Joint density function

#1
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.
 

Martingale

TS Contributor
#2
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.
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.
 

Martingale

TS Contributor
#3
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)?
 
#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?
 

Martingale

TS Contributor
#10
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 ≤ ?
 

Martingale

TS Contributor
#12
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?


also the answer isn't 2/3