Monotone convergence theorem of random variables and its example Announcing the arrival of...
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Monotone convergence theorem of random variables and its example
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Convergence of random variables (Durrett: Probability Theory and Examples)Supremum of Expectation of a Sequence of Random VariablesAn inequality in a step in the proof of monotone convergence for random variablesAlmost sure convergence of random variablesconvergence of expectation sum of infinite random variablesProperty derived form Monotone Convergence TheoremMoment generating function and convergent random variablesExample of sequence of random variables, that almost surely converge but,but doesn't converge in quadratic mean.Convergence results for conditional random random variablesLebesgue convergence theorem extension (random variables)
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Monotone Convergence Theorem of random variables is stated as below:
Assume that there is a sequence of random variables (r.v.) satisfing $0leq X_1leq X_2 leq cdots$ and $X_nto X$ a.s.(almost surely), then
$$E[X_n]to E[X]$$
where E is expectation.
I read about a example using the Monotone Convergence Theorem on the text book and found some problems. The example is stated as follows:
Assume r.v. X is non-negative, and denote $mu=EX$, define sequence of r.v.:
$$X_n=min{X,n},nin mathbb{N}$$
Then as $ntoinfty$, $EX_n=mu$ because $X_n$ is monotone.
My question is, why they say $X_n$ is monotone when there are posotive probability of $X_{k-1}>X_{k}$? For example, as $X_n$ are independent,
$$P(X_{k-1}= k-1, X_k<k-1)=P(X_{k-1}= k-1)P(X_k<k-1)$$
$$=P(Xgeq k-1)P( X<k-1)>0.$$
real-analysis probability-theory
asked Mar 25 at 9:48
ZishuoZishuo
163
$endgroup$
add a comment |
$begingroup$
Monotone Convergence Theorem of random variables is stated as below:
Assume that there is a sequence of random variables (r.v.) satisfing $0leq X_1leq X_2 leq cdots$ and $X_nto X$ a.s.(almost surely), then
$$E[X_n]to E[X]$$
where E is expectation.
I read about a example using the Monotone Convergence Theorem on the text book and found some problems. The example is stated as follows:
Assume r.v. X is non-negative, and denote $mu=EX$, define sequence of r.v.:
$$X_n=min{X,n},nin mathbb{N}$$
Then as $ntoinfty$, $EX_n=mu$ because $X_n$ is monotone.
My question is, why they say $X_n$ is monotone when there are posotive probability of $X_{k-1}>X_{k}$? For example, as $X_n$ are independent,
$$P(X_{k-1}= k-1, X_k<k-1)=P(X_{k-1}= k-1)P(X_k<k-1)$$
$$=P(Xgeq k-1)P( X<k-1)>0.$$
real-analysis probability-theory
asked Mar 25 at 9:48
ZishuoZishuo
163
$endgroup$
add a comment |
$begingroup$
Monotone Convergence Theorem of random variables is stated as below:
Assume that there is a sequence of random variables (r.v.) satisfing $0leq X_1leq X_2 leq cdots$ and $X_nto X$ a.s.(almost surely), then
$$E[X_n]to E[X]$$
where E is expectation.
I read about a example using the Monotone Convergence Theorem on the text book and found some problems. The example is stated as follows:
Assume r.v. X is non-negative, and denote $mu=EX$, define sequence of r.v.:
$$X_n=min{X,n},nin mathbb{N}$$
Then as $ntoinfty$, $EX_n=mu$ because $X_n$ is monotone.
My question is, why they say $X_n$ is monotone when there are posotive probability of $X_{k-1}>X_{k}$? For example, as $X_n$ are independent,
$$P(X_{k-1}= k-1, X_k<k-1)=P(X_{k-1}= k-1)P(X_k<k-1)$$
$$=P(Xgeq k-1)P( X<k-1)>0.$$
real-analysis probability-theory
asked Mar 25 at 9:48
ZishuoZishuo
163
$endgroup$
Monotone Convergence Theorem of random variables is stated as below:
Assume that there is a sequence of random variables (r.v.) satisfing $0leq X_1leq X_2 leq cdots$ and $X_nto X$ a.s.(almost surely), then
$$E[X_n]to E[X]$$
where E is expectation.
I read about a example using the Monotone Convergence Theorem on the text book and found some problems. The example is stated as follows:
Assume r.v. X is non-negative, and denote $mu=EX$, define sequence of r.v.:
$$X_n=min{X,n},nin mathbb{N}$$
Then as $ntoinfty$, $EX_n=mu$ because $X_n$ is monotone.
