Discussion Problem_(21)

[duck]

Suppose, joint probability density fucntion (pdf) of is given by:
f(x,y) = 2 for 0<x<y<1.
         = 0 otherwise.

Find P(X+Y≤1).

[Sinistral]

let u=x+y; w=y
==> x=u-w & y=w
|J|=1

f(u,w)=2|J|=2 , 0<w<u<2w<2
        =  0 , e.w.

f_u(u)= integral (from u/2 to u) f(u,w) dw     (as w lies between u/2 and u)  where f_u(u) is the marginal pdf of u
        =2u/2 =u          0<u<2

P(X+Y≤1)= P(U<=1)= integral (from 0 to 1) f_u(u) du
                           =  integral (from 0 to 1)   u    du
                           = 1/2

[Sinistral]

I just thought it can be done in a simpler way:

P(X+Y<=1) = Integral (from x=0 to 1/2) Integral (from y=x to 1-x) f(x,y) dy dx  
                                 Since:  (double integration limits can be seen by superimposing x<y graph on x+y<=1 graph and hashing the required area)
               = 1/2

[Mauli]

what will the limits of the integral be if we fix y instead of y.?

[Mauli]

i am sorry .
i mean the limits when we fix y instead of x .?

[anon_econ]

0 to 1-y for 0.5<y<1 and 0 to y for 0<y<0.5. you'll have a sum of two integrals in this case.

[duck]

Great!
Correct Answer = 1/2

@Mauli:
If we fix y then, there will be two triangle over which you'll integrate.
Limits are clearly mentioned by Vasudha in this case.. :)

[Erika]

Im not getting the ans! x to 1-x, what value do we out in the integral?

[laracroft]

hey sinistral.... i did not understand this one.... kindly explain.... how exactly did u arrive at the solution !!???

[Sinistral]

find the area of the shaded triangle (figure below) and multiply it by f(x,y) ie 2. u will get 1/2

or u can integrate as done here
Integration is done over shaded area as only in this region x<y and x+y<1
it can be seen from the graph that as x goes from 0 to 1/2 => y goes from blue line to red line. blue line is y=x and red line is y=1-x. so limit of y will go from x to 1-x

[laracroft]

i know this will seem silly, but plz tell me how did u arrive at this limit of 1/2 for x ??

[Sinistral]

beyond x=1/2 we will go outside the required triangle. So x=1/2.
x=1/2 is the point where blue line (y=x) and red line (y=1-x) intersect.
equating the above 2:
x=1-x ==> x=1/2

[laracroft]

oh yeah.... thanku sooo much !

i am so sorry to have bothered u with such a silly question... plz excuse me..

bt thanx, nevertheless!