is this P(X<Y | Y = y) = P(X<y) true
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Is this formula true? $P(X<Y | Y = y) = P(X<y)$
I am trying to solve this with conditional formula, but don't know how
$P(X<Y | Y= y) = frac{P(X<Y, Y= y)}{P(Y=y)} = frac{P(X<y)}{P(Y=y)}. $
conditional-probability
edited Nov 13 at 17:10
Crazy for maths
5109
asked Nov 13 at 16:56
Zewen Huang
61
add a comment |
up vote
0
down vote
favorite
Is this formula true? $P(X<Y | Y = y) = P(X<y)$
I am trying to solve this with conditional formula, but don't know how
$P(X<Y | Y= y) = frac{P(X<Y, Y= y)}{P(Y=y)} = frac{P(X<y)}{P(Y=y)}. $
conditional-probability
edited Nov 13 at 17:10
Crazy for maths
5109
asked Nov 13 at 16:56
Zewen Huang
61
1
Are $X$ and $Y$ independent?
– Mankind
Nov 13 at 17:12
suppose it is independent, then P(X<Y,Y =y) = P(X<Y)P(Y=y) gives P(X<Y) which is still not P(X<y)
– Zewen Huang
Nov 15 at 6:19
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is this formula true? $P(X<Y | Y = y) = P(X<y)$
I am trying to solve this with conditional formula, but don't know how
$P(X<Y | Y= y) = frac{P(X<Y, Y= y)}{P(Y=y)} = frac{P(X<y)}{P(Y=y)}. $
conditional-probability
edited Nov 13 at 17:10
Crazy for maths
5109
asked Nov 13 at 16:56
Zewen Huang
61
Is this formula true? $P(X<Y | Y = y) = P(X<y)$
I am trying to solve this with conditional formula, but don't know how
$P(X<Y | Y= y) = frac{P(X<Y, Y= y)}{P(Y=y)} = frac{P(X<y)}{P(Y=y)}. $
conditional-probability
conditional-probability
edited Nov 13 at 17:10
Crazy for maths
5109
asked Nov 13 at 16:56
Zewen Huang
61
edited Nov 13 at 17:10
Crazy for maths
5109
asked Nov 13 at 16:56
Zewen Huang
61
edited Nov 13 at 17:10
Crazy for maths
5109
edited Nov 13 at 17:10
Crazy for maths
5109
edited Nov 13 at 17:10
Crazy for maths
5109
5109
asked Nov 13 at 16:56
Zewen Huang
61
asked Nov 13 at 16:56
Zewen Huang
61
asked Nov 13 at 16:56
Zewen Huang
61
61
1
Are $X$ and $Y$ independent?
– Mankind
Nov 13 at 17:12
suppose it is independent, then P(X<Y,Y =y) = P(X<Y)P(Y=y) gives P(X<Y) which is still not P(X<y)
– Zewen Huang
Nov 15 at 6:19
add a comment |
1
Are $X$ and $Y$ independent?
– Mankind
Nov 13 at 17:12
suppose it is independent, then P(X<Y,Y =y) = P(X<Y)P(Y=y) gives P(X<Y) which is still not P(X<y)
– Zewen Huang
Nov 15 at 6:19
1
1
Are $X$ and $Y$ independent?
– Mankind
Nov 13 at 17:12
Are $X$ and $Y$ independent?
– Mankind
Nov 13 at 17:12
suppose it is independent, then P(X<Y,Y =y) = P(X<Y)P(Y=y) gives P(X<Y) which is still not P(X<y)
– Zewen Huang
Nov 15 at 6:19
suppose it is independent, then P(X<Y,Y =y) = P(X<Y)P(Y=y) gives P(X<Y) which is still not P(X<y)
– Zewen Huang
Nov 15 at 6:19
add a comment |
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Are $X$ and $Y$ independent?
– Mankind
Nov 13 at 17:12
suppose it is independent, then P(X<Y,Y =y) = P(X<Y)P(Y=y) gives P(X<Y) which is still not P(X<y)
– Zewen Huang
Nov 15 at 6:19