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$P(Y le X)=int_0^infty P(Y le X | X=x)f_X(x)dx$

0

1

$begingroup$

I was looking at a solution of a probability exercise and the author of the solution uses the formula $$P(Y le X)=int_0^infty P(Y le X | X=x)f_X(x)dx$$ where $X$, $Y$ are the random variables $f_X$ is the density function of the random variable $X$. From where does this result come from?

probability probability-theory random-variables conditional-probability density-function

share|cite|improve this question

edited Mar 10 at 20:05

gt6989b

36k22557

asked Mar 10 at 19:59

roi_saumonroi_saumon

69138

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It is the Law of total probability for random variables.
  $endgroup$
  – StubbornAtom
  Mar 10 at 20:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thanks, I didn't know!
  $endgroup$
  – roi_saumon
  Mar 11 at 16:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you should write $P(Y le X)=int_0^infty P(Y le X | X=x)f_X(x)dx=int_0^infty P(Y le x | X=x)f_X(x)dx$ when $X=x$ so $Yleq X=Yleq x$ ($Yleq X=Yleq x$)
  $endgroup$
  – masoud
  Apr 1 at 23:14
  
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I was looking at a solution of a probability exercise and the author of the solution uses the formula $$P(Y le X)=int_0^infty P(Y le X | X=x)f_X(x)dx$$ where $X$, $Y$ are the random variables $f_X$ is the density function of the random variable $X$. From where does this result come from?

probability probability-theory random-variables conditional-probability density-function

share|cite|improve this question

edited Mar 10 at 20:05

gt6989b

36k22557

asked Mar 10 at 19:59

roi_saumonroi_saumon

69138

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  It is the Law of total probability for random variables.
  $endgroup$
  – StubbornAtom
  Mar 10 at 20:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thanks, I didn't know!
  $endgroup$
  – roi_saumon
  Mar 11 at 16:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you should write $P(Y le X)=int_0^infty P(Y le X | X=x)f_X(x)dx=int_0^infty P(Y le x | X=x)f_X(x)dx$ when $X=x$ so $Yleq X=Yleq x$ ($Yleq X=Yleq x$)
  $endgroup$
  – masoud
  Apr 1 at 23:14
  
  

  
  
  
  

add a comment | 

$begingroup$

I was looking at a solution of a probability exercise and the author of the solution uses the formula $$P(Y le X)=int_0^infty P(Y le X | X=x)f_X(x)dx$$ where $X$, $Y$ are the random variables $f_X$ is the density function of the random variable $X$. From where does this result come from?

probability probability-theory random-variables conditional-probability density-function

share|cite|improve this question

edited Mar 10 at 20:05

gt6989b

36k22557

asked Mar 10 at 19:59

roi_saumonroi_saumon

69138

$endgroup$

share|cite|improve this question

edited Mar 10 at 20:05

gt6989b

36k22557

asked Mar 10 at 19:59

roi_saumonroi_saumon

69138

share|cite|improve this question

edited Mar 10 at 20:05

gt6989b

36k22557

asked Mar 10 at 19:59

roi_saumonroi_saumon

69138

edited Mar 10 at 20:05

gt6989b

36k22557

edited Mar 10 at 20:05

gt6989b

36k22557

asked Mar 10 at 19:59

roi_saumonroi_saumon

69138

asked Mar 10 at 19:59

roi_saumonroi_saumon

69138

2 Answers
2

active

oldest

votes

2

$begingroup$

This is conditional probability. Remember that
$$
mathbbP[A|B] = fracmathbbP[Acap B]mathbbP[B] iff mathbbP[Acap B] = mathbbP[A|B]mathbbP[B]
$$
but if $X$ is continuous, $mathbbP[X=x]=0$, the correct analog being
$$
lim_h to 0^+ mathbbP[x-hle X le x+h] = f_X(x).
$$

