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Does for two equivalent probability measures $Q approx P$ boundedness in $L^1(Q)$ imply boundedness in $L^0(P)$?
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Assume we have a measurable space $(Omega,mathcalF)$ with two probability measures $Q$ and $P$, which are equivalent, i.e. for all $A in mathcalF$ we have
$$Q(A)=0 iff P(A)=0.$$
I will denote expectations with respect to $P$ as $E_P[cdot]$ and expectations with respect to $Q$ as $E_Q[cdot]$.
Let $(f_n)_n in mathbbN$ be a sequence of non-negative random variables such that $sup_n in mathbbNE_Q[f_n] < infty$, i.e. $(f_n)_n in mathbbN$ is bounded in $L^1(Q)$. Assume that de Radon-Nikodyim derivative satisfies $$fracdQdP in L^infty(P).$$
I want to show that the sequence $(f_n)_n in mathbbN$ is also bounded in $L^0(P)$, i.e. that the following holds
$$sup_n in mathbbNP[f_n > K] rightarrow 0, text for K rightarrow infty.$$
I cannot seem to get to this conclusion. Does anyone have a hint or a complete proof?
Thanks a lot in advance!
real-analysis probability probability-theory measure-theory radon-nikodym
asked Aug 3 at 16:39
vaoy
45228
add a comment |Â
up vote
3
down vote
favorite
Assume we have a measurable space $(Omega,mathcalF)$ with two probability measures $Q$ and $P$, which are equivalent, i.e. for all $A in mathcalF$ we have
$$Q(A)=0 iff P(A)=0.$$
I will denote expectations with respect to $P$ as $E_P[cdot]$ and expectations with respect to $Q$ as $E_Q[cdot]$.
Let $(f_n)_n in mathbbN$ be a sequence of non-negative random variables such that $sup_n in mathbbNE_Q[f_n] < infty$, i.e. $(f_n)_n in mathbbN$ is bounded in $L^1(Q)$. Assume that de Radon-Nikodyim derivative satisfies $$fracdQdP in L^infty(P).$$
I want to show that the sequence $(f_n)_n in mathbbN$ is also bounded in $L^0(P)$, i.e. that the following holds
$$sup_n in mathbbNP[f_n > K] rightarrow 0, text for K rightarrow infty.$$
I cannot seem to get to this conclusion. Does anyone have a hint or a complete proof?
Thanks a lot in advance!
real-analysis probability probability-theory measure-theory radon-nikodym
asked Aug 3 at 16:39
vaoy
45228
Use the Radon-Nikodym derivative to rewrite the $L^p$ norm on $P$ in terms of the one on $Q$.
– anomaly
Aug 3 at 16:58
I tried that, but I only got this $$int f dP = int f fracdPdQ dQ$$ and this I cannot estimate, since I only have that $fracdQdP$ is in $L^infty(P)$. Or did you have something else in mind?
– vaoy
Aug 3 at 17:01
Estimate the latter using the fact that $int f geq int_f > K fgeq K, P(f > K)$ for any $K$ (with $fgeq 0$).
– anomaly
Aug 3 at 17:04
I don't see how that helps unfortunately. I don't see how I could estimate the latter.
– vaoy
Aug 3 at 17:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Assume we have a measurable space $(Omega,mathcalF)$ with two probability measures $Q$ and $P$, which are equivalent, i.e. for all $A in mathcalF$ we have
$$Q(A)=0 iff P(A)=0.$$
I will denote expectations with respect to $P$ as $E_P[cdot]$ and expectations with respect to $Q$ as $E_Q[cdot]$.
Let $(f_n)_n in mathbbN$ be a sequence of non-negative random variables such that $sup_n in mathbbNE_Q[f_n] < infty$, i.e. $(f_n)_n in mathbbN$ is bounded in $L^1(Q)$. Assume that de Radon-Nikodyim derivative satisfies $$fracdQdP in L^infty(P).$$
I want to show that the sequence $(f_n)_n in mathbbN$ is also bounded in $L^0(P)$, i.e. that the following holds
$$sup_n in mathbbNP[f_n > K] rightarrow 0, text for K rightarrow infty.$$
I cannot seem to get to this conclusion. Does anyone have a hint or a complete proof?
Thanks a lot in advance!
real-analysis probability probability-theory measure-theory radon-nikodym
asked Aug 3 at 16:39
vaoy
45228
Assume we have a measurable space $(Omega,mathcalF)$ with two probability measures $Q$ and $P$, which are equivalent, i.e. for all $A in mathcalF$ we have
$$Q(A)=0 iff P(A)=0.$$
I will denote expectations with respect to $P$ as $E_P[cdot]$ and expectations with respect to $Q$ as $E_Q[cdot]$.
Let $(f_n)_n in mathbbN$ be a sequence of non-negative random variables such that $sup_n in mathbbNE_Q[f_n] < infty$, i.e. $(f_n)_n in mathbbN$ is bounded in $L^1(Q)$. Assume that de Radon-Nikodyim derivative satisfies $$fracdQdP in L^infty(P).$$
I want to show that the sequence $(f_n)_n in mathbbN$ is also bounded in $L^0(P)$, i.e. that the following holds
$$sup_n in mathbbNP[f_n > K] rightarrow 0, text for K rightarrow infty.$$
I cannot seem to get to this conclusion. Does anyone have a hint or a complete proof?
