Conditioning random variables
up vote
0
down vote
favorite
I have two issues that seem to be related.
1) Suppose we have two random variables $A$ and $B$ that are conditioned by $X$. If we want to calculate $$ frac{p(A, B | X)}{p(B | X)},$$ do we always need to use Bayes rule and get $p(A|X, B)$? Since both are conditioned on $X$, why can't we say $p(A|X)$?
2) Suppose $Y sim rm{Normal}(0,1)$. I think $mathbb{E}[Y]$ is 0, but does it change the value if we calculate $mathbb{E}[Y|Y]$, is it $mathbb{E}[Y|Y] = Y$? I cannot understand why conditioning changes its expectation.
probability conditional-probability
asked Nov 27 at 3:53
user51966
14710
add a comment |
up vote
0
down vote
favorite
I have two issues that seem to be related.
1) Suppose we have two random variables $A$ and $B$ that are conditioned by $X$. If we want to calculate $$ frac{p(A, B | X)}{p(B | X)},$$ do we always need to use Bayes rule and get $p(A|X, B)$? Since both are conditioned on $X$, why can't we say $p(A|X)$?
2) Suppose $Y sim rm{Normal}(0,1)$. I think $mathbb{E}[Y]$ is 0, but does it change the value if we calculate $mathbb{E}[Y|Y]$, is it $mathbb{E}[Y|Y] = Y$? I cannot understand why conditioning changes its expectation.
probability conditional-probability
asked Nov 27 at 3:53
user51966
14710
1
For #2, you may know that E[Y] has mean zero based on the distribution, but conditioning will change the expectation. If you know that $Y=c$ for some c, then of course, $E[Y|Y=c]=E[c]=c$. In a sense, it is trivial. Knowing that $Y=c$, you wouldn't expect Y to be zero any more.
– bob
Nov 27 at 4:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have two issues that seem to be related.
1) Suppose we have two random variables $A$ and $B$ that are conditioned by $X$. If we want to calculate $$ frac{p(A, B | X)}{p(B | X)},$$ do we always need to use Bayes rule and get $p(A|X, B)$? Since both are conditioned on $X$, why can't we say $p(A|X)$?
2) Suppose $Y sim rm{Normal}(0,1)$. I think $mathbb{E}[Y]$ is 0, but does it change the value if we calculate $mathbb{E}[Y|Y]$, is it $mathbb{E}[Y|Y] = Y$? I cannot understand why conditioning changes its expectation.
probability conditional-probability
asked Nov 27 at 3:53
user51966
14710
I have two issues that seem to be related.
1) Suppose we have two random variables $A$ and $B$ that are conditioned by $X$. If we want to calculate $$ frac{p(A, B | X)}{p(B | X)},$$ do we always need to use Bayes rule and get $p(A|X, B)$? Since both are conditioned on $X$, why can't we say $p(A|X)$?
2) Suppose $Y sim rm{Normal}(0,1)$. I think $mathbb{E}[Y]$ is 0, but does it change the value if we calculate $mathbb{E}[Y|Y]$, is it $mathbb{E}[Y|Y] = Y$? I cannot understand why conditioning changes its expectation.
probability conditional-probability
probability conditional-probability
asked Nov 27 at 3:53
user51966
14710
asked Nov 27 at 3:53
user51966
14710
asked Nov 27 at 3:53
user51966
14710
asked Nov 27 at 3:53
user51966
14710
asked Nov 27 at 3:53
user51966
14710
14710
1
For #2, you may know that E[Y] has mean zero based on the distribution, but conditioning will change the expectation. If you know that $Y=c$ for some c, then of course, $E[Y|Y=c]=E[c]=c$. In a sense, it is trivial. Knowing that $Y=c$, you wouldn't expect Y to be zero any more.
– bob
Nov 27 at 4:55
add a comment |
1
For #2, you may know that E[Y] has mean zero based on the distribution, but conditioning will change the expectation. If you know that $Y=c$ for some c, then of course, $E[Y|Y=c]=E[c]=c$. In a sense, it is trivial. Knowing that $Y=c$, you wouldn't expect Y to be zero any more.
– bob
Nov 27 at 4:55
1
1
For #2, you may know that E[Y] has mean zero based on the distribution, but conditioning will change the expectation. If you know that $Y=c$ for some c, then of course, $E[Y|Y=c]=E[c]=c$. In a sense, it is trivial. Knowing that $Y=c$, you wouldn't expect Y to be zero any more.
