Construction of a “density” or a Radon-Nikodym for a Semicontinuous Distribution
I just started self-studying measure theory and had a question about the "generalised density" (as it appears on Wikipedia):
A random variable X with values in a measurable space $({mathcal{X}},{mathcal {A}})$ (usually $mathbb {R} ^{n}$ with the Borel sets
as measurable subsets) has as probability distribution the measure $X_*P$ on $({mathcal {X}},{mathcal {A}})$: the density of $X$ with
respect to a reference measure $mu$ on $({mathcal {X}},{mathcal{A}})$ is the Radon–Nikodym derivative:
$$f={frac {dX_{*}P}{dmu }}.$$
For a random variable $X$, if we had a cumulative distribution function:
$$F(X) = begin{cases} 0 & if ; X < 0 \ X^2 & if ; 0 leq X < 0.8 \ 1 & if ; X geq 0.8 end{cases}$$
... that looks like:
Then, does a "generalised density" or a Radon-Nikodym derivative as defined above exist?
I think that it does... If $mathcal X = [0, 0.8) cup {0.8}$, for an appropriate $sigma$-algebra $mathcal A$, could one use the "reference measure":
$$mu = l + delta_{0.8}$$
... where $l$ is the Lebesgue measure and $delta_{0.8}$ is the Dirac measure on ${0.8}$? I'd also love some help clarifying what $X_*P$ is (is this a "pushforward measure", and if so, is this the same as the "measure induced by the CDF"?)
If one can use the reference measure above, why can we use it? And would the Radon-Nikodym derivative be:
$$f = frac{dX_*P}{dmu} = 2X * mathcal I(X in [0, 0.8)) + 0.2 * mathcal I(X = 0.8)$$
where $mathcal I$ is the indicator function?
probability probability-theory measure-theory lebesgue-measure
asked Nov 25 at 2:52
Aditya Ravuri
546
add a comment |
I just started self-studying measure theory and had a question about the "generalised density" (as it appears on Wikipedia):
A random variable X with values in a measurable space $({mathcal{X}},{mathcal {A}})$ (usually $mathbb {R} ^{n}$ with the Borel sets
as measurable subsets) has as probability distribution the measure $X_*P$ on $({mathcal {X}},{mathcal {A}})$: the density of $X$ with
respect to a reference measure $mu$ on $({mathcal {X}},{mathcal{A}})$ is the Radon–Nikodym derivative:
$$f={frac {dX_{*}P}{dmu }}.$$
For a random variable $X$, if we had a cumulative distribution function:
$$F(X) = begin{cases} 0 & if ; X < 0 \ X^2 & if ; 0 leq X < 0.8 \ 1 & if ; X geq 0.8 end{cases}$$
... that looks like:
Then, does a "generalised density" or a Radon-Nikodym derivative as defined above exist?
I think that it does... If $mathcal X = [0, 0.8) cup {0.8}$, for an appropriate $sigma$-algebra $mathcal A$, could one use the "reference measure":
$$mu = l + delta_{0.8}$$
... where $l$ is the Lebesgue measure and $delta_{0.8}$ is the Dirac measure on ${0.8}$? I'd also love some help clarifying what $X_*P$ is (is this a "pushforward measure", and if so, is this the same as the "measure induced by the CDF"?)
If one can use the reference measure above, why can we use it? And would the Radon-Nikodym derivative be:
$$f = frac{dX_*P}{dmu} = 2X * mathcal I(X in [0, 0.8)) + 0.2 * mathcal I(X = 0.8)$$
where $mathcal I$ is the indicator function?
probability probability-theory measure-theory lebesgue-measure
asked Nov 25 at 2:52
Aditya Ravuri
546
You can only ask about existence of RND w.r.t. a given reference measure. Every measure is absolutely continuous w.r.t. itself with density $1$ so your question does not make sense.Also do not use the same symbols $X$ for a random variable and a real variable used to define its distribution function.
– Kavi Rama Murthy
Nov 25 at 5:14
Thanks for the reply, what if the reference measure was the sum of lebesgue measure and a dirac measure on the set ${0.8}$?
– Aditya Ravuri
Nov 25 at 10:40
If that is the reference measure then we do have absolute continuity and hence RND exists.
– Kavi Rama Murthy
Nov 25 at 11:34
Could you please post a proof or a very rough sketch and what the RND looks like? I'd be really grateful!!
