Does there exist any probability density function $f:mathbb{R}tomathbb{R}$ which is not...
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be a probability density function. Can the following be happened for $f$?
(1) $f$ is not integrable on an (some) interval of $mathbb{R}$.
(2) $f$ is not integrable on every closed interval of $mathbb{R}$.
I know that if $f$ is a probability density function then
(1) $f(x)geq0 quadtext{for all} ; x$,
(2) $int_{-infty}^{+infty}f(x),dx=1$.
but here we have Lebesgue integral not Riemann integral. Moreover if $f$ wants to be Riemann integrable on the whole $mathbb{R}$, it must hold in the following conditions
(a) $f$ is integrable on every closed interval of $mathbb{R},$
(b) the following integral is convergent
$$int_{-infty}^{+infty}f(x),dx=int_{-infty}^{0}f(x),dx+int_{0}^{+infty}f(x),dx.$$
According the mentioned things, the most pdf are Riemann integrable, and I could not find any example as I asked. Would anyone help me to find that. thanks a lot.
probability integration probability-theory probability-distributions
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be a probability density function. Can the following be happened for $f$?
(1) $f$ is not integrable on an (some) interval of $mathbb{R}$.
(2) $f$ is not integrable on every closed interval of $mathbb{R}$.
I know that if $f$ is a probability density function then
(1) $f(x)geq0 quadtext{for all} ; x$,
(2) $int_{-infty}^{+infty}f(x),dx=1$.
but here we have Lebesgue integral not Riemann integral. Moreover if $f$ wants to be Riemann integrable on the whole $mathbb{R}$, it must hold in the following conditions
(a) $f$ is integrable on every closed interval of $mathbb{R},$
(b) the following integral is convergent
$$int_{-infty}^{+infty}f(x),dx=int_{-infty}^{0}f(x),dx+int_{0}^{+infty}f(x),dx.$$
According the mentioned things, the most pdf are Riemann integrable, and I could not find any example as I asked. Would anyone help me to find that. thanks a lot.
probability integration probability-theory probability-distributions
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
$endgroup$
$begingroup$
Take $Xsim U[0,1]$. Then $g=1-1_{mathbb{Q}}$ is a version of $f_X$...
$endgroup$
– d.k.o.
Dec 20 '18 at 7:37
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be a probability density function. Can the following be happened for $f$?
(1) $f$ is not integrable on an (some) interval of $mathbb{R}$.
(2) $f$ is not integrable on every closed interval of $mathbb{R}$.
I know that if $f$ is a probability density function then
(1) $f(x)geq0 quadtext{for all} ; x$,
(2) $int_{-infty}^{+infty}f(x),dx=1$.
but here we have Lebesgue integral not Riemann integral. Moreover if $f$ wants to be Riemann integrable on the whole $mathbb{R}$, it must hold in the following conditions
(a) $f$ is integrable on every closed interval of $mathbb{R},$
(b) the following integral is convergent
$$int_{-infty}^{+infty}f(x),dx=int_{-infty}^{0}f(x),dx+int_{0}^{+infty}f(x),dx.$$
According the mentioned things, the most pdf are Riemann integrable, and I could not find any example as I asked. Would anyone help me to find that. thanks a lot.
probability integration probability-theory probability-distributions
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
$endgroup$
Let $f:mathbb{R}tomathbb{R}$ be a probability density function. Can the following be happened for $f$?
(1) $f$ is not integrable on an (some) interval of $mathbb{R}$.
(2) $f$ is not integrable on every closed interval of $mathbb{R}$.
I know that if $f$ is a probability density function then
(1) $f(x)geq0 quadtext{for all} ; x$,
(2) $int_{-infty}^{+infty}f(x),dx=1$.
but here we have Lebesgue integral not Riemann integral. Moreover if $f$ wants to be Riemann integrable on the whole $mathbb{R}$, it must hold in the following conditions
(a) $f$ is integrable on every closed interval of $mathbb{R},$
(b) the following integral is convergent
$$int_{-infty}^{+infty}f(x),dx=int_{-infty}^{0}f(x),dx+int_{0}^{+infty}f(x),dx.$$
According the mentioned things, the most pdf are Riemann integrable, and I could not find any example as I asked. Would anyone help me to find that. thanks a lot.
probability integration probability-theory probability-distributions
probability integration probability-theory probability-distributions
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
edited Dec 20 '18 at 7:39
Lord Shark the Unknown
105k1160133
105k1160133
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
asked Dec 20 '18 at 7:25
soodehMehboodisoodehMehboodi
62138
62138
$begingroup$
Take $Xsim U[0,1]$. Then $g=1-1_{mathbb{Q}}$ is a version of $f_X$...