Thanks a lot in advance!
real-analysis probability probability-theory measure-theory radon-nikodym
asked Aug 3 at 16:39
vaoy
45228
asked Aug 3 at 16:39
vaoy
45228
asked Aug 3 at 16:39
vaoy
45228
asked Aug 3 at 16:39
vaoy
45228
45228
Use the Radon-Nikodym derivative to rewrite the $L^p$ norm on $P$ in terms of the one on $Q$.
– anomaly
Aug 3 at 16:58
I tried that, but I only got this $$int f dP = int f fracdPdQ dQ$$ and this I cannot estimate, since I only have that $fracdQdP$ is in $L^infty(P)$. Or did you have something else in mind?
– vaoy
Aug 3 at 17:01
Estimate the latter using the fact that $int f geq int_f > K fgeq K, P(f > K)$ for any $K$ (with $fgeq 0$).
– anomaly
Aug 3 at 17:04
I don't see how that helps unfortunately. I don't see how I could estimate the latter.
– vaoy
Aug 3 at 17:36
add a comment |Â
Use the Radon-Nikodym derivative to rewrite the $L^p$ norm on $P$ in terms of the one on $Q$.
– anomaly
Aug 3 at 16:58
I tried that, but I only got this $$int f dP = int f fracdPdQ dQ$$ and this I cannot estimate, since I only have that $fracdQdP$ is in $L^infty(P)$. Or did you have something else in mind?
– vaoy
Aug 3 at 17:01
Estimate the latter using the fact that $int f geq int_f > K fgeq K, P(f > K)$ for any $K$ (with $fgeq 0$).
– anomaly
Aug 3 at 17:04
I don't see how that helps unfortunately. I don't see how I could estimate the latter.
– vaoy
Aug 3 at 17:36
Use the Radon-Nikodym derivative to rewrite the $L^p$ norm on $P$ in terms of the one on $Q$.
– anomaly
Aug 3 at 16:58
Use the Radon-Nikodym derivative to rewrite the $L^p$ norm on $P$ in terms of the one on $Q$.
– anomaly
Aug 3 at 16:58
I tried that, but I only got this $$int f dP = int f fracdPdQ dQ$$ and this I cannot estimate, since I only have that $fracdQdP$ is in $L^infty(P)$. Or did you have something else in mind?
– vaoy
Aug 3 at 17:01
I tried that, but I only got this $$int f dP = int f fracdPdQ dQ$$ and this I cannot estimate, since I only have that $fracdQdP$ is in $L^infty(P)$. Or did you have something else in mind?
– vaoy
Aug 3 at 17:01
Estimate the latter using the fact that $int f geq int_f > K fgeq K, P(f > K)$ for any $K$ (with $fgeq 0$).
– anomaly
Aug 3 at 17:04
Estimate the latter using the fact that $int f geq int_f > K fgeq K, P(f > K)$ for any $K$ (with $fgeq 0$).
– anomaly
Aug 3 at 17:04
I don't see how that helps unfortunately. I don't see how I could estimate the latter.
– vaoy
Aug 3 at 17:36
I don't see how that helps unfortunately. I don't see how I could estimate the latter.
– vaoy
Aug 3 at 17:36
add a comment |Â
1 Answer
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up vote
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Since $mathbbP$ and $mathbbQ$ are equivalent probability measures, there exists $g in L^1(mathbbQ)$, $g geq 0$, such that $dmathbbP = g , dmathbbQ$. The integrability of $g$ implies that $g$ is uniformly integrable in the sense that for any $epsilon>0$ there exists $delta>0$ such that
$$A in mathcalF, mathbbQ(A) leq delta implies int_A g , dmathbbQ leq epsilon.tag1$$
Fix some $epsilon>0$ and choose $delta>0$ according to $(1)$. Since
$$sup_n in mathbbN mathbbQ(f_n>K) leq frac1K sup_n in mathbbN mathbbE_mathbbQ(f_n)$$
we can choose $K>0$ sufficiently large such that
$$sup_n in mathbbN mathbbQ(f_n>K) leq delta$$
which implies by $(1)$ that
$$sup_n in mathbbN mathbbP(f_n>K) = sup_n in mathbbN int_f_n>K g , dmathbbQ leq epsilon.$$
As $epsilon>0$ is arbitrary, this proves the assertion.
