1. ## Joint density function

This is from Exercise 5.26 from Mathematical Statistics with Applications v6.

f (y1, y2) = {2, 0 <= y1 <= 1, 0 <= y2 <= 1, 0 <= y1 + y2 <= 1
{0, elsewhere

Find :
P ( Y1 ≥ ½ | Y2 = ¼ ) =
P ( Y1 ≥ ½ | Y2 ≤ ¼ ) =

So I know I have to do something with the f(y1, y2) = f(y1, y2) / f (y2) but I really am not sure how to get there. Any help would be ace. Cheers.

2. Originally Posted by statsdud
This is from Exercise 5.26 from Mathematical Statistics with Applications v6.

f (y1, y2) = {2, 0 <= y1 <= 1, 0 <= y2 <= 1, 0 <= y1 + y2 <= 1
{0, elsewhere

Find :
P ( Y1 ≥ &#189; | Y2 = &#188; ) =
P ( Y1 ≥ &#189; | Y2 ≤ &#188; ) =

So I know I have to do something with the f(y1, y2) = f(y1, y2) / f (y2) but I really am not sure how to get there. Any help would be ace. Cheers.
you mean : f(y1|y2) = f(y1, y2) / f (y2)

you can use that or just draw a picture and calculate the ration of the areas (since the density is constant on its support) at least for the first one.

3. Originally Posted by statsdud
This is from Exercise 5.26 from Mathematical Statistics with Applications v6.

f (y1, y2) = {2, 0 <= y1 <= 1, 0 <= y2 <= 1, 0 <= y1 + y2 <= 1
{0, elsewhere

Find :
P ( Y1 ≥ ½ | Y2 = ¼ ) =
P ( Y1 ≥ ½ | Y2 ≤ ¼ ) =

So I know I have to do something with the f(y1, y2) = f(y1, y2) / f (y2) but I really am not sure how to get there. Any help would be ace. Cheers.
Do you know how to calculate f_Y2(y2)?

4. f_Y2(y2) - Wouldn't it just stay at 2?

5. Originally Posted by statsdud
f_Y2(y2) - Wouldn't it just stay at 2?
No

...........

6. Do I have to sub y2 into it? So something like 2y2? Or do I have to integrate with respect to y2 to get there?

7. Originally Posted by statsdud
Do I have to sub y2 into it? So something like 2y2? Or do I have to integrate with respect to y2 to get there?
you have to integrate w.r.t. y1 (you are finding the marginal distribution)

8. Ok so f(y1|y2) = f(y1, y2) / f (y2) would be 2 / 2y1 = 1/y1 ?

9. Ok so then it integrates between 0 and y2 so it would be 1/y2 instead yeah?

So I'd sub in y2 = 1/4 and then do the integration between 0 and 1/2 to get to the probability?

10. Originally Posted by statsdud
Ok so then it integrates between 0 and y2 so it would be 1/y2 instead yeah?

So I'd sub in y2 = 1/4 and then do the integration between 0 and 1/2 to get to the probability?
no...

since

0 <= y1 + y2 <= 1

and y1>=0 and y2>=0
we have that

0<=y1<=1-y2

so integrate from 0 to 1-y2

11. So:

f( Y1 | Y2 ) = f( Y1 , Y2 ) / f2( Y2 )
= 2 / 2 ( 1 – y2 )
= 1 / 1 – y2

P ( Y1 ≥ ½ | Y2 = ¼ ) = ∫ 1 / ( 1 – ¼ ) dy1
= 1/21 ∫ 1 / ( 1 – ¼ ) dy1
= 1/21 ∫ 4/3 dy1
= [ 4/3 y1 ] 1/21
= 4/3 - 2/3
= 2/3 ?

but is it different for the ≤ ?

12. Originally Posted by statsdud
So:

f( Y1 | Y2 ) = f( Y1 , Y2 ) / f2( Y2 )
= 2 / 2 ( 1 – y2 )
= 1 / 1 – y2

P ( Y1 ≥ ½ | Y2 = ¼ ) = ∫ 1 / ( 1 – ¼ ) dy1
= 1/21 ∫ 1 / ( 1 – ¼ ) dy1
= 1/21 ∫ 4/3 dy1
= [ 4/3 y1 ] 1/21
= 4/3 - 2/3
= 2/3 ?

but is it different for the ≤ ?
what is 1/21 in 1/21 ∫ 1 / ( 1 – ¼ ) dy1?

13. The intergratal between 1 and 1/2 of 1/(1-1/4)

14. Originally Posted by statsdud
The intergratal between 1 and 1/2 of 1/(1-1/4)
you cant go all the way to 1

remember we have 0<=y1+y2<=1

and y2=1/4 so the largest y1 can be is 3/4

15. Ahhh ok, so if I change that in the integral, everything else is ok?