$endgroup$
– d.k.o.
Dec 20 '18 at 7:37
add a comment |
$begingroup$
Take $Xsim U[0,1]$. Then $g=1-1_{mathbb{Q}}$ is a version of $f_X$...
$endgroup$
– d.k.o.
Dec 20 '18 at 7:37
$begingroup$
Take $Xsim U[0,1]$. Then $g=1-1_{mathbb{Q}}$ is a version of $f_X$...
$endgroup$
– d.k.o.
Dec 20 '18 at 7:37
$begingroup$
Take $Xsim U[0,1]$. Then $g=1-1_{mathbb{Q}}$ is a version of $f_X$...
$endgroup$
– d.k.o.
Dec 20 '18 at 7:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is known that there exists a measurable set $E$ in $mathbb R$ such that $0<m(Ecap I) <m(I)$ for every open interval $I$. If $f=frac {I_E} {m(E)}$ then $f$ is a density function but it is not continuous at any point so it is not Riemann integrable on any interval.
For the construction of such a set $E$ see Creating a Lebesgue measurable set with peculiar property.
It is easy give simpler examples where $f$ is almost everywhere equal to a Riemann integrable function but it is not itself Riemann integarble. In probability theory density function which are equal almost everywhere lead to the same distribution, so I tried to give a example which is not almost everywhere equal to a Riemann integrable function.
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
$endgroup$
$begingroup$
This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 20 '18 at 7:44
$begingroup$
This is a very interesting example. Is the existence of such a set $E$ easy to prove?
$endgroup$
– BigbearZzz
Dec 20 '18 at 8:27
$begingroup$
I have added a reference for that construction. @BigbearZzz
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 8:51
add a comment |
$begingroup$
$$
f(x) = begin{cases}
0 &; xinBbb Q cap[0,1] \
1 &; xinBbb Q^c cap[0,1]
end{cases}
$$
is a probability density function that is not Riemann integrable.
If you want a density function that is not Riemann integrable on any interval in $Bbb R$ then you can take $f$ to be the (normalized) Gaussian distribution and then $f+mathbf 1_{Bbb Q}$ is the desired probability density function with the properties you want.
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is known that there exists a measurable set $E$ in $mathbb R$ such that $0<m(Ecap I) <m(I)$ for every open interval $I$. If $f=frac {I_E} {m(E)}$ then $f$ is a density function but it is not continuous at any point so it is not Riemann integrable on any interval.
For the construction of such a set $E$ see Creating a Lebesgue measurable set with peculiar property.
It is easy give simpler examples where $f$ is almost everywhere equal to a Riemann integrable function but it is not itself Riemann integarble. In probability theory density function which are equal almost everywhere lead to the same distribution, so I tried to give a example which is not almost everywhere equal to a Riemann integrable function.
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
$endgroup$
$begingroup$
This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 20 '18 at 7:44
$begingroup$
This is a very interesting example. Is the existence of such a set $E$ easy to prove?
$endgroup$
– BigbearZzz
Dec 20 '18 at 8:27
$begingroup$
I have added a reference for that construction. @BigbearZzz
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 8:51
add a comment |
$begingroup$
It is known that there exists a measurable set $E$ in $mathbb R$ such that $0<m(Ecap I) <m(I)$ for every open interval $I$. If $f=frac {I_E} {m(E)}$ then $f$ is a density function but it is not continuous at any point so it is not Riemann integrable on any interval.
For the construction of such a set $E$ see Creating a Lebesgue measurable set with peculiar property.
It is easy give simpler examples where $f$ is almost everywhere equal to a Riemann integrable function but it is not itself Riemann integarble. In probability theory density function which are equal almost everywhere lead to the same distribution, so I tried to give a example which is not almost everywhere equal to a Riemann integrable function.
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
$endgroup$
$begingroup$
This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 20 '18 at 7:44
$begingroup$
This is a very interesting example. Is the existence of such a set $E$ easy to prove?
$endgroup$
– BigbearZzz
Dec 20 '18 at 8:27
$begingroup$
I have added a reference for that construction. @BigbearZzz
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 8:51
add a comment |
$begingroup$
It is known that there exists a measurable set $E$ in $mathbb R$ such that $0<m(Ecap I) <m(I)$ for every open interval $I$. If $f=frac {I_E} {m(E)}$ then $f$ is a density function but it is not continuous at any point so it is not Riemann integrable on any interval.
For the construction of such a set $E$ see Creating a Lebesgue measurable set with peculiar property.
