Notation and absolute continuity for probability measures
$begingroup$
A Question on Notation:
I've realized that different notations are used in probability theory when evaluating an integral, and I am unsure as to how they "work" together, whether they're completely equiavalent or whether they do have subtle difference, and in which situations they should be used.
Let $(Omega, mathcal{F}, P)$ be a probability space and $X$ a real random variable that describes the distribution with pdf $f$.
We are told that
$mathbb E[X]=int_{mathbb R}xf(x)dx$
and
$mathbb E[X]=int_{Omega}XdP$
but that
$int_{Omega}XdP=int_{mathbb R}xf(x)dx$ only if $P_{X}$ is absolute continuous w.r.t $P$
Can anyone elaborate on this further. It still does not sit well with me.
probability probability-theory measure-theory expected-value
asked Jan 14 at 14:51
SABOYSABOY
679311
$endgroup$
add a comment |
$begingroup$
A Question on Notation:
I've realized that different notations are used in probability theory when evaluating an integral, and I am unsure as to how they "work" together, whether they're completely equiavalent or whether they do have subtle difference, and in which situations they should be used.
Let $(Omega, mathcal{F}, P)$ be a probability space and $X$ a real random variable that describes the distribution with pdf $f$.
We are told that
$mathbb E[X]=int_{mathbb R}xf(x)dx$
and
$mathbb E[X]=int_{Omega}XdP$
but that
$int_{Omega}XdP=int_{mathbb R}xf(x)dx$ only if $P_{X}$ is absolute continuous w.r.t $P$
Can anyone elaborate on this further. It still does not sit well with me.
probability probability-theory measure-theory expected-value
asked Jan 14 at 14:51
SABOYSABOY
679311
$endgroup$
add a comment |
$begingroup$
A Question on Notation:
I've realized that different notations are used in probability theory when evaluating an integral, and I am unsure as to how they "work" together, whether they're completely equiavalent or whether they do have subtle difference, and in which situations they should be used.
Let $(Omega, mathcal{F}, P)$ be a probability space and $X$ a real random variable that describes the distribution with pdf $f$.
We are told that
$mathbb E[X]=int_{mathbb R}xf(x)dx$
and
$mathbb E[X]=int_{Omega}XdP$
but that
$int_{Omega}XdP=int_{mathbb R}xf(x)dx$ only if $P_{X}$ is absolute continuous w.r.t $P$
Can anyone elaborate on this further. It still does not sit well with me.
probability probability-theory measure-theory expected-value
asked Jan 14 at 14:51
SABOYSABOY
679311
$endgroup$
A Question on Notation:
I've realized that different notations are used in probability theory when evaluating an integral, and I am unsure as to how they "work" together, whether they're completely equiavalent or whether they do have subtle difference, and in which situations they should be used.
Let $(Omega, mathcal{F}, P)$ be a probability space and $X$ a real random variable that describes the distribution with pdf $f$.
We are told that
$mathbb E[X]=int_{mathbb R}xf(x)dx$
and
$mathbb E[X]=int_{Omega}XdP$
but that
$int_{Omega}XdP=int_{mathbb R}xf(x)dx$ only if $P_{X}$ is absolute continuous w.r.t $P$
Can anyone elaborate on this further. It still does not sit well with me.
probability probability-theory measure-theory expected-value
probability probability-theory measure-theory expected-value
asked Jan 14 at 14:51
SABOYSABOY
679311
asked Jan 14 at 14:51
SABOYSABOY
679311
asked Jan 14 at 14:51
SABOYSABOY
679311
asked Jan 14 at 14:51
SABOYSABOY
679311
asked Jan 14 at 14:51
SABOYSABOY
679311
679311
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The definition of the expectation of random variable $X$ is (if it exists):$$mathbb EX=int X;dPtag1$$so is an integral on the original probability space.
A random variable $X$ induces a probability measure on $(mathbb R,mathcal B)$ where $mathcal B$ denotes the $sigma$-algebra of Borel subsets of $mathbb R$.
This probability measure is often denoted as $P_X$ and it is prescribed by:$$Bmapsto P({Xin B})$$
A new probability space is introduced now: $(mathbb R,mathcal B,P_X)$.
Further there is a theorem that says that:$$int X;dP=int x;dP_Xtag2$$
For a probability measure $mu$ on $(mathbb R,mathcal B)$ there might exist a non-negative Borel-measurable function $f:mathbb Rtomathbb R$ that satisfies:$$int_Bf(x)dx=mu(B)$$
If this is the case then it can be proved that:$$int g(x)f(x);dx=int g(x);dmutag3$$for suitable functions $g$.
