$int x^k dmu(x)geq0$ for $k$ even?
I am studying convex optimization, and the truncated moment problem is being discussed. I have no background in measure theory so I don't understand things taken for granted such as $$int x^k dmu(x)geq0 text{ for }ktext{ even}.$$
Could someone explain how this is true? I attempt to evaluate by doing:
$$
begin{align}
int x^k dmu(x)&=
int x^kmu'(x) dx
end{align}
$$
which seems like I could pick any $mu'(x)$ and force the integral to be $<0$. Is there perhaps some restraint on $mu$ that can be explained?
measure-theory convex-optimization moment-problem
asked Dec 3 '18 at 20:51
Dan
276
|
show 1 more comment
I am studying convex optimization, and the truncated moment problem is being discussed. I have no background in measure theory so I don't understand things taken for granted such as $$int x^k dmu(x)geq0 text{ for }ktext{ even}.$$
Could someone explain how this is true? I attempt to evaluate by doing:
$$
begin{align}
int x^k dmu(x)&=
int x^kmu'(x) dx
end{align}
$$
which seems like I could pick any $mu'(x)$ and force the integral to be $<0$. Is there perhaps some restraint on $mu$ that can be explained?
measure-theory convex-optimization moment-problem
asked Dec 3 '18 at 20:51
Dan
276
The integral of a non-negative function is non-negative, i.e $$f geq 0 implies int f(x) , dmu(x) geq 0.$$ Note that $f(x) := x^k$ is non-negative for even $k$.
– saz
Dec 3 '18 at 20:59
@saz what if we let $mu(x)=-x$? Because then $$int f(x) dmu(x) = int -f(x) dx leq 0$$
– Dan
Dec 3 '18 at 21:07
Well, I was indeed assuming that you are talking about non-negative measures
– saz
Dec 3 '18 at 21:08
@saz I don't know anything about measure theory but I assume you are correct. Are there typical such restrictions on $mu(x)$? For example, if we have $mu(x)=x^{k}$, then $$int x^k dmu(x)=int x^k k x^{k-1} dxpropto int x^{2k-1} dx,$$ which is an integral of an odd function.
– Dan
Dec 3 '18 at 21:12
You will need to specify the domain of integration. The statement $dmu(x) = k x^{k-1} , dx$ is only true for $x geq 0$.
– saz
Dec 3 '18 at 21:17
|
show 1 more comment
I am studying convex optimization, and the truncated moment problem is being discussed. I have no background in measure theory so I don't understand things taken for granted such as $$int x^k dmu(x)geq0 text{ for }ktext{ even}.$$
Could someone explain how this is true? I attempt to evaluate by doing:
$$
begin{align}
int x^k dmu(x)&=
int x^kmu'(x) dx
end{align}
$$
which seems like I could pick any $mu'(x)$ and force the integral to be $<0$. Is there perhaps some restraint on $mu$ that can be explained?
measure-theory convex-optimization moment-problem
asked Dec 3 '18 at 20:51
Dan
276
I am studying convex optimization, and the truncated moment problem is being discussed. I have no background in measure theory so I don't understand things taken for granted such as $$int x^k dmu(x)geq0 text{ for }ktext{ even}.$$
Could someone explain how this is true? I attempt to evaluate by doing:
$$
begin{align}
int x^k dmu(x)&=
int x^kmu'(x) dx
end{align}
$$
which seems like I could pick any $mu'(x)$ and force the integral to be $<0$. Is there perhaps some restraint on $mu$ that can be explained?
measure-theory convex-optimization moment-problem
measure-theory convex-optimization moment-problem
asked Dec 3 '18 at 20:51
Dan
276
asked Dec 3 '18 at 20:51
Dan
276
asked Dec 3 '18 at 20:51
Dan
276
asked Dec 3 '18 at 20:51
Dan
276
asked Dec 3 '18 at 20:51
Dan
276
276
The integral of a non-negative function is non-negative, i.e $$f geq 0 implies int f(x) , dmu(x) geq 0.$$ Note that $f(x) := x^k$ is non-negative for even $k$.
– saz
Dec 3 '18 at 20:59
@saz what if we let $mu(x)=-x$? Because then $$int f(x) dmu(x) = int -f(x) dx leq 0$$
– Dan
Dec 3 '18 at 21:07
Well, I was indeed assuming that you are talking about non-negative measures
– saz
Dec 3 '18 at 21:08
@saz I don't know anything about measure theory but I assume you are correct. Are there typical such restrictions on $mu(x)$? For example, if we have $mu(x)=x^{k}$, then $$int x^k dmu(x)=int x^k k x^{k-1} dxpropto int x^{2k-1} dx,$$ which is an integral of an odd function.
