Mixing
Conditional probability where conditioned event is an inequality
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Let $X$ and $Y$ be two random variables. My objective is to calculate the seemingly simple probability:
$$
mathbbP(aleq Xleq Y),
$$
where $a>0$ is a constant. Thus, from the definition of distribution functions this quantity is non-negative as long as $aleq Y$. Expanding the above expression we have that
beginalign
mathbbP(aleq Xleq Y)
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) + underbracemathbbP(aleq Xleq Ymid anleq Y)mathbbP(anleq Y)_=0 \
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) \
&= (mathbbP(Xleq Y)-mathbbP(Xleq a)) mathbbP(aleq Y) \
&= (mathbbE_Y[mathbbP(Xleq ymid Y=y)]-mathbbP(Xleq a)) mathbbP(aleq Y).
endalign
Is this correct? My doubt comes from the fact that we condition on the event $aleq Y$ rather than an event on the form $Z=z$.
probability
asked Aug 1 at 12:41
index
8810
add a comment |Â
up vote
2
down vote
favorite
Let $X$ and $Y$ be two random variables. My objective is to calculate the seemingly simple probability:
$$
mathbbP(aleq Xleq Y),
$$
where $a>0$ is a constant. Thus, from the definition of distribution functions this quantity is non-negative as long as $aleq Y$. Expanding the above expression we have that
beginalign
mathbbP(aleq Xleq Y)
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) + underbracemathbbP(aleq Xleq Ymid anleq Y)mathbbP(anleq Y)_=0 \
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) \
&= (mathbbP(Xleq Y)-mathbbP(Xleq a)) mathbbP(aleq Y) \
&= (mathbbE_Y[mathbbP(Xleq ymid Y=y)]-mathbbP(Xleq a)) mathbbP(aleq Y).
endalign
Is this correct? My doubt comes from the fact that we condition on the event $aleq Y$ rather than an event on the form $Z=z$.
probability
asked Aug 1 at 12:41
index
8810
Do you have any assumptions concerning the random variables like independence or so?
– user190080
Aug 1 at 13:12
There is an error in how you go from the 2nd line to the 3rd line. How did you turn a conditional probability into a difference of 2 unconditional probabilities?
– yurnero
Aug 1 at 13:23
@user190080 $X$ and $Y$ are independent.
– index
Aug 1 at 13:25
@yurnero For two constants $a,b$ where $a<b$ and a random variable $Z$, I was applying the definition of a distribution function $mathbbP(aleq Z<b)=F_Z(b)-F_Z(a)=mathbbP(Z<b)-mathbbP(Zleq a)$.
– index
Aug 1 at 13:28
2
Sure, but there is a conditioning event, so you have $Pr(aleq Xleq Y|E)= Pr[(Xleq Y)cap(aleq X)|E]$ where $E$ is the event $Ygeq a$. It's not clear how you get rid of $E$.
– yurnero
Aug 1 at 14:07
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X$ and $Y$ be two random variables. My objective is to calculate the seemingly simple probability:
$$
mathbbP(aleq Xleq Y),
$$
where $a>0$ is a constant. Thus, from the definition of distribution functions this quantity is non-negative as long as $aleq Y$. Expanding the above expression we have that
beginalign
mathbbP(aleq Xleq Y)
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) + underbracemathbbP(aleq Xleq Ymid anleq Y)mathbbP(anleq Y)_=0 \
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) \
&= (mathbbP(Xleq Y)-mathbbP(Xleq a)) mathbbP(aleq Y) \
&= (mathbbE_Y[mathbbP(Xleq ymid Y=y)]-mathbbP(Xleq a)) mathbbP(aleq Y).
endalign
Is this correct? My doubt comes from the fact that we condition on the event $aleq Y$ rather than an event on the form $Z=z$.
probability
asked Aug 1 at 12:41
index
8810
Let $X$ and $Y$ be two random variables. My objective is to calculate the seemingly simple probability:
$$
mathbbP(aleq Xleq Y),
$$
where $a>0$ is a constant. Thus, from the definition of distribution functions this quantity is non-negative as long as $aleq Y$. Expanding the above expression we have that
beginalign
mathbbP(aleq Xleq Y)
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) + underbracemathbbP(aleq Xleq Ymid anleq Y)mathbbP(anleq Y)_=0 \
&= mathbbP(aleq Xleq Ymid aleq Y)mathbbP(aleq Y) \
&= (mathbbP(Xleq Y)-mathbbP(Xleq a)) mathbbP(aleq Y) \
&= (mathbbE_Y[mathbbP(Xleq ymid Y=y)]-mathbbP(Xleq a)) mathbbP(aleq Y).
endalign
Is this correct? My doubt comes from the fact that we condition on the event $aleq Y$ rather than an event on the form $Z=z$.
probability
asked Aug 1 at 12:41
index
8810
edited Aug 1 at 13:52
asked Aug 1 at 12:41
index
8810
asked Aug 1 at 12:41
index
8810
asked Aug 1 at 12:41
index
8810
8810
Do you have any assumptions concerning the random variables like independence or so?
