Monotonic operations and integralsDouble IntegralsWhen are sums and integrals “identical” in form?Difficult Indefinite IntegralImproper integrals and right-hand Riemann sumsinterpreting triple integralsTranslating integrals by scalar premultiplicationRegularity of Cramer-Rao inequality for unbiased estimatorsCramer-Rao Lower Bound when the support for $x$ depends on $theta$?Why does the parameter estimator not depend on the parameter?Find $int^pi/2_0 operatornamearccot(1-x+x^2),dx $
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Monotonic operations and integrals
Double IntegralsWhen are sums and integrals “identical” in form?Difficult Indefinite IntegralImproper integrals and right-hand Riemann sumsinterpreting triple integralsTranslating integrals by scalar premultiplicationRegularity of Cramer-Rao inequality for unbiased estimatorsCramer-Rao Lower Bound when the support for $x$ depends on $theta$?Why does the parameter estimator not depend on the parameter?Find $int^pi/2_0 operatornamearccot(1-x+x^2),dx $
$begingroup$
If I have a monotonic function, say ln, can I bring it inside an integral?
in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$
My limits of integration don't depend on $x,$ so I think I can.
Could I move expectation inside the integral?
(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)
calculus integration statistics expected-value
edited 8 hours ago
Adrian Keister
5,56872133
asked 8 hours ago
JessJess
135111
$endgroup$
add a comment |
$begingroup$
If I have a monotonic function, say ln, can I bring it inside an integral?
in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$
My limits of integration don't depend on $x,$ so I think I can.
Could I move expectation inside the integral?
(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)
calculus integration statistics expected-value
edited 8 hours ago
Adrian Keister
5,56872133
asked 8 hours ago
JessJess
135111
$endgroup$
$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago
add a comment |
$begingroup$
If I have a monotonic function, say ln, can I bring it inside an integral?
in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$
My limits of integration don't depend on $x,$ so I think I can.
Could I move expectation inside the integral?
(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)
calculus integration statistics expected-value
edited 8 hours ago
Adrian Keister
5,56872133
asked 8 hours ago
JessJess
135111
$endgroup$
If I have a monotonic function, say ln, can I bring it inside an integral?
in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$
My limits of integration don't depend on $x,$ so I think I can.
Could I move expectation inside the integral?
(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)
calculus integration statistics expected-value
calculus integration statistics expected-value
edited 8 hours ago
Adrian Keister
5,56872133
asked 8 hours ago
JessJess
135111
edited 8 hours ago
Adrian Keister
5,56872133
asked 8 hours ago
JessJess
135111
edited 8 hours ago
Adrian Keister
5,56872133
edited 8 hours ago
Adrian Keister
5,56872133
edited 8 hours ago
Adrian Keister
5,56872133
5,56872133
asked 8 hours ago
JessJess
135111
asked 8 hours ago
JessJess
135111
asked 8 hours ago
JessJess
135111
135111
$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago
add a comment |
$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago
$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago
$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.
edited 8 hours ago
Adrian Keister
5,56872133
answered 8 hours ago
MarkMark
13.3k1825
$endgroup$
add a comment |
$begingroup$
No. As a counterexample, take $f(x)=e^x$.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.
edited 8 hours ago
Adrian Keister
5,56872133
answered 8 hours ago
MarkMark
13.3k1825
$endgroup$
add a comment |
$begingroup$
$lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.
edited 8 hours ago
Adrian Keister
5,56872133
answered 8 hours ago
MarkMark
13.3k1825
$endgroup$
add a comment |
$begingroup$
$lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.
edited 8 hours ago
Adrian Keister
5,56872133
answered 8 hours ago
MarkMark
13.3k1825
$endgroup$
$lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.
edited 8 hours ago
Adrian Keister
5,56872133
answered 8 hours ago
MarkMark
13.3k1825
edited 8 hours ago
Adrian Keister
5,56872133
edited 8 hours ago
Adrian Keister
5,56872133
edited 8 hours ago
Adrian Keister
5,56872133
5,56872133
answered 8 hours ago
MarkMark
13.3k1825
answered 8 hours ago
MarkMark
13.3k1825
answered 8 hours ago
MarkMark
13.3k1825
13.3k1825
add a comment |
add a comment |
$begingroup$
No. As a counterexample, take $f(x)=e^x$.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
$endgroup$
add a comment |
$begingroup$
No. As a counterexample, take $f(x)=e^x$.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
$endgroup$
add a comment |
$begingroup$
No. As a counterexample, take $f(x)=e^x$.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
$endgroup$
No. As a counterexample, take $f(x)=e^x$.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,9941723
7,9941723
add a comment |
add a comment |
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$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago