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Three quick beginner questions about Conditional Expectation
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Newbie learning about conditional expectation in Sheldon Ross's book and I had some basic questions to help my understanding.
Question 1
For a discrete or continuous RV $X$, is it always true that $E[X|X] = X$?
My attempt: $E[X|X] = sum_x x P(X=x | X=x_0)= x_0 $ and as a result, the value that $E[X|X]$ takes will be one of the possible values of $X$ with probability based on the pmf of $X$, therefore $E[X|X] = X$
Question 2
Let's say $X$ represents a dice roll. Then how would you describe in your own words what the $E[X|X]$ is saying? Besides saying that it is $X$.
Question 3
What is the diference between $E[X|Y]$ and $E[X|Y=y]$ ? Is it just a notation or convention thing or is there a fundamental difference? I want to say the former always represents a RV and the latter could be either constant or a RV (which is confusing).
Thank you for your help and patience.
probability
asked Jul 25 at 23:26
HJ_beginner
42315
add a comment |Â
up vote
0
down vote
favorite
Newbie learning about conditional expectation in Sheldon Ross's book and I had some basic questions to help my understanding.
Question 1
For a discrete or continuous RV $X$, is it always true that $E[X|X] = X$?
My attempt: $E[X|X] = sum_x x P(X=x | X=x_0)= x_0 $ and as a result, the value that $E[X|X]$ takes will be one of the possible values of $X$ with probability based on the pmf of $X$, therefore $E[X|X] = X$
Question 2
Let's say $X$ represents a dice roll. Then how would you describe in your own words what the $E[X|X]$ is saying? Besides saying that it is $X$.
Question 3
What is the diference between $E[X|Y]$ and $E[X|Y=y]$ ? Is it just a notation or convention thing or is there a fundamental difference? I want to say the former always represents a RV and the latter could be either constant or a RV (which is confusing).
Thank you for your help and patience.
probability
asked Jul 25 at 23:26
HJ_beginner
42315
I think you can refer to the formal definition of conditional expectation. You can find it on Wikipedia
– Edward Wang
Jul 25 at 23:28
I think $E[X|X] = X$ is maybe too trivial for the wiki... I admit these are very beginner questions. Sometimes it just helps me to write things in my own words and get confirmation from experts on this site.
– HJ_beginner
Jul 25 at 23:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Newbie learning about conditional expectation in Sheldon Ross's book and I had some basic questions to help my understanding.
Question 1
For a discrete or continuous RV $X$, is it always true that $E[X|X] = X$?
My attempt: $E[X|X] = sum_x x P(X=x | X=x_0)= x_0 $ and as a result, the value that $E[X|X]$ takes will be one of the possible values of $X$ with probability based on the pmf of $X$, therefore $E[X|X] = X$
Question 2
Let's say $X$ represents a dice roll. Then how would you describe in your own words what the $E[X|X]$ is saying? Besides saying that it is $X$.
Question 3
What is the diference between $E[X|Y]$ and $E[X|Y=y]$ ? Is it just a notation or convention thing or is there a fundamental difference? I want to say the former always represents a RV and the latter could be either constant or a RV (which is confusing).
Thank you for your help and patience.
probability
asked Jul 25 at 23:26
HJ_beginner
42315
Newbie learning about conditional expectation in Sheldon Ross's book and I had some basic questions to help my understanding.
Question 1
For a discrete or continuous RV $X$, is it always true that $E[X|X] = X$?
My attempt: $E[X|X] = sum_x x P(X=x | X=x_0)= x_0 $ and as a result, the value that $E[X|X]$ takes will be one of the possible values of $X$ with probability based on the pmf of $X$, therefore $E[X|X] = X$
Question 2
Let's say $X$ represents a dice roll. Then how would you describe in your own words what the $E[X|X]$ is saying? Besides saying that it is $X$.
Question 3
What is the diference between $E[X|Y]$ and $E[X|Y=y]$ ? Is it just a notation or convention thing or is there a fundamental difference? I want to say the former always represents a RV and the latter could be either constant or a RV (which is confusing).
Thank you for your help and patience.
probability
asked Jul 25 at 23:26
HJ_beginner
42315
asked Jul 25 at 23:26
HJ_beginner
42315
asked Jul 25 at 23:26
HJ_beginner
42315
asked Jul 25 at 23:26
HJ_beginner
42315
42315
I think you can refer to the formal definition of conditional expectation. You can find it on Wikipedia
– Edward Wang
Jul 25 at 23:28
I think $E[X|X] = X$ is maybe too trivial for the wiki... I admit these are very beginner questions. Sometimes it just helps me to write things in my own words and get confirmation from experts on this site.
– HJ_beginner
Jul 25 at 23:32
add a comment |Â
I think you can refer to the formal definition of conditional expectation. You can find it on Wikipedia
– Edward Wang
Jul 25 at 23:28
I think $E[X|X] = X$ is maybe too trivial for the wiki... I admit these are very beginner questions. Sometimes it just helps me to write things in my own words and get confirmation from experts on this site.
– HJ_beginner
Jul 25 at 23:32
I think you can refer to the formal definition of conditional expectation. You can find it on Wikipedia
– Edward Wang
Jul 25 at 23:28
I think you can refer to the formal definition of conditional expectation. You can find it on Wikipedia
– Edward Wang
Jul 25 at 23:28
I think $E[X|X] = X$ is maybe too trivial for the wiki... I admit these are very beginner questions. Sometimes it just helps me to write things in my own words and get confirmation from experts on this site.
– HJ_beginner
Jul 25 at 23:32
I think $E[X|X] = X$ is maybe too trivial for the wiki... I admit these are very beginner questions. Sometimes it just helps me to write things in my own words and get confirmation from experts on this site.
– HJ_beginner
Jul 25 at 23:32
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let's try Question 3 first:
$E[Xmid Y=y]$ is just a value, which may change with $y$ if $X$ and $Y$ are not independent, and it may only make sense if $y$ is in the support of $Y$. So you might say $E[Xmid Y=y]$ is a function of $y$, say $h(y)$
If we apply this function to $Y$, in the same way we would might with any other function of a random variable, we get another random variable $h(Y)$ which in a sense is $E[Xmid Y]$
In summary, $E[Xmid Y=y]$ is a value which may depend on $y$, while $E[Xmid Y]$ is a random variable. They are not the same
This makes Question 1 easier:
Clearly $E[X mid X=x]=x$, providing that $x$ is in the support of $X$.
So the function is the identity function and $E[X mid X]=X$
It does not help much with Question 2, mainly because $E[Xmid Y]$ is not a particularly enlightening concept when $X$ is completely determined by $Y$, and even less so when $Y=X$
answered Jul 25 at 23:50
Henry
92.9k469147
Hmm interesting, thanks for your help. I know I'm wrong, but I have this feeling that the book I'm reading is being sloppy about using $E[X|Y]$ vs $E[X|Y=y]$ but I will go back and think through it more carefully.
– HJ_beginner
Jul 26 at 0:00
For question 2, does this maybe work? If $X$ is the outcome of a dice roll, then the expected value of a dice roll given that you know the outcome of the dice roll can take on one of 6 different values with equal probability, which is equivalent to the oucome of a dice roll.
– HJ_beginner
Jul 26 at 0:18
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let's try Question 3 first:
$E[Xmid Y=y]$ is just a value, which may change with $y$ if $X$ and $Y$ are not independent, and it may only make sense if $y$ is in the support of $Y$. So you might say $E[Xmid Y=y]$ is a function of $y$, say $h(y)$
If we apply this function to $Y$, in the same way we would might with any other function of a random variable, we get another random variable $h(Y)$ which in a sense is $E[Xmid Y]$
In summary, $E[Xmid Y=y]$ is a value which may depend on $y$, while $E[Xmid Y]$ is a random variable. They are not the same
This makes Question 1 easier:
Clearly $E[X mid X=x]=x$, providing that $x$ is in the support of $X$.
So the function is the identity function and $E[X mid X]=X$
It does not help much with Question 2, mainly because $E[Xmid Y]$ is not a particularly enlightening concept when $X$ is completely determined by $Y$, and even less so when $Y=X$
answered Jul 25 at 23:50
Henry
92.9k469147
Hmm interesting, thanks for your help. I know I'm wrong, but I have this feeling that the book I'm reading is being sloppy about using $E[X|Y]$ vs $E[X|Y=y]$ but I will go back and think through it more carefully.
– HJ_beginner
Jul 26 at 0:00
For question 2, does this maybe work? If $X$ is the outcome of a dice roll, then the expected value of a dice roll given that you know the outcome of the dice roll can take on one of 6 different values with equal probability, which is equivalent to the oucome of a dice roll.
– HJ_beginner
Jul 26 at 0:18
add a comment |Â
up vote
2
down vote
accepted
Let's try Question 3 first:
$E[Xmid Y=y]$ is just a value, which may change with $y$ if $X$ and $Y$ are not independent, and it may only make sense if $y$ is in the support of $Y$. So you might say $E[Xmid Y=y]$ is a function of $y$, say $h(y)$
If we apply this function to $Y$, in the same way we would might with any other function of a random variable, we get another random variable $h(Y)$ which in a sense is $E[Xmid Y]$
In summary, $E[Xmid Y=y]$ is a value which may depend on $y$, while $E[Xmid Y]$ is a random variable. They are not the same
This makes Question 1 easier:
Clearly $E[X mid X=x]=x$, providing that $x$ is in the support of $X$.
So the function is the identity function and $E[X mid X]=X$
It does not help much with Question 2, mainly because $E[Xmid Y]$ is not a particularly enlightening concept when $X$ is completely determined by $Y$, and even less so when $Y=X$
answered Jul 25 at 23:50
Henry
92.9k469147
Hmm interesting, thanks for your help. I know I'm wrong, but I have this feeling that the book I'm reading is being sloppy about using $E[X|Y]$ vs $E[X|Y=y]$ but I will go back and think through it more carefully.