– user190080
Aug 1 at 13:12
There is an error in how you go from the 2nd line to the 3rd line. How did you turn a conditional probability into a difference of 2 unconditional probabilities?
– yurnero
Aug 1 at 13:23
@user190080 $X$ and $Y$ are independent.
– index
Aug 1 at 13:25
@yurnero For two constants $a,b$ where $a<b$ and a random variable $Z$, I was applying the definition of a distribution function $mathbbP(aleq Z<b)=F_Z(b)-F_Z(a)=mathbbP(Z<b)-mathbbP(Zleq a)$.
– index
Aug 1 at 13:28
2
Sure, but there is a conditioning event, so you have $Pr(aleq Xleq Y|E)= Pr[(Xleq Y)cap(aleq X)|E]$ where $E$ is the event $Ygeq a$. It's not clear how you get rid of $E$.
– yurnero
Aug 1 at 14:07
add a comment |Â
Do you have any assumptions concerning the random variables like independence or so?
– user190080
Aug 1 at 13:12
There is an error in how you go from the 2nd line to the 3rd line. How did you turn a conditional probability into a difference of 2 unconditional probabilities?
– yurnero
Aug 1 at 13:23
@user190080 $X$ and $Y$ are independent.
– index
Aug 1 at 13:25
@yurnero For two constants $a,b$ where $a<b$ and a random variable $Z$, I was applying the definition of a distribution function $mathbbP(aleq Z<b)=F_Z(b)-F_Z(a)=mathbbP(Z<b)-mathbbP(Zleq a)$.
– index
Aug 1 at 13:28
2
Sure, but there is a conditioning event, so you have $Pr(aleq Xleq Y|E)= Pr[(Xleq Y)cap(aleq X)|E]$ where $E$ is the event $Ygeq a$. It's not clear how you get rid of $E$.
– yurnero
Aug 1 at 14:07
Do you have any assumptions concerning the random variables like independence or so?
– user190080
Aug 1 at 13:12
Do you have any assumptions concerning the random variables like independence or so?
– user190080
Aug 1 at 13:12
There is an error in how you go from the 2nd line to the 3rd line. How did you turn a conditional probability into a difference of 2 unconditional probabilities?
– yurnero
Aug 1 at 13:23
There is an error in how you go from the 2nd line to the 3rd line. How did you turn a conditional probability into a difference of 2 unconditional probabilities?
– yurnero
Aug 1 at 13:23
@user190080 $X$ and $Y$ are independent.
– index
Aug 1 at 13:25
@user190080 $X$ and $Y$ are independent.
– index
Aug 1 at 13:25
@yurnero For two constants $a,b$ where $a<b$ and a random variable $Z$, I was applying the definition of a distribution function $mathbbP(aleq Z<b)=F_Z(b)-F_Z(a)=mathbbP(Z<b)-mathbbP(Zleq a)$.
– index
Aug 1 at 13:28
@yurnero For two constants $a,b$ where $a<b$ and a random variable $Z$, I was applying the definition of a distribution function $mathbbP(aleq Z<b)=F_Z(b)-F_Z(a)=mathbbP(Z<b)-mathbbP(Zleq a)$.
– index
Aug 1 at 13:28
2
2
Sure, but there is a conditioning event, so you have $Pr(aleq Xleq Y|E)= Pr[(Xleq Y)cap(aleq X)|E]$ where $E$ is the event $Ygeq a$. It's not clear how you get rid of $E$.
– yurnero
Aug 1 at 14:07
Sure, but there is a conditioning event, so you have $Pr(aleq Xleq Y|E)= Pr[(Xleq Y)cap(aleq X)|E]$ where $E$ is the event $Ygeq a$. It's not clear how you get rid of $E$.
– yurnero
Aug 1 at 14:07
add a comment |Â
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Do you have any assumptions concerning the random variables like independence or so?
– user190080
Aug 1 at 13:12
There is an error in how you go from the 2nd line to the 3rd line. How did you turn a conditional probability into a difference of 2 unconditional probabilities?
– yurnero
Aug 1 at 13:23
@user190080 $X$ and $Y$ are independent.
– index
Aug 1 at 13:25
@yurnero For two constants $a,b$ where $a<b$ and a random variable $Z$, I was applying the definition of a distribution function $mathbbP(aleq Z<b)=F_Z(b)-F_Z(a)=mathbbP(Z<b)-mathbbP(Zleq a)$.
– index
Aug 1 at 13:28
2
Sure, but there is a conditioning event, so you have $Pr(aleq Xleq Y|E)= Pr[(Xleq Y)cap(aleq X)|E]$ where $E$ is the event $Ygeq a$. It's not clear how you get rid of $E$.
– yurnero
Aug 1 at 14:07