# Joint probability density function

• Jul 22nd 2009, 04:24 AM
roshanhero
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
Find:(i)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
• Jul 22nd 2009, 05:24 AM
mr fantastic
Quote:

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
Find:(i)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:

$\displaystyle \Pr(X + Y < 3) = \int_{x = 0}^{x = 1} \int_{y = 2}^{y = 3 - x} f(x, y) \, dy \, dx$.

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

(ii) You should know that $\displaystyle \Pr(X < 3/2 | Y < 5/2) = \frac{\Pr(X < 3/2 \cap Y < 5/2)}{\Pr(Y < 5/2)}$.

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

$\displaystyle \Pr(X < 3/2 | Y < 5/2) = \int_{0}^{3/2} \int_{0}^{5/2} f(x, y) \, dx \, dy$.

$\displaystyle \Pr(Y < 5/2) = \int_{0}^{2} \int_{0}^{5/2} f(x, y) \, dx \, dy$.
• Jul 24th 2009, 06:08 AM
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.
• Jul 24th 2009, 10:00 PM
mr fantastic
Quote:

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 $\displaystyle x + y < 3 \Rightarrow y < -x + 3$ 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.