– HJ_beginner
Jul 26 at 0:00
For question 2, does this maybe work? If $X$ is the outcome of a dice roll, then the expected value of a dice roll given that you know the outcome of the dice roll can take on one of 6 different values with equal probability, which is equivalent to the oucome of a dice roll.
– HJ_beginner
Jul 26 at 0:18
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let's try Question 3 first:
$E[Xmid Y=y]$ is just a value, which may change with $y$ if $X$ and $Y$ are not independent, and it may only make sense if $y$ is in the support of $Y$. So you might say $E[Xmid Y=y]$ is a function of $y$, say $h(y)$
If we apply this function to $Y$, in the same way we would might with any other function of a random variable, we get another random variable $h(Y)$ which in a sense is $E[Xmid Y]$
In summary, $E[Xmid Y=y]$ is a value which may depend on $y$, while $E[Xmid Y]$ is a random variable. They are not the same
This makes Question 1 easier:
Clearly $E[X mid X=x]=x$, providing that $x$ is in the support of $X$.
So the function is the identity function and $E[X mid X]=X$
It does not help much with Question 2, mainly because $E[Xmid Y]$ is not a particularly enlightening concept when $X$ is completely determined by $Y$, and even less so when $Y=X$
answered Jul 25 at 23:50
Henry
92.9k469147
Let's try Question 3 first:
$E[Xmid Y=y]$ is just a value, which may change with $y$ if $X$ and $Y$ are not independent, and it may only make sense if $y$ is in the support of $Y$. So you might say $E[Xmid Y=y]$ is a function of $y$, say $h(y)$
If we apply this function to $Y$, in the same way we would might with any other function of a random variable, we get another random variable $h(Y)$ which in a sense is $E[Xmid Y]$
In summary, $E[Xmid Y=y]$ is a value which may depend on $y$, while $E[Xmid Y]$ is a random variable. They are not the same
This makes Question 1 easier:
Clearly $E[X mid X=x]=x$, providing that $x$ is in the support of $X$.
So the function is the identity function and $E[X mid X]=X$
It does not help much with Question 2, mainly because $E[Xmid Y]$ is not a particularly enlightening concept when $X$ is completely determined by $Y$, and even less so when $Y=X$
answered Jul 25 at 23:50
Henry
92.9k469147
answered Jul 25 at 23:50
Henry
92.9k469147
answered Jul 25 at 23:50
Henry
92.9k469147
answered Jul 25 at 23:50
Henry
92.9k469147
92.9k469147
Hmm interesting, thanks for your help. I know I'm wrong, but I have this feeling that the book I'm reading is being sloppy about using $E[X|Y]$ vs $E[X|Y=y]$ but I will go back and think through it more carefully.
– HJ_beginner
Jul 26 at 0:00
For question 2, does this maybe work? If $X$ is the outcome of a dice roll, then the expected value of a dice roll given that you know the outcome of the dice roll can take on one of 6 different values with equal probability, which is equivalent to the oucome of a dice roll.
– HJ_beginner
Jul 26 at 0:18
add a comment |Â
Hmm interesting, thanks for your help. I know I'm wrong, but I have this feeling that the book I'm reading is being sloppy about using $E[X|Y]$ vs $E[X|Y=y]$ but I will go back and think through it more carefully.
– HJ_beginner
Jul 26 at 0:00
For question 2, does this maybe work? If $X$ is the outcome of a dice roll, then the expected value of a dice roll given that you know the outcome of the dice roll can take on one of 6 different values with equal probability, which is equivalent to the oucome of a dice roll.
– HJ_beginner
Jul 26 at 0:18
Hmm interesting, thanks for your help. I know I'm wrong, but I have this feeling that the book I'm reading is being sloppy about using $E[X|Y]$ vs $E[X|Y=y]$ but I will go back and think through it more carefully.
– HJ_beginner
Jul 26 at 0:00
Hmm interesting, thanks for your help. I know I'm wrong, but I have this feeling that the book I'm reading is being sloppy about using $E[X|Y]$ vs $E[X|Y=y]$ but I will go back and think through it more carefully.
– HJ_beginner
Jul 26 at 0:00
For question 2, does this maybe work? If $X$ is the outcome of a dice roll, then the expected value of a dice roll given that you know the outcome of the dice roll can take on one of 6 different values with equal probability, which is equivalent to the oucome of a dice roll.
– HJ_beginner
Jul 26 at 0:18
For question 2, does this maybe work? If $X$ is the outcome of a dice roll, then the expected value of a dice roll given that you know the outcome of the dice roll can take on one of 6 different values with equal probability, which is equivalent to the oucome of a dice roll.
– HJ_beginner
Jul 26 at 0:18
add a comment |Â
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I think you can refer to the formal definition of conditional expectation. You can find it on Wikipedia
– Edward Wang
Jul 25 at 23:28
I think $E[X|X] = X$ is maybe too trivial for the wiki... I admit these are very beginner questions. Sometimes it just helps me to write things in my own words and get confirmation from experts on this site.
– HJ_beginner
Jul 25 at 23:32