Integral problem about probability
0
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Let $h(x)=frac{f'(x)^2}{f(x)},$ and $f(x)$ is the probability density function of r.v. $X$.
If $int_{-infty}^{infty}(1+x^2)h(x)dx<infty$, do we have $int_{-infty}^{infty} xh(x)dx=0$?
Many thanks
real-analysis calculus probability integration probability-theory
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
$endgroup$
add a comment |
0
$begingroup$
Let $h(x)=frac{f'(x)^2}{f(x)},$ and $f(x)$ is the probability density function of r.v. $X$.
If $int_{-infty}^{infty}(1+x^2)h(x)dx<infty$, do we have $int_{-infty}^{infty} xh(x)dx=0$?
Many thanks
real-analysis calculus probability integration probability-theory
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
$endgroup$
1
$begingroup$
I am not sure whether one could say an indefinite integral is bounded or not, i.e. I suppose there are some borders of integration missing.
$endgroup$
– mrtaurho
Dec 23 '18 at 12:49
1
$begingroup$
The integration could be regarded as expectation. I should put the borders from $-infty$ to $infty$
$endgroup$
– FoxFlu
Dec 23 '18 at 12:54
$begingroup$
Note that $int_{-infty}^infty h(x),mathrm{d}xgt0$. Furthermore, if $f(x)$ is a probability density function, then $f(x-1)$ is also. If $int_{-infty}^infty x h(x),mathrm{d}x=0$, then $int_{-infty}^infty x h(x-1),mathrm{d}x=int_{-infty}^infty h(x),mathrm{d}xgt0$. Thus, it is not true for all probability densities.
$endgroup$
– robjohn♦
Dec 23 '18 at 13:39
$begingroup$
For exponential distribution with pdf $f(x)=e^{-x}$ ($x>0$) we get $h(x)=f(x)$ and $int_{-infty}^infty xh(x)dx=1$.
$endgroup$
– NCh
Dec 24 '18 at 3:25
add a comment |
0
0
0
$begingroup$
Let $h(x)=frac{f'(x)^2}{f(x)},$ and $f(x)$ is the probability density function of r.v. $X$.
If $int_{-infty}^{infty}(1+x^2)h(x)dx<infty$, do we have $int_{-infty}^{infty} xh(x)dx=0$?
Many thanks
real-analysis calculus probability integration probability-theory
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
$endgroup$
Let $h(x)=frac{f'(x)^2}{f(x)},$ and $f(x)$ is the probability density function of r.v. $X$.
If $int_{-infty}^{infty}(1+x^2)h(x)dx<infty$, do we have $int_{-infty}^{infty} xh(x)dx=0$?
Many thanks
real-analysis calculus probability integration probability-theory
real-analysis calculus probability integration probability-theory
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
edited Dec 23 '18 at 12:55
FoxFlu
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
asked Dec 23 '18 at 12:47
FoxFluFoxFlu
12
12
1
$begingroup$
I am not sure whether one could say an indefinite integral is bounded or not, i.e. I suppose there are some borders of integration missing.
$endgroup$
– mrtaurho
Dec 23 '18 at 12:49
1
$begingroup$
The integration could be regarded as expectation. I should put the borders from $-infty$ to $infty$
$endgroup$
– FoxFlu
Dec 23 '18 at 12:54
$begingroup$
Note that $int_{-infty}^infty h(x),mathrm{d}xgt0$. Furthermore, if $f(x)$ is a probability density function, then $f(x-1)$ is also. If $int_{-infty}^infty x h(x),mathrm{d}x=0$, then $int_{-infty}^infty x h(x-1),mathrm{d}x=int_{-infty}^infty h(x),mathrm{d}xgt0$. Thus, it is not true for all probability densities.
$endgroup$
– robjohn♦
Dec 23 '18 at 13:39
$begingroup$
For exponential distribution with pdf $f(x)=e^{-x}$ ($x>0$) we get $h(x)=f(x)$ and $int_{-infty}^infty xh(x)dx=1$.
$endgroup$
– NCh
Dec 24 '18 at 3:25
add a comment |
1
$begingroup$
I am not sure whether one could say an indefinite integral is bounded or not, i.e. I suppose there are some borders of integration missing.
