What's the difference between the expectation of a function of a random variable and the law of the...
$begingroup$
Given a random variable $X$, some function $g(X)$, and $X$'s pdf $p_x(X)$ I know from probability
that:
$$mathbb{E}(g(X)) = int_x g(X)p_x(X) dx$$
In my reading, the Law of the Unconscious Statistician (LotUS) came up as a reason for one of steps of a proof in an academic paper. When I looked into the wiki link above, it seems to say the same thing as the equation above.
My question is, is there a difference between the two? Or is the LotUS just a formalism or a nickname for the expectation of a function of a random variable?
statistics expected-value
asked Dec 27 '18 at 6:38
RVCRVC
30818
$endgroup$
add a comment |
$begingroup$
Given a random variable $X$, some function $g(X)$, and $X$'s pdf $p_x(X)$ I know from probability
that:
$$mathbb{E}(g(X)) = int_x g(X)p_x(X) dx$$
In my reading, the Law of the Unconscious Statistician (LotUS) came up as a reason for one of steps of a proof in an academic paper. When I looked into the wiki link above, it seems to say the same thing as the equation above.
My question is, is there a difference between the two? Or is the LotUS just a formalism or a nickname for the expectation of a function of a random variable?
statistics expected-value
asked Dec 27 '18 at 6:38
RVCRVC
30818
$endgroup$
$begingroup$
Actually doesnt the wikipedia page you mentioned answer this? Maybe in the etymology section?
$endgroup$
– Ben
Dec 27 '18 at 6:48
add a comment |
$begingroup$
Given a random variable $X$, some function $g(X)$, and $X$'s pdf $p_x(X)$ I know from probability
that:
$$mathbb{E}(g(X)) = int_x g(X)p_x(X) dx$$
In my reading, the Law of the Unconscious Statistician (LotUS) came up as a reason for one of steps of a proof in an academic paper. When I looked into the wiki link above, it seems to say the same thing as the equation above.
My question is, is there a difference between the two? Or is the LotUS just a formalism or a nickname for the expectation of a function of a random variable?
statistics expected-value
asked Dec 27 '18 at 6:38
RVCRVC
30818
$endgroup$
Given a random variable $X$, some function $g(X)$, and $X$'s pdf $p_x(X)$ I know from probability
that:
$$mathbb{E}(g(X)) = int_x g(X)p_x(X) dx$$
In my reading, the Law of the Unconscious Statistician (LotUS) came up as a reason for one of steps of a proof in an academic paper. When I looked into the wiki link above, it seems to say the same thing as the equation above.
My question is, is there a difference between the two? Or is the LotUS just a formalism or a nickname for the expectation of a function of a random variable?
statistics expected-value
statistics expected-value
asked Dec 27 '18 at 6:38
RVCRVC
30818
asked Dec 27 '18 at 6:38
RVCRVC
30818
asked Dec 27 '18 at 6:38
RVCRVC
30818
asked Dec 27 '18 at 6:38
RVCRVC
30818
asked Dec 27 '18 at 6:38
RVCRVC
30818
30818
$begingroup$
Actually doesnt the wikipedia page you mentioned answer this? Maybe in the etymology section?
$endgroup$
– Ben
Dec 27 '18 at 6:48
add a comment |
$begingroup$
Actually doesnt the wikipedia page you mentioned answer this? Maybe in the etymology section?
$endgroup$
– Ben
Dec 27 '18 at 6:48
$begingroup$
Actually doesnt the wikipedia page you mentioned answer this? Maybe in the etymology section?
$endgroup$
– Ben
Dec 27 '18 at 6:48
$begingroup$
Actually doesnt the wikipedia page you mentioned answer this? Maybe in the etymology section?
$endgroup$
– Ben
Dec 27 '18 at 6:48
add a comment |
2 Answers
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The "Law of the Unconscious Statistician" is just a name for the fact that $E(g(X))$ is given by the formula you wrote. There is no difference.
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
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$begingroup$
Finding $mathbb Eg(X)$ can be done on several ways.
One of them is $$mathbb Eg(X)=int y;dF_Y(y)$$where $Y:=g(X)$ and $F_Y$ denotes the CDF of $Y$.
Another (in almost all cases more convenient) is practicizing LOTUS:$$mathbb Eg(X)=int g(x);dF_X(x)$$where $F_X$ denotes the CDF of $X$.
