# Joint density function

#### statsdud

##### New Member
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
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
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)?

#### statsdud

##### New Member
f_Y2(y2) - Wouldn't it just stay at 2?

#### statsdud

##### New Member
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?

#### Martingale

##### TS Contributor
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)

#### statsdud

##### New Member
Ok so f(y1|y2) = f(y1, y2) / f (y2) would be 2 / 2y1 = 1/y1 ?

#### statsdud

##### New Member
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
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

#### statsdud

##### New Member
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
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?

#### statsdud

##### New Member
The intergratal between 1 and 1/2 of 1/(1-1/4)

#### Martingale

##### TS Contributor
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

#### statsdud

##### New Member
Ahhh ok, so if I change that in the integral, everything else is ok?

#### statsdud

##### New Member
Ok cool. Thanks heaps for your help.

So I just do the same for the ≤ ?

#### Martingale

##### TS Contributor
...

but is it different for the ≤ ?
yes. you are going to have a double integral.

Be careful with those limits of integration. Try drawing a picture...that might help.