share|cite|improve this answer

answered Mar 10 at 20:04

gt6989bgt6989b

36k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thanks. But what does $P(Y le X vert X=x)$ mean? I know that by definition $P(A vert B) = P(A cap B) /P(B)$ but then this would not make sense since as you say, $P(X=x)=0$
  $endgroup$
  – roi_saumon
  Mar 11 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @roi_saumon You interpret $X=x$ as the limit I explained in the answer
  $endgroup$
  – gt6989b
  Mar 11 at 17:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $P(Y≤X|X=x)=P(Y≤X, X=x) f_X(x)$ so that $P(Y le X)=int_0^infty P(Y le X, X=x)(f_X(x))^2dx$? Isn't that wrong?
  $endgroup$
  – roi_saumon
  Mar 11 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @roi_saumon no, what you want to say is $$mathbbP[Y le X] = int_mathbbR mathbbP[Y le X cap X=x] dx = int_mathbbR mathbbP[Y le X | X=x] f_X(x) dx$$
  $endgroup$
  – gt6989b
  Mar 11 at 19:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how you guarantee that limit is exist? is any theorem show that always that limit exist ?
  $endgroup$
  – masoud
  Mar 22 at 11:31
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

conditional probability is special case of conditional expectation(definition of conditional probability is based on conditional expectation ).

define:

$A=omega in Omega$

$p(Yleq X)= p(A)=E(I_A)=E(E(I_A|X))$

$=E(g(X))$ (by definition of conditional expectation ,$E(I_A|X)$ is a function of $X$)

$=int g(x) f_X(x) dx=int E(I_A|X=x) f_X(x) dx=int p(A|X=x) f_X(x) dx=int p(Y leq X|X=x) f_X(x) dx=int p(Y leq x|X=x) f_X(x) dx$

so you pass the problem with, $p(X=x)=0$ in $p(Yleq X |X=x)$. since the defination of conditional expectation is based on projection property and not based on $p(A|B)=fracp(A cap B)p(B)$

share|cite|improve this answer

edited Apr 1 at 23:10

answered Mar 15 at 10:15

masoudmasoud

34718

$endgroup$

add a comment | 

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2

$begingroup$

This is conditional probability. Remember that
$$
mathbbP[A|B] = fracmathbbP[Acap B]mathbbP[B] iff mathbbP[Acap B] = mathbbP[A|B]mathbbP[B]
$$
but if $X$ is continuous, $mathbbP[X=x]=0$, the correct analog being
$$
lim_h to 0^+ mathbbP[x-hle X le x+h] = f_X(x).
$$

share|cite|improve this answer

answered Mar 10 at 20:04

gt6989bgt6989b

36k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thanks. But what does $P(Y le X vert X=x)$ mean? I know that by definition $P(A vert B) = P(A cap B) /P(B)$ but then this would not make sense since as you say, $P(X=x)=0$
  $endgroup$
  – roi_saumon
  Mar 11 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @roi_saumon You interpret $X=x$ as the limit I explained in the answer
  $endgroup$
  – gt6989b
  Mar 11 at 17:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $P(Y≤X|X=x)=P(Y≤X, X=x) f_X(x)$ so that $P(Y le X)=int_0^infty P(Y le X, X=x)(f_X(x))^2dx$? Isn't that wrong?
  $endgroup$
  – roi_saumon
  Mar 11 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @roi_saumon no, what you want to say is $$mathbbP[Y le X] = int_mathbbR mathbbP[Y le X cap X=x] dx = int_mathbbR mathbbP[Y le X | X=x] f_X(x) dx$$
  $endgroup$
  – gt6989b
  Mar 11 at 19:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how you guarantee that limit is exist? is any theorem show that always that limit exist ?
  $endgroup$
  – masoud
  Mar 22 at 11:31
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

This is conditional probability. Remember that
$$
mathbbP[A|B] = fracmathbbP[Acap B]mathbbP[B] iff mathbbP[Acap B] = mathbbP[A|B]mathbbP[B]
$$
but if $X$ is continuous, $mathbbP[X=x]=0$, the correct analog being
$$
lim_h to 0^+ mathbbP[x-hle X le x+h] = f_X(x).
$$

share|cite|improve this answer

answered Mar 10 at 20:04

gt6989bgt6989b

36k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thanks. But what does $P(Y le X vert X=x)$ mean? I know that by definition $P(A vert B) = P(A cap B) /P(B)$ but then this would not make sense since as you say, $P(X=x)=0$
  $endgroup$
  – roi_saumon
  Mar 11 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @roi_saumon You interpret $X=x$ as the limit I explained in the answer
  $endgroup$
  – gt6989b
  Mar 11 at 17:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $P(Y≤X|X=x)=P(Y≤X, X=x) f_X(x)$ so that $P(Y le X)=int_0^infty P(Y le X, X=x)(f_X(x))^2dx$? Isn't that wrong?
  $endgroup$
  – roi_saumon
  Mar 11 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @roi_saumon no, what you want to say is $$mathbbP[Y le X] = int_mathbbR mathbbP[Y le X cap X=x] dx = int_mathbbR mathbbP[Y le X | X=x] f_X(x) dx$$
  $endgroup$
  – gt6989b
  Mar 11 at 19:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how you guarantee that limit is exist? is any theorem show that always that limit exist ?
  $endgroup$
  – masoud
  Mar 22 at 11:31
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

