Confusion when trying to compute a conditional expectation
$begingroup$
If $X$ and $Y$ are independent uniform rv on $(0,1)$, I want to Compute $E(X mid X < Y )$.
Try
By definition, what I am looking for is
$$ E(X mid X < Y) = int_0^1 x f_{X mid X<Y} (x|y) dx $$
now, we have
$$ f_{X mid X<Y} (x|y) = frac{ P(X < x, X<Y) }{P(X<Y)} = 2 P(X <x, X<Y)$$
Since $P(X<Y) = 1/2$. Now, here where I find really troublesome the notation to work with. how we compute expression of the form $P(X<x, X<Y)$? Now, this is supposed to be adouble integral where we integrate over the region ${ (x,y): x<x, x<y } $? but this region is triangular with area $x^2/2$ and so the required value is
$$ int_0^1 x x^2 = x^4/4 |_0^1 = 1/4 $$
but, the answer is supposed to be $1/6$. what is wrong in my calculation?
probability
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
$endgroup$
|
show 3 more comments
$begingroup$
If $X$ and $Y$ are independent uniform rv on $(0,1)$, I want to Compute $E(X mid X < Y )$.
Try
By definition, what I am looking for is
$$ E(X mid X < Y) = int_0^1 x f_{X mid X<Y} (x|y) dx $$
now, we have
$$ f_{X mid X<Y} (x|y) = frac{ P(X < x, X<Y) }{P(X<Y)} = 2 P(X <x, X<Y)$$
Since $P(X<Y) = 1/2$. Now, here where I find really troublesome the notation to work with. how we compute expression of the form $P(X<x, X<Y)$? Now, this is supposed to be adouble integral where we integrate over the region ${ (x,y): x<x, x<y } $? but this region is triangular with area $x^2/2$ and so the required value is
$$ int_0^1 x x^2 = x^4/4 |_0^1 = 1/4 $$
but, the answer is supposed to be $1/6$. what is wrong in my calculation?
probability
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
$endgroup$
$begingroup$
Note that a PDF is different from a CDF. I think you mean $F_{X|X<Y}(x)= frac{P[Xleq x, X<Y]}{P[X<Y]}$ (and note this has no dependence on a variable "$y$"). Its derivative $f_{X|X<Y}(x)$ also has no dependence on a variable "$y$." You may also want to double-check by drawing the (trapezoid) region in the unit square of all $(X,Y)$ associated with ${X<x, X<Y}$. This region depends on a value "$x$."
$endgroup$
– Michael
Dec 17 '18 at 4:31
1
$begingroup$
Weird, I get an answer of 1/3.
$endgroup$
– maridia
Dec 17 '18 at 4:39
$begingroup$
@norfair : The answer is $1/3$ but that is not why the above calculation is wrong.
$endgroup$
– Michael
Dec 17 '18 at 4:43
$begingroup$
The OP claims that the answer is 1/6.
$endgroup$
– maridia
Dec 17 '18 at 4:45
2
$begingroup$
Possible duplicate of Find conditional expectation
$endgroup$
– StubbornAtom
Dec 17 '18 at 5:30
|
show 3 more comments
$begingroup$
If $X$ and $Y$ are independent uniform rv on $(0,1)$, I want to Compute $E(X mid X < Y )$.
Try
By definition, what I am looking for is
$$ E(X mid X < Y) = int_0^1 x f_{X mid X<Y} (x|y) dx $$
now, we have
$$ f_{X mid X<Y} (x|y) = frac{ P(X < x, X<Y) }{P(X<Y)} = 2 P(X <x, X<Y)$$
Since $P(X<Y) = 1/2$. Now, here where I find really troublesome the notation to work with. how we compute expression of the form $P(X<x, X<Y)$? Now, this is supposed to be adouble integral where we integrate over the region ${ (x,y): x<x, x<y } $? but this region is triangular with area $x^2/2$ and so the required value is
$$ int_0^1 x x^2 = x^4/4 |_0^1 = 1/4 $$
but, the answer is supposed to be $1/6$. what is wrong in my calculation?
probability
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
$endgroup$
If $X$ and $Y$ are independent uniform rv on $(0,1)$, I want to Compute $E(X mid X < Y )$.
Try
By definition, what I am looking for is
$$ E(X mid X < Y) = int_0^1 x f_{X mid X<Y} (x|y) dx $$
now, we have
$$ f_{X mid X<Y} (x|y) = frac{ P(X < x, X<Y) }{P(X<Y)} = 2 P(X <x, X<Y)$$
Since $P(X<Y) = 1/2$. Now, here where I find really troublesome the notation to work with. how we compute expression of the form $P(X<x, X<Y)$? Now, this is supposed to be adouble integral where we integrate over the region ${ (x,y): x<x, x<y } $? but this region is triangular with area $x^2/2$ and so the required value is
$$ int_0^1 x x^2 = x^4/4 |_0^1 = 1/4 $$
but, the answer is supposed to be $1/6$. what is wrong in my calculation?
probability
probability
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
asked Dec 17 '18 at 4:08
Jimmy SabaterJimmy Sabater
2,226319
2,226319
$begingroup$
Note that a PDF is different from a CDF. I think you mean $F_{X|X<Y}(x)= frac{P[Xleq x, X<Y]}{P[X<Y]}$ (and note this has no dependence on a variable "$y$"). Its derivative $f_{X|X<Y}(x)$ also has no dependence on a variable "$y$." You may also want to double-check by drawing the (trapezoid) region in the unit square of all $(X,Y)$ associated with ${X<x, X<Y}$. This region depends on a value "$x$."