If for $P_X$ such a function exists then the function serves a so-called PDF (probability density function) and is mostly denoted as $f_X$.
Application of $(2),(3)$ on the identity function then results in:$$int g(x)f_X(x);dx=int g(x);dP_X=int g(X);dP$$
Doing this for the identity function gives:$$int xf_X(x);dx=int x;dP_X=int X;dP=mathbb EX$$
answered Jan 14 at 15:31
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you! Just to make sure In your statement $(1)$ it would be $int_{Omega}XdP=int_{mathbb R}xdP_{x}$ , correct?
$endgroup$
– SABOY
Jan 14 at 16:12
$begingroup$
Yes, that's correct. You are welcome.
$endgroup$
– drhab
Jan 14 at 16:35
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
The definition of the expectation of random variable $X$ is (if it exists):$$mathbb EX=int X;dPtag1$$so is an integral on the original probability space.
A random variable $X$ induces a probability measure on $(mathbb R,mathcal B)$ where $mathcal B$ denotes the $sigma$-algebra of Borel subsets of $mathbb R$.
This probability measure is often denoted as $P_X$ and it is prescribed by:$$Bmapsto P({Xin B})$$
A new probability space is introduced now: $(mathbb R,mathcal B,P_X)$.
Further there is a theorem that says that:$$int X;dP=int x;dP_Xtag2$$
For a probability measure $mu$ on $(mathbb R,mathcal B)$ there might exist a non-negative Borel-measurable function $f:mathbb Rtomathbb R$ that satisfies:$$int_Bf(x)dx=mu(B)$$
If this is the case then it can be proved that:$$int g(x)f(x);dx=int g(x);dmutag3$$for suitable functions $g$.
If for $P_X$ such a function exists then the function serves a so-called PDF (probability density function) and is mostly denoted as $f_X$.
Application of $(2),(3)$ on the identity function then results in:$$int g(x)f_X(x);dx=int g(x);dP_X=int g(X);dP$$
Doing this for the identity function gives:$$int xf_X(x);dx=int x;dP_X=int X;dP=mathbb EX$$
answered Jan 14 at 15:31
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you! Just to make sure In your statement $(1)$ it would be $int_{Omega}XdP=int_{mathbb R}xdP_{x}$ , correct?
$endgroup$
– SABOY
Jan 14 at 16:12
$begingroup$
Yes, that's correct. You are welcome.
$endgroup$
– drhab
Jan 14 at 16:35
add a comment |
$begingroup$
The definition of the expectation of random variable $X$ is (if it exists):$$mathbb EX=int X;dPtag1$$so is an integral on the original probability space.
A random variable $X$ induces a probability measure on $(mathbb R,mathcal B)$ where $mathcal B$ denotes the $sigma$-algebra of Borel subsets of $mathbb R$.
This probability measure is often denoted as $P_X$ and it is prescribed by:$$Bmapsto P({Xin B})$$
A new probability space is introduced now: $(mathbb R,mathcal B,P_X)$.
Further there is a theorem that says that:$$int X;dP=int x;dP_Xtag2$$
For a probability measure $mu$ on $(mathbb R,mathcal B)$ there might exist a non-negative Borel-measurable function $f:mathbb Rtomathbb R$ that satisfies:$$int_Bf(x)dx=mu(B)$$
If this is the case then it can be proved that:$$int g(x)f(x);dx=int g(x);dmutag3$$for suitable functions $g$.
If for $P_X$ such a function exists then the function serves a so-called PDF (probability density function) and is mostly denoted as $f_X$.
Application of $(2),(3)$ on the identity function then results in:$$int g(x)f_X(x);dx=int g(x);dP_X=int g(X);dP$$
Doing this for the identity function gives:$$int xf_X(x);dx=int x;dP_X=int X;dP=mathbb EX$$
answered Jan 14 at 15:31
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you! Just to make sure In your statement $(1)$ it would be $int_{Omega}XdP=int_{mathbb R}xdP_{x}$ , correct?
$endgroup$
– SABOY
Jan 14 at 16:12
$begingroup$
Yes, that's correct. You are welcome.
$endgroup$
– drhab
Jan 14 at 16:35
add a comment |
$begingroup$
The definition of the expectation of random variable $X$ is (if it exists):$$mathbb EX=int X;dPtag1$$so is an integral on the original probability space.