$endgroup$
– mrtaurho
Dec 23 '18 at 12:49
1
$begingroup$
The integration could be regarded as expectation. I should put the borders from $-infty$ to $infty$
$endgroup$
– FoxFlu
Dec 23 '18 at 12:54
$begingroup$
Note that $int_{-infty}^infty h(x),mathrm{d}xgt0$. Furthermore, if $f(x)$ is a probability density function, then $f(x-1)$ is also. If $int_{-infty}^infty x h(x),mathrm{d}x=0$, then $int_{-infty}^infty x h(x-1),mathrm{d}x=int_{-infty}^infty h(x),mathrm{d}xgt0$. Thus, it is not true for all probability densities.
$endgroup$
– robjohn♦
Dec 23 '18 at 13:39
$begingroup$
For exponential distribution with pdf $f(x)=e^{-x}$ ($x>0$) we get $h(x)=f(x)$ and $int_{-infty}^infty xh(x)dx=1$.
$endgroup$
– NCh
Dec 24 '18 at 3:25
1
1
$begingroup$
I am not sure whether one could say an indefinite integral is bounded or not, i.e. I suppose there are some borders of integration missing.
$endgroup$
– mrtaurho
Dec 23 '18 at 12:49
$begingroup$
I am not sure whether one could say an indefinite integral is bounded or not, i.e. I suppose there are some borders of integration missing.
$endgroup$
– mrtaurho
Dec 23 '18 at 12:49
1
1
$begingroup$
The integration could be regarded as expectation. I should put the borders from $-infty$ to $infty$
$endgroup$
– FoxFlu
Dec 23 '18 at 12:54
$begingroup$
The integration could be regarded as expectation. I should put the borders from $-infty$ to $infty$
$endgroup$
– FoxFlu
Dec 23 '18 at 12:54
$begingroup$
Note that $int_{-infty}^infty h(x),mathrm{d}xgt0$. Furthermore, if $f(x)$ is a probability density function, then $f(x-1)$ is also. If $int_{-infty}^infty x h(x),mathrm{d}x=0$, then $int_{-infty}^infty x h(x-1),mathrm{d}x=int_{-infty}^infty h(x),mathrm{d}xgt0$. Thus, it is not true for all probability densities.
$endgroup$
– robjohn♦
Dec 23 '18 at 13:39
$begingroup$
Note that $int_{-infty}^infty h(x),mathrm{d}xgt0$. Furthermore, if $f(x)$ is a probability density function, then $f(x-1)$ is also. If $int_{-infty}^infty x h(x),mathrm{d}x=0$, then $int_{-infty}^infty x h(x-1),mathrm{d}x=int_{-infty}^infty h(x),mathrm{d}xgt0$. Thus, it is not true for all probability densities.
$endgroup$
– robjohn♦
Dec 23 '18 at 13:39
$begingroup$
For exponential distribution with pdf $f(x)=e^{-x}$ ($x>0$) we get $h(x)=f(x)$ and $int_{-infty}^infty xh(x)dx=1$.
$endgroup$
– NCh
Dec 24 '18 at 3:25
$begingroup$
For exponential distribution with pdf $f(x)=e^{-x}$ ($x>0$) we get $h(x)=f(x)$ and $int_{-infty}^infty xh(x)dx=1$.
$endgroup$
– NCh
Dec 24 '18 at 3:25
add a comment |
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1
$begingroup$
I am not sure whether one could say an indefinite integral is bounded or not, i.e. I suppose there are some borders of integration missing.
$endgroup$
– mrtaurho
Dec 23 '18 at 12:49
1
$begingroup$
The integration could be regarded as expectation. I should put the borders from $-infty$ to $infty$
$endgroup$
– FoxFlu
Dec 23 '18 at 12:54
$begingroup$
Note that $int_{-infty}^infty h(x),mathrm{d}xgt0$. Furthermore, if $f(x)$ is a probability density function, then $f(x-1)$ is also. If $int_{-infty}^infty x h(x),mathrm{d}x=0$, then $int_{-infty}^infty x h(x-1),mathrm{d}x=int_{-infty}^infty h(x),mathrm{d}xgt0$. Thus, it is not true for all probability densities.
$endgroup$
– robjohn♦
Dec 23 '18 at 13:39
$begingroup$
For exponential distribution with pdf $f(x)=e^{-x}$ ($x>0$) we get $h(x)=f(x)$ and $int_{-infty}^infty xh(x)dx=1$.
$endgroup$
– NCh
Dec 24 '18 at 3:25