Lebesgue convergence theorem extension (random variables)Martingale convergence almost surelyLebesgue Convergence TheoremNon-negativity in Fatou's Lemma and Lebesgue Dominated Convergence TheoremAbout Vitali Convergence Theorem and Uniform IntegrabilityProve the monotone convergence theorem for sequences of Lebesgue-integrable functionsTheoretical Advantages of Lebesgue IntegrationFunctions involving expectations - The dominated/monotone convergence theoremAn counterexample for the monotone convergence theorem and dominated convergence theoremConvergence of Riemann integralUsing the Lebesgue dominated convergence theorem
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Lebesgue convergence theorem extension (random variables)
Martingale convergence almost surelyLebesgue Convergence TheoremNon-negativity in Fatou's Lemma and Lebesgue Dominated Convergence TheoremAbout Vitali Convergence Theorem and Uniform IntegrabilityProve the monotone convergence theorem for sequences of Lebesgue-integrable functionsTheoretical Advantages of Lebesgue IntegrationFunctions involving expectations - The dominated/monotone convergence theoremAn counterexample for the monotone convergence theorem and dominated convergence theoremConvergence of Riemann integralUsing the Lebesgue dominated convergence theorem
$begingroup$
Let $Z$ be a lower semi integrable r.v. and $W_n$ a sequence of bounded r.v increasing and dominated by another bounded r.v from above $W_n leq W$, why is true that $E[ZW_n]$ converges to $E[ZW]$?
The negative part $Z^-$ is taken care of by the standard Lebesgue convergence theorem, but the positive part $Z^+W_n$ I don't know how to make it converge since it may be negative.
This was used in the book of Paolo Baldi: Stochastic Calculus page 88. He uses the monotone class theorem to prove that to find the conditional expectation it is sufficient to check for a class generating the sigma algebra, stable with respect to finite intersections.
Thanks.
Taking the lebesgue measure in $[0,1]$, $lambda$, $Z=dfrac1x$, $W_n = -dfrac1n$, one can see that the integral sequence doesn't even exist.
real-analysis probability-theory convergence lebesgue-integral
asked Mar 20 at 18:37
lucmobzlucmobz
404
$endgroup$
add a comment |
$begingroup$
Let $Z$ be a lower semi integrable r.v. and $W_n$ a sequence of bounded r.v increasing and dominated by another bounded r.v from above $W_n leq W$, why is true that $E[ZW_n]$ converges to $E[ZW]$?
The negative part $Z^-$ is taken care of by the standard Lebesgue convergence theorem, but the positive part $Z^+W_n$ I don't know how to make it converge since it may be negative.
This was used in the book of Paolo Baldi: Stochastic Calculus page 88. He uses the monotone class theorem to prove that to find the conditional expectation it is sufficient to check for a class generating the sigma algebra, stable with respect to finite intersections.
Thanks.
Taking the lebesgue measure in $[0,1]$, $lambda$, $Z=dfrac1x$, $W_n = -dfrac1n$, one can see that the integral sequence doesn't even exist.
real-analysis probability-theory convergence lebesgue-integral
asked Mar 20 at 18:37
lucmobzlucmobz
404
$endgroup$
add a comment |
$begingroup$
Let $Z$ be a lower semi integrable r.v. and $W_n$ a sequence of bounded r.v increasing and dominated by another bounded r.v from above $W_n leq W$, why is true that $E[ZW_n]$ converges to $E[ZW]$?
The negative part $Z^-$ is taken care of by the standard Lebesgue convergence theorem, but the positive part $Z^+W_n$ I don't know how to make it converge since it may be negative.
This was used in the book of Paolo Baldi: Stochastic Calculus page 88. He uses the monotone class theorem to prove that to find the conditional expectation it is sufficient to check for a class generating the sigma algebra, stable with respect to finite intersections.
Thanks.
Taking the lebesgue measure in $[0,1]$, $lambda$, $Z=dfrac1x$, $W_n = -dfrac1n$, one can see that the integral sequence doesn't even exist.
real-analysis probability-theory convergence lebesgue-integral
asked Mar 20 at 18:37
lucmobzlucmobz
404
$endgroup$
Let $Z$ be a lower semi integrable r.v. and $W_n$ a sequence of bounded r.v increasing and dominated by another bounded r.v from above $W_n leq W$, why is true that $E[ZW_n]$ converges to $E[ZW]$?
The negative part $Z^-$ is taken care of by the standard Lebesgue convergence theorem, but the positive part $Z^+W_n$ I don't know how to make it converge since it may be negative.
This was used in the book of Paolo Baldi: Stochastic Calculus page 88. He uses the monotone class theorem to prove that to find the conditional expectation it is sufficient to check for a class generating the sigma algebra, stable with respect to finite intersections.
Thanks.