$endgroup$
– Michael
Dec 17 '18 at 4:31
1
$begingroup$
Weird, I get an answer of 1/3.
$endgroup$
– maridia
Dec 17 '18 at 4:39
$begingroup$
@norfair : The answer is $1/3$ but that is not why the above calculation is wrong.
$endgroup$
– Michael
Dec 17 '18 at 4:43
$begingroup$
The OP claims that the answer is 1/6.
$endgroup$
– maridia
Dec 17 '18 at 4:45
2
$begingroup$
Possible duplicate of Find conditional expectation
$endgroup$
– StubbornAtom
Dec 17 '18 at 5:30
|
show 3 more comments
$begingroup$
Note that a PDF is different from a CDF. I think you mean $F_{X|X<Y}(x)= frac{P[Xleq x, X<Y]}{P[X<Y]}$ (and note this has no dependence on a variable "$y$"). Its derivative $f_{X|X<Y}(x)$ also has no dependence on a variable "$y$." You may also want to double-check by drawing the (trapezoid) region in the unit square of all $(X,Y)$ associated with ${X<x, X<Y}$. This region depends on a value "$x$."
$endgroup$
– Michael
Dec 17 '18 at 4:31
1
$begingroup$
Weird, I get an answer of 1/3.
$endgroup$
– maridia
Dec 17 '18 at 4:39
$begingroup$
@norfair : The answer is $1/3$ but that is not why the above calculation is wrong.
$endgroup$
– Michael
Dec 17 '18 at 4:43
$begingroup$
The OP claims that the answer is 1/6.
$endgroup$
– maridia
Dec 17 '18 at 4:45
2
$begingroup$
Possible duplicate of Find conditional expectation
$endgroup$
– StubbornAtom
Dec 17 '18 at 5:30
$begingroup$
Note that a PDF is different from a CDF. I think you mean $F_{X|X<Y}(x)= frac{P[Xleq x, X<Y]}{P[X<Y]}$ (and note this has no dependence on a variable "$y$"). Its derivative $f_{X|X<Y}(x)$ also has no dependence on a variable "$y$." You may also want to double-check by drawing the (trapezoid) region in the unit square of all $(X,Y)$ associated with ${X<x, X<Y}$. This region depends on a value "$x$."
$endgroup$
– Michael
Dec 17 '18 at 4:31
$begingroup$
Note that a PDF is different from a CDF. I think you mean $F_{X|X<Y}(x)= frac{P[Xleq x, X<Y]}{P[X<Y]}$ (and note this has no dependence on a variable "$y$"). Its derivative $f_{X|X<Y}(x)$ also has no dependence on a variable "$y$." You may also want to double-check by drawing the (trapezoid) region in the unit square of all $(X,Y)$ associated with ${X<x, X<Y}$. This region depends on a value "$x$."
$endgroup$
– Michael
Dec 17 '18 at 4:31
1
1
$begingroup$
Weird, I get an answer of 1/3.
$endgroup$
– maridia
Dec 17 '18 at 4:39
$begingroup$
Weird, I get an answer of 1/3.
$endgroup$
– maridia
Dec 17 '18 at 4:39
$begingroup$
@norfair : The answer is $1/3$ but that is not why the above calculation is wrong.
$endgroup$
– Michael
Dec 17 '18 at 4:43
$begingroup$
@norfair : The answer is $1/3$ but that is not why the above calculation is wrong.
$endgroup$
– Michael
Dec 17 '18 at 4:43
$begingroup$
The OP claims that the answer is 1/6.
$endgroup$
– maridia
Dec 17 '18 at 4:45
$begingroup$
The OP claims that the answer is 1/6.
$endgroup$
– maridia
Dec 17 '18 at 4:45
2
2
$begingroup$
Possible duplicate of Find conditional expectation
$endgroup$
– StubbornAtom
Dec 17 '18 at 5:30
$begingroup$
Possible duplicate of Find conditional expectation
$endgroup$
– StubbornAtom
Dec 17 '18 at 5:30
|
show 3 more comments
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$begingroup$
Note that a PDF is different from a CDF. I think you mean $F_{X|X<Y}(x)= frac{P[Xleq x, X<Y]}{P[X<Y]}$ (and note this has no dependence on a variable "$y$"). Its derivative $f_{X|X<Y}(x)$ also has no dependence on a variable "$y$." You may also want to double-check by drawing the (trapezoid) region in the unit square of all $(X,Y)$ associated with ${X<x, X<Y}$. This region depends on a value "$x$."
$endgroup$
– Michael
Dec 17 '18 at 4:31
1
$begingroup$
Weird, I get an answer of 1/3.
$endgroup$
– maridia
Dec 17 '18 at 4:39
$begingroup$
@norfair : The answer is $1/3$ but that is not why the above calculation is wrong.
$endgroup$
– Michael
Dec 17 '18 at 4:43
$begingroup$
The OP claims that the answer is 1/6.
$endgroup$
– maridia
Dec 17 '18 at 4:45
2
$begingroup$
Possible duplicate of Find conditional expectation
$endgroup$
– StubbornAtom
Dec 17 '18 at 5:30