– Aditya Ravuri
Nov 25 at 11:35
1
The RND is $0$ for $x<0$ as well as $x>0.8$, $2x$ for $0leq x <0.8$ and $[1-(0,8)^{2}]$ for $x=0.8$
– Kavi Rama Murthy
Nov 25 at 11:43
add a comment |
I just started self-studying measure theory and had a question about the "generalised density" (as it appears on Wikipedia):
A random variable X with values in a measurable space $({mathcal{X}},{mathcal {A}})$ (usually $mathbb {R} ^{n}$ with the Borel sets
as measurable subsets) has as probability distribution the measure $X_*P$ on $({mathcal {X}},{mathcal {A}})$: the density of $X$ with
respect to a reference measure $mu$ on $({mathcal {X}},{mathcal{A}})$ is the Radon–Nikodym derivative:
$$f={frac {dX_{*}P}{dmu }}.$$
For a random variable $X$, if we had a cumulative distribution function:
$$F(X) = begin{cases} 0 & if ; X < 0 \ X^2 & if ; 0 leq X < 0.8 \ 1 & if ; X geq 0.8 end{cases}$$
... that looks like:
Then, does a "generalised density" or a Radon-Nikodym derivative as defined above exist?
I think that it does... If $mathcal X = [0, 0.8) cup {0.8}$, for an appropriate $sigma$-algebra $mathcal A$, could one use the "reference measure":
$$mu = l + delta_{0.8}$$
... where $l$ is the Lebesgue measure and $delta_{0.8}$ is the Dirac measure on ${0.8}$? I'd also love some help clarifying what $X_*P$ is (is this a "pushforward measure", and if so, is this the same as the "measure induced by the CDF"?)
If one can use the reference measure above, why can we use it? And would the Radon-Nikodym derivative be:
$$f = frac{dX_*P}{dmu} = 2X * mathcal I(X in [0, 0.8)) + 0.2 * mathcal I(X = 0.8)$$
where $mathcal I$ is the indicator function?
probability probability-theory measure-theory lebesgue-measure
asked Nov 25 at 2:52
Aditya Ravuri
546
I just started self-studying measure theory and had a question about the "generalised density" (as it appears on Wikipedia):
A random variable X with values in a measurable space $({mathcal{X}},{mathcal {A}})$ (usually $mathbb {R} ^{n}$ with the Borel sets
as measurable subsets) has as probability distribution the measure $X_*P$ on $({mathcal {X}},{mathcal {A}})$: the density of $X$ with
respect to a reference measure $mu$ on $({mathcal {X}},{mathcal{A}})$ is the Radon–Nikodym derivative:
$$f={frac {dX_{*}P}{dmu }}.$$
For a random variable $X$, if we had a cumulative distribution function:
$$F(X) = begin{cases} 0 & if ; X < 0 \ X^2 & if ; 0 leq X < 0.8 \ 1 & if ; X geq 0.8 end{cases}$$
... that looks like:
Then, does a "generalised density" or a Radon-Nikodym derivative as defined above exist?
I think that it does... If $mathcal X = [0, 0.8) cup {0.8}$, for an appropriate $sigma$-algebra $mathcal A$, could one use the "reference measure":
$$mu = l + delta_{0.8}$$
... where $l$ is the Lebesgue measure and $delta_{0.8}$ is the Dirac measure on ${0.8}$? I'd also love some help clarifying what $X_*P$ is (is this a "pushforward measure", and if so, is this the same as the "measure induced by the CDF"?)
If one can use the reference measure above, why can we use it? And would the Radon-Nikodym derivative be:
$$f = frac{dX_*P}{dmu} = 2X * mathcal I(X in [0, 0.8)) + 0.2 * mathcal I(X = 0.8)$$
where $mathcal I$ is the indicator function?
probability probability-theory measure-theory lebesgue-measure
probability probability-theory measure-theory lebesgue-measure
asked Nov 25 at 2:52
Aditya Ravuri
546
asked Nov 25 at 2:52
Aditya Ravuri
546
edited Nov 25 at 10:42
asked Nov 25 at 2:52
Aditya Ravuri
546
asked Nov 25 at 2:52
Aditya Ravuri
546
asked Nov 25 at 2:52
Aditya Ravuri
546
546
You can only ask about existence of RND w.r.t. a given reference measure. Every measure is absolutely continuous w.r.t. itself with density $1$ so your question does not make sense.Also do not use the same symbols $X$ for a random variable and a real variable used to define its distribution function.
– Kavi Rama Murthy
Nov 25 at 5:14
Thanks for the reply, what if the reference measure was the sum of lebesgue measure and a dirac measure on the set ${0.8}$?