Taking the lebesgue measure in $[0,1]$, $lambda$, $Z=dfrac1x$, $W_n = -dfrac1n$, one can see that the integral sequence doesn't even exist.
real-analysis probability-theory convergence lebesgue-integral
real-analysis probability-theory convergence lebesgue-integral
asked Mar 20 at 18:37
lucmobzlucmobz
404
asked Mar 20 at 18:37
lucmobzlucmobz
404
edited Mar 20 at 18:55
lucmobz
asked Mar 20 at 18:37
lucmobzlucmobz
404
asked Mar 20 at 18:37
lucmobzlucmobz
404
asked Mar 20 at 18:37
lucmobzlucmobz
404
404
add a comment |
add a comment |
1 Answer
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$begingroup$
It seems that Remark 4.2 refers to integrable random variables $X$ and $Z$. Proceeding with your example, where $([0,1],mathcalB_[0,1],lambda)$ is the prob. space and $Z(omega)=omega^-1times 1omega>0$, take $W_n(omega)=-1_[0,n^-1)(omega)$. Then $Z$ is l.s.i., $-1le W_nle 0$, and $W_nnearrow 0$. However,
$$
0=mathsfEWZnelim_ntoinftymathsfEW_nZ=-infty.
$$
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
For the same reason one can only extend the $E[ZW]=E[XW]$ property to all bounded measurable functions when $X$ is integrable, and also when he talks about the expectation function given a random variable he assumes integrability when he says "for all bounded (l.s.i. is just bounded positive) measurable"? Thanks for clarifying, the book was a bit unclear there.
$endgroup$
– lucmobz
Mar 20 at 22:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It seems that Remark 4.2 refers to integrable random variables $X$ and $Z$. Proceeding with your example, where $([0,1],mathcalB_[0,1],lambda)$ is the prob. space and $Z(omega)=omega^-1times 1omega>0$, take $W_n(omega)=-1_[0,n^-1)(omega)$. Then $Z$ is l.s.i., $-1le W_nle 0$, and $W_nnearrow 0$. However,
$$
0=mathsfEWZnelim_ntoinftymathsfEW_nZ=-infty.
$$
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
For the same reason one can only extend the $E[ZW]=E[XW]$ property to all bounded measurable functions when $X$ is integrable, and also when he talks about the expectation function given a random variable he assumes integrability when he says "for all bounded (l.s.i. is just bounded positive) measurable"? Thanks for clarifying, the book was a bit unclear there.
$endgroup$
– lucmobz
Mar 20 at 22:31
add a comment |
$begingroup$
It seems that Remark 4.2 refers to integrable random variables $X$ and $Z$. Proceeding with your example, where $([0,1],mathcalB_[0,1],lambda)$ is the prob. space and $Z(omega)=omega^-1times 1omega>0$, take $W_n(omega)=-1_[0,n^-1)(omega)$. Then $Z$ is l.s.i., $-1le W_nle 0$, and $W_nnearrow 0$. However,
$$
0=mathsfEWZnelim_ntoinftymathsfEW_nZ=-infty.
$$
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
For the same reason one can only extend the $E[ZW]=E[XW]$ property to all bounded measurable functions when $X$ is integrable, and also when he talks about the expectation function given a random variable he assumes integrability when he says "for all bounded (l.s.i. is just bounded positive) measurable"? Thanks for clarifying, the book was a bit unclear there.
$endgroup$
– lucmobz
Mar 20 at 22:31
add a comment |
$begingroup$
It seems that Remark 4.2 refers to integrable random variables $X$ and $Z$. Proceeding with your example, where $([0,1],mathcalB_[0,1],lambda)$ is the prob. space and $Z(omega)=omega^-1times 1omega>0$, take $W_n(omega)=-1_[0,n^-1)(omega)$. Then $Z$ is l.s.i., $-1le W_nle 0$, and $W_nnearrow 0$. However,
$$
0=mathsfEWZnelim_ntoinftymathsfEW_nZ=-infty.
$$
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
$endgroup$
It seems that Remark 4.2 refers to integrable random variables $X$ and $Z$. Proceeding with your example, where $([0,1],mathcalB_[0,1],lambda)$ is the prob. space and $Z(omega)=omega^-1times 1omega>0$, take $W_n(omega)=-1_[0,n^-1)(omega)$. Then $Z$ is l.s.i., $-1le W_nle 0$, and $W_nnearrow 0$. However,
$$
0=mathsfEWZnelim_ntoinftymathsfEW_nZ=-infty.
$$
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
answered Mar 20 at 21:12
d.k.o.d.k.o.
10.5k630
10.5k630
$begingroup$
For the same reason one can only extend the $E[ZW]=E[XW]$ property to all bounded measurable functions when $X$ is integrable, and also when he talks about the expectation function given a random variable he assumes integrability when he says "for all bounded (l.s.i. is just bounded positive) measurable"? Thanks for clarifying, the book was a bit unclear there.
$endgroup$
– lucmobz
Mar 20 at 22:31
add a comment |
$begingroup$
For the same reason one can only extend the $E[ZW]=E[XW]$ property to all bounded measurable functions when $X$ is integrable, and also when he talks about the expectation function given a random variable he assumes integrability when he says "for all bounded (l.s.i. is just bounded positive) measurable"? Thanks for clarifying, the book was a bit unclear there.
$endgroup$
– lucmobz
Mar 20 at 22:31
$begingroup$
For the same reason one can only extend the $E[ZW]=E[XW]$ property to all bounded measurable functions when $X$ is integrable, and also when he talks about the expectation function given a random variable he assumes integrability when he says "for all bounded (l.s.i. is just bounded positive) measurable"? Thanks for clarifying, the book was a bit unclear there.
$endgroup$
– lucmobz
Mar 20 at 22:31
$begingroup$
For the same reason one can only extend the $E[ZW]=E[XW]$ property to all bounded measurable functions when $X$ is integrable, and also when he talks about the expectation function given a random variable he assumes integrability when he says "for all bounded (l.s.i. is just bounded positive) measurable"? Thanks for clarifying, the book was a bit unclear there.
$endgroup$
– lucmobz
Mar 20 at 22:31
add a comment |
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