expected value of deterministic function
$begingroup$
I am intuitively convinced that $E[f(X)] = f(X)$, if $f$ is deterministic, but I cannot derive it using the integral definition of expectation.
Assuming $X sim unif(0,1)$ I get:
$$
E[f(X)] = int^{1}_{0}{f(x)p(x)dx} = int^{1}_{0}{f(x)dx} = overline{f}
$$
which is not necessarily $f$.
Appreciate any help!
Edit: Thank you all, I made a mess with the notations, and somehow my question is not a question anymore
expected-value
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
$endgroup$
|
show 3 more comments
$begingroup$
I am intuitively convinced that $E[f(X)] = f(X)$, if $f$ is deterministic, but I cannot derive it using the integral definition of expectation.
Assuming $X sim unif(0,1)$ I get:
$$
E[f(X)] = int^{1}_{0}{f(x)p(x)dx} = int^{1}_{0}{f(x)dx} = overline{f}
$$
which is not necessarily $f$.
Appreciate any help!
Edit: Thank you all, I made a mess with the notations, and somehow my question is not a question anymore
expected-value
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
$endgroup$
$begingroup$
No, it will happen when $X$ is also deterministic or when $f$ is constant on the range of $X$ but not otherwise. (In these cases the whole quantity $f(X)$ is deterministic.)
$endgroup$
– Ian
Dec 8 '18 at 5:35
$begingroup$
Do you mean that $X$ is deterministic? Otherwise I’m not sure what you mean by a deterministic function.
$endgroup$
– platty
Dec 8 '18 at 5:35
$begingroup$
@platty maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
@lan maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
If you mean that $f$ is a constant function, then this holds (plug in $f(X) = c$ for all $x$ and integrate the constant). Otherwise, it’s not clear what you mean
$endgroup$
– platty
Dec 8 '18 at 5:47
|
show 3 more comments
$begingroup$
I am intuitively convinced that $E[f(X)] = f(X)$, if $f$ is deterministic, but I cannot derive it using the integral definition of expectation.
Assuming $X sim unif(0,1)$ I get:
$$
E[f(X)] = int^{1}_{0}{f(x)p(x)dx} = int^{1}_{0}{f(x)dx} = overline{f}
$$
which is not necessarily $f$.
Appreciate any help!
Edit: Thank you all, I made a mess with the notations, and somehow my question is not a question anymore
expected-value
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
$endgroup$
I am intuitively convinced that $E[f(X)] = f(X)$, if $f$ is deterministic, but I cannot derive it using the integral definition of expectation.
Assuming $X sim unif(0,1)$ I get:
$$
E[f(X)] = int^{1}_{0}{f(x)p(x)dx} = int^{1}_{0}{f(x)dx} = overline{f}
$$
which is not necessarily $f$.
Appreciate any help!
Edit: Thank you all, I made a mess with the notations, and somehow my question is not a question anymore
expected-value
expected-value
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
edited Dec 9 '18 at 1:38
Reza Abdolhakim
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
asked Dec 8 '18 at 5:33
Reza AbdolhakimReza Abdolhakim
11
11
$begingroup$
No, it will happen when $X$ is also deterministic or when $f$ is constant on the range of $X$ but not otherwise. (In these cases the whole quantity $f(X)$ is deterministic.)
$endgroup$
– Ian
Dec 8 '18 at 5:35
$begingroup$
Do you mean that $X$ is deterministic? Otherwise I’m not sure what you mean by a deterministic function.
$endgroup$
– platty
Dec 8 '18 at 5:35
$begingroup$
@platty maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
@lan maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
If you mean that $f$ is a constant function, then this holds (plug in $f(X) = c$ for all $x$ and integrate the constant). Otherwise, it’s not clear what you mean
$endgroup$
– platty
Dec 8 '18 at 5:47
|
show 3 more comments
$begingroup$
No, it will happen when $X$ is also deterministic or when $f$ is constant on the range of $X$ but not otherwise. (In these cases the whole quantity $f(X)$ is deterministic.)
$endgroup$
– Ian
Dec 8 '18 at 5:35
$begingroup$
Do you mean that $X$ is deterministic? Otherwise I’m not sure what you mean by a deterministic function.
