Thread: Joint probability density function

If the joint probability density function of X and Y is,
f(x,y)=1/8(6-x-y);0<x<2;2<y<4 or 0 otherwise
Findi)P(X+Y<3)
(ii)P(X<3/2 /Y<5/2)
In the first part I am completely lost whereas in the second part I am lost in finding limits for the required integrals.Pls help me through it

Originally Posted by roshanhero

If the joint probability density function of X and Y is,
f(x,y)=1/8(6-x-y);0<x<2;2<y<4 or 0 otherwise
Findi)P(X+Y<3)
(ii)P(X<3/2 /Y<5/2)
In the first part I am completely lost whereas in the second part I am lost in finding limits for the required integrals.Pls help me through it

Draw the rectangular region defined by 0<x<2 and 2<y<4. Then:

(i) Draw the line x + y = 3. Shade the region of the rectangle enclosed by the line. Integrate the joint pdf over that region. From the sketch it should be clear that:

.

Note: You should not be attempting questions like this until you have learnt how to evaluate double integrals.



(ii) You should know that .

Draw the horizontal line y = 5/2. Draw the vertical line x = 3/2. Then it should be clear that:

.

.

I draw the graph which might be wrong,from the graph I got the limits of x and y opposite to the values given by mr fantastic.Please clear me on it.

Originally Posted by roshanhero

I draw the graph which might be wrong,from the graph I got the limits of x and y opposite to the values given by mr fantastic.Please clear me on it.

The support of the joint pdf is the interior of the rectangle defined by 0<x<2 and 2<y<4. Have you drawn this? Now draw the line x + y = 3. Have you done this? Labels all the points where the line intersects the rectangle.

Now note that so shade the region of the rectangle that lies below the line x + y = 3. Integrate the given joint pdf over that shaded region. I have set the double integral up for you in my previous post.

How much experience do you have in calculating double integrals? I think you will benefit greatly by extensively reviewing that part of calculus.