Finding a probability given joint p.d.f of the continuous random variables

[nabilsaleh]

I'm having a trouble doing this kind of problems :S

Lets try this for example:

The joint p.d.f of the continuous random variable X and Y is:

f(x,y)= (2y+x)/8 for 0<x<2 ; 1<y<2

now we're asked to find a probability, say P(X+Y<2)

I know i have to double integrate but how do I choose my integration limits and what do I do after I finish integrating?

My Attempt at the solution is:

P(X+Y<2) = \int_ \int (2y+x)/8 = 1 = \int [y^2/8 + xy/ 8]

 

[mathman]

Inner integral of x, (0,2-y), outer integral of y (1,2).

or

Inner integral of y, (1,2-x), outer integral of x (0,1). (For x > 1, y < 1).

 

[LCKurtz]

I'm having a trouble doing this kind of problems :S

Lets try this for example:

The joint p.d.f of the continuous random variable X and Y is:

f(x,y)= (2y+x)/8 for 0<x<2 ; 1<y<2

now we're asked to find a probability, say P(X+Y<2)

I know i have to double integrate but how do I choose my integration limits and what do I do after I finish integrating?

My Attempt at the solution is:

P(X+Y<2) = \int_ \int (2y+x)/8 = 1 = \int [y^2/8 + xy/ 8]

The trick to solving this kind of problem is to draw a picture of the relevant area. If you were in Calc III doing a double integral for area, that is what you would do to help you figure out the limits of integration. Same thing here. In Calc III if you were calculating the area of a region R you would do this integral:
$$\iint_R 1\,dxdy$$
with appropriate limits for dxdy or dydx, whichever is easier. The only difference is that instead of integrating 1 and getting area, you integrate your joint density f(x,y) over the region. In your example the appropriate region is a triangle. More interesting would be if you wanted P(X+Y<3). Do you see that in that case you would to need to break the region into two parts as a dydx integral but not as a dxdy integral? Draw a picture and see.