It is easy give simpler examples where $f$ is almost everywhere equal to a Riemann integrable function but it is not itself Riemann integarble. In probability theory density function which are equal almost everywhere lead to the same distribution, so I tried to give a example which is not almost everywhere equal to a Riemann integrable function.
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
$endgroup$
It is known that there exists a measurable set $E$ in $mathbb R$ such that $0<m(Ecap I) <m(I)$ for every open interval $I$. If $f=frac {I_E} {m(E)}$ then $f$ is a density function but it is not continuous at any point so it is not Riemann integrable on any interval.
For the construction of such a set $E$ see Creating a Lebesgue measurable set with peculiar property.
It is easy give simpler examples where $f$ is almost everywhere equal to a Riemann integrable function but it is not itself Riemann integarble. In probability theory density function which are equal almost everywhere lead to the same distribution, so I tried to give a example which is not almost everywhere equal to a Riemann integrable function.
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
edited Dec 20 '18 at 8:50
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
answered Dec 20 '18 at 7:38
Kavi Rama MurthyKavi Rama Murthy
64.1k42464
64.1k42464
$begingroup$
This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 20 '18 at 7:44
$begingroup$
This is a very interesting example. Is the existence of such a set $E$ easy to prove?
$endgroup$
– BigbearZzz
Dec 20 '18 at 8:27
$begingroup$
I have added a reference for that construction. @BigbearZzz
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 8:51
add a comment |
$begingroup$
This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 20 '18 at 7:44
$begingroup$
This is a very interesting example. Is the existence of such a set $E$ easy to prove?
$endgroup$
– BigbearZzz
Dec 20 '18 at 8:27
$begingroup$
I have added a reference for that construction. @BigbearZzz
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 8:51
$begingroup$
This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 20 '18 at 7:44
$begingroup$
This is very nice since, also, $f$ is not a.e. equal to a Riemann integrable function.
$endgroup$
– Bjørn Kjos-Hanssen
Dec 20 '18 at 7:44
$begingroup$
This is a very interesting example. Is the existence of such a set $E$ easy to prove?
$endgroup$
– BigbearZzz
Dec 20 '18 at 8:27
$begingroup$
This is a very interesting example. Is the existence of such a set $E$ easy to prove?
$endgroup$
– BigbearZzz
Dec 20 '18 at 8:27
$begingroup$
I have added a reference for that construction. @BigbearZzz
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 8:51
$begingroup$
I have added a reference for that construction. @BigbearZzz
$endgroup$
– Kavi Rama Murthy
Dec 20 '18 at 8:51
add a comment |
$begingroup$
$$
f(x) = begin{cases}
0 &; xinBbb Q cap[0,1] \
1 &; xinBbb Q^c cap[0,1]
end{cases}
$$
is a probability density function that is not Riemann integrable.
If you want a density function that is not Riemann integrable on any interval in $Bbb R$ then you can take $f$ to be the (normalized) Gaussian distribution and then $f+mathbf 1_{Bbb Q}$ is the desired probability density function with the properties you want.
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
$begingroup$
$$
f(x) = begin{cases}
0 &; xinBbb Q cap[0,1] \
1 &; xinBbb Q^c cap[0,1]
end{cases}
$$
is a probability density function that is not Riemann integrable.
If you want a density function that is not Riemann integrable on any interval in $Bbb R$ then you can take $f$ to be the (normalized) Gaussian distribution and then $f+mathbf 1_{Bbb Q}$ is the desired probability density function with the properties you want.
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
$begingroup$
$$
f(x) = begin{cases}
0 &; xinBbb Q cap[0,1] \
1 &; xinBbb Q^c cap[0,1]
end{cases}
$$
is a probability density function that is not Riemann integrable.
If you want a density function that is not Riemann integrable on any interval in $Bbb R$ then you can take $f$ to be the (normalized) Gaussian distribution and then $f+mathbf 1_{Bbb Q}$ is the desired probability density function with the properties you want.
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
$endgroup$
$$
f(x) = begin{cases}
0 &; xinBbb Q cap[0,1] \
1 &; xinBbb Q^c cap[0,1]
end{cases}
$$
is a probability density function that is not Riemann integrable.
If you want a density function that is not Riemann integrable on any interval in $Bbb R$ then you can take $f$ to be the (normalized) Gaussian distribution and then $f+mathbf 1_{Bbb Q}$ is the desired probability density function with the properties you want.
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
answered Dec 20 '18 at 7:40
BigbearZzzBigbearZzz
8,88821652
8,88821652
add a comment |
add a comment |
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$begingroup$
Take $Xsim U[0,1]$. Then $g=1-1_{mathbb{Q}}$ is a version of $f_X$...
$endgroup$
– d.k.o.
Dec 20 '18 at 7:37