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.
f_Y2(y2) - Wouldn't it just stay at 2?
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?
Ok so f(y1|y2) = f(y1, y2) / f (y2) would be 2 / 2y1 = 1/y1 ?
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?
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 ≤ ?
The intergratal between 1 and 1/2 of 1/(1-1/4)
Ahhh ok, so if I change that in the integral, everything else is ok?
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