This works if you know the distribution of $X$ and you can hold your shoulders for the distribution of $g(X)$.
In both cases a PDF or PMF might exist leading to variants.
One of them is mentioned by you:$$mathbb Eg(X)=int g(x)p_X(x);dx$$where $p_X$ denotes a PDF of $X$.
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
The "Law of the Unconscious Statistician" is just a name for the fact that $E(g(X))$ is given by the formula you wrote. There is no difference.
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
$endgroup$
add a comment |
$begingroup$
The "Law of the Unconscious Statistician" is just a name for the fact that $E(g(X))$ is given by the formula you wrote. There is no difference.
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
$endgroup$
add a comment |
$begingroup$
The "Law of the Unconscious Statistician" is just a name for the fact that $E(g(X))$ is given by the formula you wrote. There is no difference.
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
$endgroup$
The "Law of the Unconscious Statistician" is just a name for the fact that $E(g(X))$ is given by the formula you wrote. There is no difference.
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
answered Dec 27 '18 at 6:55
littleOlittleO
30.1k647110
30.1k647110
add a comment |
add a comment |
$begingroup$
Finding $mathbb Eg(X)$ can be done on several ways.
One of them is $$mathbb Eg(X)=int y;dF_Y(y)$$where $Y:=g(X)$ and $F_Y$ denotes the CDF of $Y$.
Another (in almost all cases more convenient) is practicizing LOTUS:$$mathbb Eg(X)=int g(x);dF_X(x)$$where $F_X$ denotes the CDF of $X$.
This works if you know the distribution of $X$ and you can hold your shoulders for the distribution of $g(X)$.
In both cases a PDF or PMF might exist leading to variants.
One of them is mentioned by you:$$mathbb Eg(X)=int g(x)p_X(x);dx$$where $p_X$ denotes a PDF of $X$.
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Finding $mathbb Eg(X)$ can be done on several ways.
One of them is $$mathbb Eg(X)=int y;dF_Y(y)$$where $Y:=g(X)$ and $F_Y$ denotes the CDF of $Y$.
Another (in almost all cases more convenient) is practicizing LOTUS:$$mathbb Eg(X)=int g(x);dF_X(x)$$where $F_X$ denotes the CDF of $X$.
This works if you know the distribution of $X$ and you can hold your shoulders for the distribution of $g(X)$.
In both cases a PDF or PMF might exist leading to variants.
One of them is mentioned by you:$$mathbb Eg(X)=int g(x)p_X(x);dx$$where $p_X$ denotes a PDF of $X$.
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Finding $mathbb Eg(X)$ can be done on several ways.
One of them is $$mathbb Eg(X)=int y;dF_Y(y)$$where $Y:=g(X)$ and $F_Y$ denotes the CDF of $Y$.
Another (in almost all cases more convenient) is practicizing LOTUS:$$mathbb Eg(X)=int g(x);dF_X(x)$$where $F_X$ denotes the CDF of $X$.
This works if you know the distribution of $X$ and you can hold your shoulders for the distribution of $g(X)$.
In both cases a PDF or PMF might exist leading to variants.
One of them is mentioned by you:$$mathbb Eg(X)=int g(x)p_X(x);dx$$where $p_X$ denotes a PDF of $X$.
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
$endgroup$
Finding $mathbb Eg(X)$ can be done on several ways.
One of them is $$mathbb Eg(X)=int y;dF_Y(y)$$where $Y:=g(X)$ and $F_Y$ denotes the CDF of $Y$.
Another (in almost all cases more convenient) is practicizing LOTUS:$$mathbb Eg(X)=int g(x);dF_X(x)$$where $F_X$ denotes the CDF of $X$.
This works if you know the distribution of $X$ and you can hold your shoulders for the distribution of $g(X)$.
In both cases a PDF or PMF might exist leading to variants.
One of them is mentioned by you:$$mathbb Eg(X)=int g(x)p_X(x);dx$$where $p_X$ denotes a PDF of $X$.
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
answered Dec 27 '18 at 8:08
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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$begingroup$
Actually doesnt the wikipedia page you mentioned answer this? Maybe in the etymology section?
$endgroup$
– Ben
Dec 27 '18 at 6:48