This is conditional probability. Remember that
$$
mathbbP[A|B] = fracmathbbP[Acap B]mathbbP[B] iff mathbbP[Acap B] = mathbbP[A|B]mathbbP[B]
$$
but if $X$ is continuous, $mathbbP[X=x]=0$, the correct analog being
$$
lim_h to 0^+ mathbbP[x-hle X le x+h] = f_X(x).
$$

share|cite|improve this answer

answered Mar 10 at 20:04

gt6989bgt6989b

36k22557

$endgroup$

share|cite|improve this answer

answered Mar 10 at 20:04

gt6989bgt6989b

36k22557

answered Mar 10 at 20:04

gt6989bgt6989b

36k22557

answered Mar 10 at 20:04

gt6989bgt6989b

36k22557

0

$begingroup$

conditional probability is special case of conditional expectation(definition of conditional probability is based on conditional expectation ).

define:

$A=omega in Omega$

$p(Yleq X)= p(A)=E(I_A)=E(E(I_A|X))$

$=E(g(X))$ (by definition of conditional expectation ,$E(I_A|X)$ is a function of $X$)

$=int g(x) f_X(x) dx=int E(I_A|X=x) f_X(x) dx=int p(A|X=x) f_X(x) dx=int p(Y leq X|X=x) f_X(x) dx=int p(Y leq x|X=x) f_X(x) dx$

so you pass the problem with, $p(X=x)=0$ in $p(Yleq X |X=x)$. since the defination of conditional expectation is based on projection property and not based on $p(A|B)=fracp(A cap B)p(B)$

share|cite|improve this answer

edited Apr 1 at 23:10

answered Mar 15 at 10:15

masoudmasoud

34718

$endgroup$

add a comment | 

0

$begingroup$

conditional probability is special case of conditional expectation(definition of conditional probability is based on conditional expectation ).

define:

$A=omega in Omega$

$p(Yleq X)= p(A)=E(I_A)=E(E(I_A|X))$

$=E(g(X))$ (by definition of conditional expectation ,$E(I_A|X)$ is a function of $X$)

$=int g(x) f_X(x) dx=int E(I_A|X=x) f_X(x) dx=int p(A|X=x) f_X(x) dx=int p(Y leq X|X=x) f_X(x) dx=int p(Y leq x|X=x) f_X(x) dx$

so you pass the problem with, $p(X=x)=0$ in $p(Yleq X |X=x)$. since the defination of conditional expectation is based on projection property and not based on $p(A|B)=fracp(A cap B)p(B)$

share|cite|improve this answer

edited Apr 1 at 23:10

answered Mar 15 at 10:15

masoudmasoud

34718

$endgroup$

add a comment | 

$begingroup$

conditional probability is special case of conditional expectation(definition of conditional probability is based on conditional expectation ).

define:

$A=omega in Omega$

$p(Yleq X)= p(A)=E(I_A)=E(E(I_A|X))$

$=E(g(X))$ (by definition of conditional expectation ,$E(I_A|X)$ is a function of $X$)

$=int g(x) f_X(x) dx=int E(I_A|X=x) f_X(x) dx=int p(A|X=x) f_X(x) dx=int p(Y leq X|X=x) f_X(x) dx=int p(Y leq x|X=x) f_X(x) dx$

so you pass the problem with, $p(X=x)=0$ in $p(Yleq X |X=x)$. since the defination of conditional expectation is based on projection property and not based on $p(A|B)=fracp(A cap B)p(B)$

share|cite|improve this answer

edited Apr 1 at 23:10

answered Mar 15 at 10:15

masoudmasoud

34718

$endgroup$

share|cite|improve this answer

edited Apr 1 at 23:10

answered Mar 15 at 10:15

masoudmasoud

34718

answered Mar 15 at 10:15

masoudmasoud

34718

answered Mar 15 at 10:15

masoudmasoud

34718