A random variable $X$ induces a probability measure on $(mathbb R,mathcal B)$ where $mathcal B$ denotes the $sigma$-algebra of Borel subsets of $mathbb R$.
This probability measure is often denoted as $P_X$ and it is prescribed by:$$Bmapsto P({Xin B})$$
A new probability space is introduced now: $(mathbb R,mathcal B,P_X)$.
Further there is a theorem that says that:$$int X;dP=int x;dP_Xtag2$$
For a probability measure $mu$ on $(mathbb R,mathcal B)$ there might exist a non-negative Borel-measurable function $f:mathbb Rtomathbb R$ that satisfies:$$int_Bf(x)dx=mu(B)$$
If this is the case then it can be proved that:$$int g(x)f(x);dx=int g(x);dmutag3$$for suitable functions $g$.
If for $P_X$ such a function exists then the function serves a so-called PDF (probability density function) and is mostly denoted as $f_X$.
Application of $(2),(3)$ on the identity function then results in:$$int g(x)f_X(x);dx=int g(x);dP_X=int g(X);dP$$
Doing this for the identity function gives:$$int xf_X(x);dx=int x;dP_X=int X;dP=mathbb EX$$
answered Jan 14 at 15:31
drhabdrhab
103k545136
$endgroup$
The definition of the expectation of random variable $X$ is (if it exists):$$mathbb EX=int X;dPtag1$$so is an integral on the original probability space.
A random variable $X$ induces a probability measure on $(mathbb R,mathcal B)$ where $mathcal B$ denotes the $sigma$-algebra of Borel subsets of $mathbb R$.
This probability measure is often denoted as $P_X$ and it is prescribed by:$$Bmapsto P({Xin B})$$
A new probability space is introduced now: $(mathbb R,mathcal B,P_X)$.
Further there is a theorem that says that:$$int X;dP=int x;dP_Xtag2$$
For a probability measure $mu$ on $(mathbb R,mathcal B)$ there might exist a non-negative Borel-measurable function $f:mathbb Rtomathbb R$ that satisfies:$$int_Bf(x)dx=mu(B)$$
If this is the case then it can be proved that:$$int g(x)f(x);dx=int g(x);dmutag3$$for suitable functions $g$.
If for $P_X$ such a function exists then the function serves a so-called PDF (probability density function) and is mostly denoted as $f_X$.
Application of $(2),(3)$ on the identity function then results in:$$int g(x)f_X(x);dx=int g(x);dP_X=int g(X);dP$$
Doing this for the identity function gives:$$int xf_X(x);dx=int x;dP_X=int X;dP=mathbb EX$$
answered Jan 14 at 15:31
drhabdrhab
103k545136
answered Jan 14 at 15:31
drhabdrhab
103k545136
answered Jan 14 at 15:31
drhabdrhab
103k545136
answered Jan 14 at 15:31
drhabdrhab
103k545136
103k545136
$begingroup$
Thank you! Just to make sure In your statement $(1)$ it would be $int_{Omega}XdP=int_{mathbb R}xdP_{x}$ , correct?
$endgroup$
– SABOY
Jan 14 at 16:12
$begingroup$
Yes, that's correct. You are welcome.
$endgroup$
– drhab
Jan 14 at 16:35
add a comment |
$begingroup$
Thank you! Just to make sure In your statement $(1)$ it would be $int_{Omega}XdP=int_{mathbb R}xdP_{x}$ , correct?
$endgroup$
– SABOY
Jan 14 at 16:12
$begingroup$
Yes, that's correct. You are welcome.
$endgroup$
– drhab
Jan 14 at 16:35
$begingroup$
Thank you! Just to make sure In your statement $(1)$ it would be $int_{Omega}XdP=int_{mathbb R}xdP_{x}$ , correct?
$endgroup$
– SABOY
Jan 14 at 16:12
$begingroup$
Thank you! Just to make sure In your statement $(1)$ it would be $int_{Omega}XdP=int_{mathbb R}xdP_{x}$ , correct?
$endgroup$
– SABOY
Jan 14 at 16:12
$begingroup$
Yes, that's correct. You are welcome.
$endgroup$
– drhab
Jan 14 at 16:35
$begingroup$
Yes, that's correct. You are welcome.
$endgroup$
– drhab
Jan 14 at 16:35
add a comment |
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