# [SOLVED] joint density function /joint probability

• Nov 15th 2009, 03:37 AM
hazel
[SOLVED] joint density function /joint probability
3.55 walpole

Given the joint density function

f (x,y) = 6-x-y/ 8 , 0<x<2, 2<y<4
0, elsewhere,

find P (1<Y<3|X=1).

i have got g(x)=(3-x)/4 and also f(y|x) = (6-x-y) / (2(3-x))

i substitute x=1 into (6-x-y) / (2(3-x)) and i got (5-y)

so i integrate (5-y) to dy from 1 to 3.
but it is not right.

did i miss something ?
• Nov 15th 2009, 04:56 AM
awkward
Quote:

Originally Posted by hazel
3.55 walpole

Given the joint density function

f (x,y) = 6-x-y/ 8 , 0<x<2, 2<y<4
0, elsewhere,

find P (1<Y<3|X=1).

i have got g(x)=(3-x)/4 and also f(y|x) = (6-x-y) / (2(3-x))

i substitute x=1 into (6-x-y) / (2(3-x)) and i got (5-y) / 2

so i integrate (5-y) to dy from 1 to 3. Should be 2 to 3
but it is not right.

did i miss something ?

.
• Nov 15th 2009, 05:15 AM
galactus
The probability must be < 1.

Try $\int_{0}^{2}\int_{2}^{3}\frac{6-x-y}{8}dydx$
• Nov 15th 2009, 07:05 AM
HallsofIvy
Quote:

Originally Posted by hazel
3.55 walpole

Given the joint density function

f (x,y) = 6-x-y/ 8 , 0<x<2, 2<y<4
0, elsewhere,

Did you intend this to be (6- x- y)/8 or 6- x- (y/8)?
Either way it is impossible. for f(x,y) to be a joint density function, you must have $\int \int f(x,y)dx dy= 1$
but $\int_{x=0}^2 \int_{y=0}^4 \frac{6- x- y}{8}dy dx= 3$.
$\int_{x=0}^2\int_{y=0}^4 (6- x- \frac{y}{9})dy dx$ is even worse!

Quote:

find P (1<Y<3|X=1).

i have got g(x)=(3-x)/4 and also f(y|x) = (6-x-y) / (2(3-x))

i substitute x=1 into (6-x-y) / (2(3-x)) and i got (5-y)

so i integrate (5-y) to dy from 1 to 3.
but it is not right.

did i miss something ?
If f(x,y)= (6- x- y)/24 for 0< x< 2, 0< y< 4, and 0 elsewhere, then it is a joint probability density.

With x= 1, we have $P(0< y< 4|X= 1)= C\int_{y=0}^4 \frac{5- y}{24}dy= C(1/2)= 1$ so C= 2.

And, in that case, P(1< Y< 3|X= 1) is given by [tex]2\int_{y=1}^3 \frac{5- y}{24}dy
• Nov 17th 2009, 05:53 AM
hazel
sorry , guess i integrated wrongly.
it should be integrate (6-x-y) / 2(3-x) dy from 2 to 3, where x=1
and i should get :
=1/4 (5-y) dy from 2 to 3
= [5y - y^2] from 2 to 3
=5/8
• Dec 5th 2009, 11:30 PM
hazel
just some clarification: for this question , is there an error in the question ?

Original question :
(3.55 walpole)

Given the joint density function

f (x,y) = (6-x-y)/ 8 , 0<x<2, 2<y<4 , 0, elsewhere,

find P (1<Y<3|X=1).

Should it be : Find P(2<Y<3 | X=1) ? since Y<2 in the given condition. it can't be "1" in P(1<Y<3.....)

so when solving ,
g(x) = int 2-4 (6-x-y)/8 dy
=( 3-x )/4 for 0< x <2

and for P (2 <Y<3 |X=1)
= int 2-3 (6-x-y) /8 div (3-x) /4 , X=1