answered Aug 3 at 18:10
saz
72.7k552111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Since $mathbbP$ and $mathbbQ$ are equivalent probability measures, there exists $g in L^1(mathbbQ)$, $g geq 0$, such that $dmathbbP = g , dmathbbQ$. The integrability of $g$ implies that $g$ is uniformly integrable in the sense that for any $epsilon>0$ there exists $delta>0$ such that
$$A in mathcalF, mathbbQ(A) leq delta implies int_A g , dmathbbQ leq epsilon.tag1$$
Fix some $epsilon>0$ and choose $delta>0$ according to $(1)$. Since
$$sup_n in mathbbN mathbbQ(f_n>K) leq frac1K sup_n in mathbbN mathbbE_mathbbQ(f_n)$$
we can choose $K>0$ sufficiently large such that
$$sup_n in mathbbN mathbbQ(f_n>K) leq delta$$
which implies by $(1)$ that
$$sup_n in mathbbN mathbbP(f_n>K) = sup_n in mathbbN int_f_n>K g , dmathbbQ leq epsilon.$$
As $epsilon>0$ is arbitrary, this proves the assertion.
answered Aug 3 at 18:10
saz
72.7k552111
add a comment |Â
up vote
4
down vote
accepted
Since $mathbbP$ and $mathbbQ$ are equivalent probability measures, there exists $g in L^1(mathbbQ)$, $g geq 0$, such that $dmathbbP = g , dmathbbQ$. The integrability of $g$ implies that $g$ is uniformly integrable in the sense that for any $epsilon>0$ there exists $delta>0$ such that
$$A in mathcalF, mathbbQ(A) leq delta implies int_A g , dmathbbQ leq epsilon.tag1$$
Fix some $epsilon>0$ and choose $delta>0$ according to $(1)$. Since
$$sup_n in mathbbN mathbbQ(f_n>K) leq frac1K sup_n in mathbbN mathbbE_mathbbQ(f_n)$$
we can choose $K>0$ sufficiently large such that
$$sup_n in mathbbN mathbbQ(f_n>K) leq delta$$
which implies by $(1)$ that
$$sup_n in mathbbN mathbbP(f_n>K) = sup_n in mathbbN int_f_n>K g , dmathbbQ leq epsilon.$$
As $epsilon>0$ is arbitrary, this proves the assertion.
answered Aug 3 at 18:10
saz
72.7k552111
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Since $mathbbP$ and $mathbbQ$ are equivalent probability measures, there exists $g in L^1(mathbbQ)$, $g geq 0$, such that $dmathbbP = g , dmathbbQ$. The integrability of $g$ implies that $g$ is uniformly integrable in the sense that for any $epsilon>0$ there exists $delta>0$ such that
$$A in mathcalF, mathbbQ(A) leq delta implies int_A g , dmathbbQ leq epsilon.tag1$$
Fix some $epsilon>0$ and choose $delta>0$ according to $(1)$. Since
$$sup_n in mathbbN mathbbQ(f_n>K) leq frac1K sup_n in mathbbN mathbbE_mathbbQ(f_n)$$
we can choose $K>0$ sufficiently large such that
$$sup_n in mathbbN mathbbQ(f_n>K) leq delta$$
which implies by $(1)$ that
$$sup_n in mathbbN mathbbP(f_n>K) = sup_n in mathbbN int_f_n>K g , dmathbbQ leq epsilon.$$
As $epsilon>0$ is arbitrary, this proves the assertion.
answered Aug 3 at 18:10
saz
72.7k552111
Since $mathbbP$ and $mathbbQ$ are equivalent probability measures, there exists $g in L^1(mathbbQ)$, $g geq 0$, such that $dmathbbP = g , dmathbbQ$. The integrability of $g$ implies that $g$ is uniformly integrable in the sense that for any $epsilon>0$ there exists $delta>0$ such that
$$A in mathcalF, mathbbQ(A) leq delta implies int_A g , dmathbbQ leq epsilon.tag1$$
Fix some $epsilon>0$ and choose $delta>0$ according to $(1)$. Since
$$sup_n in mathbbN mathbbQ(f_n>K) leq frac1K sup_n in mathbbN mathbbE_mathbbQ(f_n)$$
we can choose $K>0$ sufficiently large such that
$$sup_n in mathbbN mathbbQ(f_n>K) leq delta$$
which implies by $(1)$ that
$$sup_n in mathbbN mathbbP(f_n>K) = sup_n in mathbbN int_f_n>K g , dmathbbQ leq epsilon.$$
As $epsilon>0$ is arbitrary, this proves the assertion.
answered Aug 3 at 18:10
saz
72.7k552111
edited Aug 3 at 19:32
answered Aug 3 at 18:10
saz
72.7k552111
answered Aug 3 at 18:10
saz
72.7k552111
answered Aug 3 at 18:10
saz
72.7k552111
72.7k552111
add a comment |Â
add a comment |Â
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Use the Radon-Nikodym derivative to rewrite the $L^p$ norm on $P$ in terms of the one on $Q$.
– anomaly
Aug 3 at 16:58
I tried that, but I only got this $$int f dP = int f fracdPdQ dQ$$ and this I cannot estimate, since I only have that $fracdQdP$ is in $L^infty(P)$. Or did you have something else in mind?
– vaoy
Aug 3 at 17:01
Estimate the latter using the fact that $int f geq int_f > K fgeq K, P(f > K)$ for any $K$ (with $fgeq 0$).
– anomaly
Aug 3 at 17:04
I don't see how that helps unfortunately. I don't see how I could estimate the latter.
– vaoy
Aug 3 at 17:36