$endgroup$
– platty
Dec 8 '18 at 5:35
$begingroup$
@platty maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
@lan maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
If you mean that $f$ is a constant function, then this holds (plug in $f(X) = c$ for all $x$ and integrate the constant). Otherwise, it’s not clear what you mean
$endgroup$
– platty
Dec 8 '18 at 5:47
$begingroup$
No, it will happen when $X$ is also deterministic or when $f$ is constant on the range of $X$ but not otherwise. (In these cases the whole quantity $f(X)$ is deterministic.)
$endgroup$
– Ian
Dec 8 '18 at 5:35
$begingroup$
No, it will happen when $X$ is also deterministic or when $f$ is constant on the range of $X$ but not otherwise. (In these cases the whole quantity $f(X)$ is deterministic.)
$endgroup$
– Ian
Dec 8 '18 at 5:35
$begingroup$
Do you mean that $X$ is deterministic? Otherwise I’m not sure what you mean by a deterministic function.
$endgroup$
– platty
Dec 8 '18 at 5:35
$begingroup$
Do you mean that $X$ is deterministic? Otherwise I’m not sure what you mean by a deterministic function.
$endgroup$
– platty
Dec 8 '18 at 5:35
$begingroup$
@platty maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
@platty maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
@lan maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
@lan maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
If you mean that $f$ is a constant function, then this holds (plug in $f(X) = c$ for all $x$ and integrate the constant). Otherwise, it’s not clear what you mean
$endgroup$
– platty
Dec 8 '18 at 5:47
$begingroup$
If you mean that $f$ is a constant function, then this holds (plug in $f(X) = c$ for all $x$ and integrate the constant). Otherwise, it’s not clear what you mean
$endgroup$
– platty
Dec 8 '18 at 5:47
|
show 3 more comments
1 Answer
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$begingroup$
If $X$ is a random variable, so is $f(X)$. Unless $f$ is a degenerate function (i.e. $f(x) = const$), it is not meaningful to compare $mathbb{E}[f(X)]$ (which is simply a number, if it exists) to $f(X)$, which is a random variable.
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
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add a comment |
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1 Answer
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$begingroup$
If $X$ is a random variable, so is $f(X)$. Unless $f$ is a degenerate function (i.e. $f(x) = const$), it is not meaningful to compare $mathbb{E}[f(X)]$ (which is simply a number, if it exists) to $f(X)$, which is a random variable.
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
$endgroup$
add a comment |
$begingroup$
If $X$ is a random variable, so is $f(X)$. Unless $f$ is a degenerate function (i.e. $f(x) = const$), it is not meaningful to compare $mathbb{E}[f(X)]$ (which is simply a number, if it exists) to $f(X)$, which is a random variable.
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
$endgroup$
add a comment |
$begingroup$
If $X$ is a random variable, so is $f(X)$. Unless $f$ is a degenerate function (i.e. $f(x) = const$), it is not meaningful to compare $mathbb{E}[f(X)]$ (which is simply a number, if it exists) to $f(X)$, which is a random variable.
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
$endgroup$
If $X$ is a random variable, so is $f(X)$. Unless $f$ is a degenerate function (i.e. $f(x) = const$), it is not meaningful to compare $mathbb{E}[f(X)]$ (which is simply a number, if it exists) to $f(X)$, which is a random variable.
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
answered Dec 8 '18 at 22:33
Aditya DuaAditya Dua
1,14418
1,14418
add a comment |
add a comment |
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$begingroup$
No, it will happen when $X$ is also deterministic or when $f$ is constant on the range of $X$ but not otherwise. (In these cases the whole quantity $f(X)$ is deterministic.)
$endgroup$
– Ian
Dec 8 '18 at 5:35
$begingroup$
Do you mean that $X$ is deterministic? Otherwise I’m not sure what you mean by a deterministic function.
$endgroup$
– platty
Dec 8 '18 at 5:35
$begingroup$
@platty maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
@lan maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$
$endgroup$
– Reza Abdolhakim
Dec 8 '18 at 5:46
$begingroup$
If you mean that $f$ is a constant function, then this holds (plug in $f(X) = c$ for all $x$ and integrate the constant). Otherwise, it’s not clear what you mean
$endgroup$
– platty
Dec 8 '18 at 5:47