| Y and Z are independent random variables, uniform distribution on [0,1].
The random variable X is max(Y, Z).
In the following, it is assumed that 0 <= a,b <= 1.
The probability that X < a, given that Y > b is:
P( X < a | Y > b ) = a(max(a,b)-b)/(1-b)
Other results:
P( X < a | Y < b ) = a*min(a,b)
P( Y < b | X > a ) = (b - a*min(a,b))/(1-a^2)
P( Y < b | X < a ) = min(a,b)/a
P( X < a ) = a^a
The expected value of X, given that Y > b is:
E( X | Y > b ) = ( b + sqrt(b^2-2b+2) ) / 2
Other results:
E( X | Y < b ) = 1/( 2 min(b,sqrt(1/2)) )
E( Y | X > a ) = sqrt( (a^2+1)/2 )
E( Y | X < a ) = a/2
|
| Is this (from .1) correct:
> The expected value of X, given that Y > b is:
> E( X | Y > b ) = ( b + sqrt(b^2-2b+2) ) / 2
? I get
E ( X | Y > 0 ) = 2 / 3
I tried it with this picture:
z
-------------------------
| | It's a strip (of W's) on
| | which the function
| | max (y, z)
|WWWWWWWWWWWWWWW | has value x. The strip is
| W | 2 x dx in area, so
| W |
| W | E(x) = integral of x 2 x dx
| W | = 2 / 3
| W |
| W |
------------------------- y
x
John
|
| Yes, you're right.
Excusing the tacky integral signs, the expected value is:
1 1 1 1
S S max(y,z) dz dy / S S dz dy
b 0 b 0
1 y 1
= S ( S max(y,z) dz + S max(y,z) dz ) dy / (1-b)
b 0 y
1 y 1
= S ( S y dz + S z dz ) dy / (1-b)
b 0 y
1 2 2
= S ( y + (1-y )/2 ) dy / (1-b)
b
1 2
= S ( (1+y )/2 ) dy / (1-b)
b
3 y=1
= (y/2 + y /6) / (1-b)
y=b
3
= ((1/2 + 1/6) - (b/2 + b /6)) / (1-b)
3
= ((1/2 + 1/6) - (b/2 + b /6)) / (1-b)
3
(4 - 3*b - b )
= --------------
6 (1-b)
2
(1-b) (4 + b + b )
= -----------------
6 (1-b)
2
= (4 + b + b ) / 6
|