– bob
Nov 27 at 4:55
For #2, you may know that E[Y] has mean zero based on the distribution, but conditioning will change the expectation. If you know that $Y=c$ for some c, then of course, $E[Y|Y=c]=E[c]=c$. In a sense, it is trivial. Knowing that $Y=c$, you wouldn't expect Y to be zero any more.
– bob
Nov 27 at 4:55
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
- Starting with
$$ pleft(Amiddle|Bright) = dfrac{pleft(A,Bright)}{pleft(Bright)} $$
and conditioning everything on $X$ gives
$$ pleft(Amiddle|B, Xright) = dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)} $$
So $pleft(Amiddle|Xright) neq dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)}$ in the same way that $ pleft(Aright) neq dfrac{pleft(A,Bright)}{pleft(Bright)} $. - For the particular case $mathbb{E}left[Ymiddle|Yright] = Y$, I like to reason about this as follows. If I gave you the value of $Y$ and asked what you expected the value of $Y$ to be, what would you answer? Well of course the expectation should be whatever the value I just gave you, which is $Y$.
answered Nov 27 at 10:36
rzch
1363
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
- Starting with
$$ pleft(Amiddle|Bright) = dfrac{pleft(A,Bright)}{pleft(Bright)} $$
and conditioning everything on $X$ gives
$$ pleft(Amiddle|B, Xright) = dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)} $$
So $pleft(Amiddle|Xright) neq dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)}$ in the same way that $ pleft(Aright) neq dfrac{pleft(A,Bright)}{pleft(Bright)} $. - For the particular case $mathbb{E}left[Ymiddle|Yright] = Y$, I like to reason about this as follows. If I gave you the value of $Y$ and asked what you expected the value of $Y$ to be, what would you answer? Well of course the expectation should be whatever the value I just gave you, which is $Y$.
answered Nov 27 at 10:36
rzch
1363
add a comment |
up vote
1
down vote
accepted
- Starting with
$$ pleft(Amiddle|Bright) = dfrac{pleft(A,Bright)}{pleft(Bright)} $$
and conditioning everything on $X$ gives
$$ pleft(Amiddle|B, Xright) = dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)} $$
So $pleft(Amiddle|Xright) neq dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)}$ in the same way that $ pleft(Aright) neq dfrac{pleft(A,Bright)}{pleft(Bright)} $. - For the particular case $mathbb{E}left[Ymiddle|Yright] = Y$, I like to reason about this as follows. If I gave you the value of $Y$ and asked what you expected the value of $Y$ to be, what would you answer? Well of course the expectation should be whatever the value I just gave you, which is $Y$.
answered Nov 27 at 10:36
rzch
1363
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
- Starting with
$$ pleft(Amiddle|Bright) = dfrac{pleft(A,Bright)}{pleft(Bright)} $$
and conditioning everything on $X$ gives
$$ pleft(Amiddle|B, Xright) = dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)} $$
So $pleft(Amiddle|Xright) neq dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)}$ in the same way that $ pleft(Aright) neq dfrac{pleft(A,Bright)}{pleft(Bright)} $. - For the particular case $mathbb{E}left[Ymiddle|Yright] = Y$, I like to reason about this as follows. If I gave you the value of $Y$ and asked what you expected the value of $Y$ to be, what would you answer? Well of course the expectation should be whatever the value I just gave you, which is $Y$.
answered Nov 27 at 10:36
rzch
1363
- Starting with
$$ pleft(Amiddle|Bright) = dfrac{pleft(A,Bright)}{pleft(Bright)} $$
and conditioning everything on $X$ gives
$$ pleft(Amiddle|B, Xright) = dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)} $$
So $pleft(Amiddle|Xright) neq dfrac{pleft(A,Bmiddle| Xright)}{pleft(Bmiddle| Xright)}$ in the same way that $ pleft(Aright) neq dfrac{pleft(A,Bright)}{pleft(Bright)} $. - For the particular case $mathbb{E}left[Ymiddle|Yright] = Y$, I like to reason about this as follows. If I gave you the value of $Y$ and asked what you expected the value of $Y$ to be, what would you answer? Well of course the expectation should be whatever the value I just gave you, which is $Y$.
answered Nov 27 at 10:36
rzch
1363
answered Nov 27 at 10:36
rzch
1363
answered Nov 27 at 10:36
rzch
1363
answered Nov 27 at 10:36
rzch
1363
1363
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015308%2fconditioning-random-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
For #2, you may know that E[Y] has mean zero based on the distribution, but conditioning will change the expectation. If you know that $Y=c$ for some c, then of course, $E[Y|Y=c]=E[c]=c$. In a sense, it is trivial. Knowing that $Y=c$, you wouldn't expect Y to be zero any more.
– bob
Nov 27 at 4:55