– Aditya Ravuri
Nov 25 at 10:40
If that is the reference measure then we do have absolute continuity and hence RND exists.
– Kavi Rama Murthy
Nov 25 at 11:34
Could you please post a proof or a very rough sketch and what the RND looks like? I'd be really grateful!!
– Aditya Ravuri
Nov 25 at 11:35
1
The RND is $0$ for $x<0$ as well as $x>0.8$, $2x$ for $0leq x <0.8$ and $[1-(0,8)^{2}]$ for $x=0.8$
– Kavi Rama Murthy
Nov 25 at 11:43
add a comment |
You can only ask about existence of RND w.r.t. a given reference measure. Every measure is absolutely continuous w.r.t. itself with density $1$ so your question does not make sense.Also do not use the same symbols $X$ for a random variable and a real variable used to define its distribution function.
– Kavi Rama Murthy
Nov 25 at 5:14
Thanks for the reply, what if the reference measure was the sum of lebesgue measure and a dirac measure on the set ${0.8}$?
– Aditya Ravuri
Nov 25 at 10:40
If that is the reference measure then we do have absolute continuity and hence RND exists.
– Kavi Rama Murthy
Nov 25 at 11:34
Could you please post a proof or a very rough sketch and what the RND looks like? I'd be really grateful!!
– Aditya Ravuri
Nov 25 at 11:35
1
The RND is $0$ for $x<0$ as well as $x>0.8$, $2x$ for $0leq x <0.8$ and $[1-(0,8)^{2}]$ for $x=0.8$
– Kavi Rama Murthy
Nov 25 at 11:43
You can only ask about existence of RND w.r.t. a given reference measure. Every measure is absolutely continuous w.r.t. itself with density $1$ so your question does not make sense.Also do not use the same symbols $X$ for a random variable and a real variable used to define its distribution function.
– Kavi Rama Murthy
Nov 25 at 5:14
You can only ask about existence of RND w.r.t. a given reference measure. Every measure is absolutely continuous w.r.t. itself with density $1$ so your question does not make sense.Also do not use the same symbols $X$ for a random variable and a real variable used to define its distribution function.
– Kavi Rama Murthy
Nov 25 at 5:14
Thanks for the reply, what if the reference measure was the sum of lebesgue measure and a dirac measure on the set ${0.8}$?
– Aditya Ravuri
Nov 25 at 10:40
Thanks for the reply, what if the reference measure was the sum of lebesgue measure and a dirac measure on the set ${0.8}$?
– Aditya Ravuri
Nov 25 at 10:40
If that is the reference measure then we do have absolute continuity and hence RND exists.
– Kavi Rama Murthy
Nov 25 at 11:34
If that is the reference measure then we do have absolute continuity and hence RND exists.
– Kavi Rama Murthy
Nov 25 at 11:34
Could you please post a proof or a very rough sketch and what the RND looks like? I'd be really grateful!!
– Aditya Ravuri
Nov 25 at 11:35
Could you please post a proof or a very rough sketch and what the RND looks like? I'd be really grateful!!
– Aditya Ravuri
Nov 25 at 11:35
1
1
The RND is $0$ for $x<0$ as well as $x>0.8$, $2x$ for $0leq x <0.8$ and $[1-(0,8)^{2}]$ for $x=0.8$
– Kavi Rama Murthy
Nov 25 at 11:43
The RND is $0$ for $x<0$ as well as $x>0.8$, $2x$ for $0leq x <0.8$ and $[1-(0,8)^{2}]$ for $x=0.8$
– Kavi Rama Murthy
Nov 25 at 11:43
add a comment |
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You can only ask about existence of RND w.r.t. a given reference measure. Every measure is absolutely continuous w.r.t. itself with density $1$ so your question does not make sense.Also do not use the same symbols $X$ for a random variable and a real variable used to define its distribution function.
– Kavi Rama Murthy
Nov 25 at 5:14
Thanks for the reply, what if the reference measure was the sum of lebesgue measure and a dirac measure on the set ${0.8}$?
– Aditya Ravuri
Nov 25 at 10:40
If that is the reference measure then we do have absolute continuity and hence RND exists.
– Kavi Rama Murthy
Nov 25 at 11:34
Could you please post a proof or a very rough sketch and what the RND looks like? I'd be really grateful!!
– Aditya Ravuri
Nov 25 at 11:35
1
The RND is $0$ for $x<0$ as well as $x>0.8$, $2x$ for $0leq x <0.8$ and $[1-(0,8)^{2}]$ for $x=0.8$
– Kavi Rama Murthy
Nov 25 at 11:43