– Dan
Dec 3 '18 at 21:12
You will need to specify the domain of integration. The statement $dmu(x) = k x^{k-1} , dx$ is only true for $x geq 0$.
– saz
Dec 3 '18 at 21:17
|
show 1 more comment
The integral of a non-negative function is non-negative, i.e $$f geq 0 implies int f(x) , dmu(x) geq 0.$$ Note that $f(x) := x^k$ is non-negative for even $k$.
– saz
Dec 3 '18 at 20:59
@saz what if we let $mu(x)=-x$? Because then $$int f(x) dmu(x) = int -f(x) dx leq 0$$
– Dan
Dec 3 '18 at 21:07
Well, I was indeed assuming that you are talking about non-negative measures
– saz
Dec 3 '18 at 21:08
@saz I don't know anything about measure theory but I assume you are correct. Are there typical such restrictions on $mu(x)$? For example, if we have $mu(x)=x^{k}$, then $$int x^k dmu(x)=int x^k k x^{k-1} dxpropto int x^{2k-1} dx,$$ which is an integral of an odd function.
– Dan
Dec 3 '18 at 21:12
You will need to specify the domain of integration. The statement $dmu(x) = k x^{k-1} , dx$ is only true for $x geq 0$.
– saz
Dec 3 '18 at 21:17
The integral of a non-negative function is non-negative, i.e $$f geq 0 implies int f(x) , dmu(x) geq 0.$$ Note that $f(x) := x^k$ is non-negative for even $k$.
– saz
Dec 3 '18 at 20:59
The integral of a non-negative function is non-negative, i.e $$f geq 0 implies int f(x) , dmu(x) geq 0.$$ Note that $f(x) := x^k$ is non-negative for even $k$.
– saz
Dec 3 '18 at 20:59
@saz what if we let $mu(x)=-x$? Because then $$int f(x) dmu(x) = int -f(x) dx leq 0$$
– Dan
Dec 3 '18 at 21:07
@saz what if we let $mu(x)=-x$? Because then $$int f(x) dmu(x) = int -f(x) dx leq 0$$
– Dan
Dec 3 '18 at 21:07
Well, I was indeed assuming that you are talking about non-negative measures
– saz
Dec 3 '18 at 21:08
Well, I was indeed assuming that you are talking about non-negative measures
– saz
Dec 3 '18 at 21:08
@saz I don't know anything about measure theory but I assume you are correct. Are there typical such restrictions on $mu(x)$? For example, if we have $mu(x)=x^{k}$, then $$int x^k dmu(x)=int x^k k x^{k-1} dxpropto int x^{2k-1} dx,$$ which is an integral of an odd function.
– Dan
Dec 3 '18 at 21:12
@saz I don't know anything about measure theory but I assume you are correct. Are there typical such restrictions on $mu(x)$? For example, if we have $mu(x)=x^{k}$, then $$int x^k dmu(x)=int x^k k x^{k-1} dxpropto int x^{2k-1} dx,$$ which is an integral of an odd function.
– Dan
Dec 3 '18 at 21:12
You will need to specify the domain of integration. The statement $dmu(x) = k x^{k-1} , dx$ is only true for $x geq 0$.
– saz
Dec 3 '18 at 21:17
You will need to specify the domain of integration. The statement $dmu(x) = k x^{k-1} , dx$ is only true for $x geq 0$.
– saz
Dec 3 '18 at 21:17
|
show 1 more comment
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The integral of a non-negative function is non-negative, i.e $$f geq 0 implies int f(x) , dmu(x) geq 0.$$ Note that $f(x) := x^k$ is non-negative for even $k$.
– saz
Dec 3 '18 at 20:59
@saz what if we let $mu(x)=-x$? Because then $$int f(x) dmu(x) = int -f(x) dx leq 0$$
– Dan
Dec 3 '18 at 21:07
Well, I was indeed assuming that you are talking about non-negative measures
– saz
Dec 3 '18 at 21:08
@saz I don't know anything about measure theory but I assume you are correct. Are there typical such restrictions on $mu(x)$? For example, if we have $mu(x)=x^{k}$, then $$int x^k dmu(x)=int x^k k x^{k-1} dxpropto int x^{2k-1} dx,$$ which is an integral of an odd function.
– Dan
Dec 3 '18 at 21:12
You will need to specify the domain of integration. The statement $dmu(x) = k x^{k-1} , dx$ is only true for $x geq 0$.
– saz
Dec 3 '18 at 21:17