\limits_{0}^{1}k\frac{y_2^3}{2}dy_2\]

\[=\left.k\frac{y_2^4}{8}\right|_{0}^{1}=\frac{k}{8}\]


set \[\frac{k}{8}=1\]

\[k=8\]

Answer 17

ah i did my second integral wrong. Thank you.

Answer 18

now back to my original question: joint distribution function

Answer 19

\[F(y_1,y_2)=P(Y_1\le y_1,Y_2\le y_2)\]