My question is, why they say $X_n$ is monotone when there are posotive probability of $X_{k-1}>X_{k}$? For example, as $X_n$ are independent,
$$P(X_{k-1}= k-1, X_k<k-1)=P(X_{k-1}= k-1)P(X_k<k-1)$$
$$=P(Xgeq k-1)P( X<k-1)>0.$$
real-analysis probability-theory
real-analysis probability-theory
asked Mar 25 at 9:48
ZishuoZishuo
163
asked Mar 25 at 9:48
ZishuoZishuo
163
asked Mar 25 at 9:48
ZishuoZishuo
163
asked Mar 25 at 9:48
ZishuoZishuo
163
asked Mar 25 at 9:48
ZishuoZishuo
163
163
add a comment |
add a comment |
1 Answer
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$begingroup$
The problem is at the understanding of the example
The example means "for the same r.v. $X$" rather than "for a sequence of i.i.d. r.v. X", which means the $X_n$ are not indepenent. For example, when $X(omega)=x$.
Those $X_n$ are $X_n=n, n< x$ and $X_n=x, ngeq x$. Obvious those $X_n$ are monotone non-decreasing and the results is correct.
answered Mar 25 at 9:48
ZishuoZishuo
163
$endgroup$
$begingroup$
In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(Xgeq k-1,X<K-1)=0$.
$endgroup$
– Zishuo
Mar 25 at 9:50
add a comment |
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$begingroup$
The problem is at the understanding of the example
The example means "for the same r.v. $X$" rather than "for a sequence of i.i.d. r.v. X", which means the $X_n$ are not indepenent. For example, when $X(omega)=x$.
Those $X_n$ are $X_n=n, n< x$ and $X_n=x, ngeq x$. Obvious those $X_n$ are monotone non-decreasing and the results is correct.
answered Mar 25 at 9:48
ZishuoZishuo
163
$endgroup$
$begingroup$
In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(Xgeq k-1,X<K-1)=0$.
$endgroup$
– Zishuo
Mar 25 at 9:50
add a comment |
$begingroup$
The problem is at the understanding of the example
The example means "for the same r.v. $X$" rather than "for a sequence of i.i.d. r.v. X", which means the $X_n$ are not indepenent. For example, when $X(omega)=x$.
Those $X_n$ are $X_n=n, n< x$ and $X_n=x, ngeq x$. Obvious those $X_n$ are monotone non-decreasing and the results is correct.
answered Mar 25 at 9:48
ZishuoZishuo
163
$endgroup$
$begingroup$
In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(Xgeq k-1,X<K-1)=0$.
$endgroup$
– Zishuo
Mar 25 at 9:50
add a comment |
$begingroup$
The problem is at the understanding of the example
The example means "for the same r.v. $X$" rather than "for a sequence of i.i.d. r.v. X", which means the $X_n$ are not indepenent. For example, when $X(omega)=x$.
Those $X_n$ are $X_n=n, n< x$ and $X_n=x, ngeq x$. Obvious those $X_n$ are monotone non-decreasing and the results is correct.
answered Mar 25 at 9:48
ZishuoZishuo
163
$endgroup$
The problem is at the understanding of the example
The example means "for the same r.v. $X$" rather than "for a sequence of i.i.d. r.v. X", which means the $X_n$ are not indepenent. For example, when $X(omega)=x$.
Those $X_n$ are $X_n=n, n< x$ and $X_n=x, ngeq x$. Obvious those $X_n$ are monotone non-decreasing and the results is correct.
answered Mar 25 at 9:48
ZishuoZishuo
163
answered Mar 25 at 9:48
ZishuoZishuo
163
answered Mar 25 at 9:48
ZishuoZishuo
163
answered Mar 25 at 9:48
ZishuoZishuo
163
163
$begingroup$
In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(Xgeq k-1,X<K-1)=0$.
$endgroup$
– Zishuo
Mar 25 at 9:50
add a comment |
$begingroup$
In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(Xgeq k-1,X<K-1)=0$.
$endgroup$
– Zishuo
Mar 25 at 9:50
$begingroup$
In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(Xgeq k-1,X<K-1)=0$.
$endgroup$
– Zishuo
Mar 25 at 9:50
$begingroup$
In this case, $P(X_{k-1}=k-1,X_k<k-1)=P(Xgeq k-1,X<K-1)=0$.
$endgroup$
– Zishuo
Mar 25 at 9:50
add a comment |
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