Do non-negative submartingales with finite second moment converge almost surely?Do non-negative submartingales with bounded second moments converge almost surely?Confusions regarding the concept of a stopping time for a martingaleConvergence of expected values as random variables converge almost surelyConvergence of ExpectationsDoes this sequence converge almost surely or not?Show that $X_n/n$ does not converge almost surelyDoes this self-conjured RV converge almost surely?Existence of a sequence of L1 random variablesAlmost surely finite stopping time for a random walkweighted sequence of arbitrary sequence of random variables can converge to zero almost surelyMaking a sequence of random variables converge almost surely to $0$
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Do non-negative submartingales with finite second moment converge almost surely?
Do non-negative submartingales with bounded second moments converge almost surely?Confusions regarding the concept of a stopping time for a martingaleConvergence of expected values as random variables converge almost surelyConvergence of ExpectationsDoes this sequence converge almost surely or not?Show that $X_n/n$ does not converge almost surelyDoes this self-conjured RV converge almost surely?Existence of a sequence of L1 random variablesAlmost surely finite stopping time for a random walkweighted sequence of arbitrary sequence of random variables can converge to zero almost surelyMaking a sequence of random variables converge almost surely to $0$
$begingroup$
Suppose $X_n_n = 1^infty$ is a discrete-time submartingale (a sequence of random variables, such that $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$), such, that $forall n in mathbbN EX_n^2 < infty text and P(X_n > 0) = 1$. Is it true, that $P(exists lim_n to infty X_n) = 1$?
I know, that as $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$, and, by Jensen inequality for conditional expectations $P(E[X_n+1^2|X_1, … X_n] geq E^2[X_n+1|X_1, … X_n]) = 1$, we have that this question can be solved by proving one of the following statements:
$$P(exists lim_n to infty E[X_n+1|X_1, … X_n]) = 1$$
$$P(exists lim_n to infty E[X_n+1^2|X_1, … X_n]) = 1$$
However, they do not seem to be any easier to prove.
probability probability-theory convergence stochastic-processes martingales
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
$endgroup$
add a comment |
$begingroup$
Suppose $X_n_n = 1^infty$ is a discrete-time submartingale (a sequence of random variables, such that $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$), such, that $forall n in mathbbN EX_n^2 < infty text and P(X_n > 0) = 1$. Is it true, that $P(exists lim_n to infty X_n) = 1$?
I know, that as $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$, and, by Jensen inequality for conditional expectations $P(E[X_n+1^2|X_1, … X_n] geq E^2[X_n+1|X_1, … X_n]) = 1$, we have that this question can be solved by proving one of the following statements:
$$P(exists lim_n to infty E[X_n+1|X_1, … X_n]) = 1$$
$$P(exists lim_n to infty E[X_n+1^2|X_1, … X_n]) = 1$$
However, they do not seem to be any easier to prove.
probability probability-theory convergence stochastic-processes martingales
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
$endgroup$
add a comment |
$begingroup$
Suppose $X_n_n = 1^infty$ is a discrete-time submartingale (a sequence of random variables, such that $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$), such, that $forall n in mathbbN EX_n^2 < infty text and P(X_n > 0) = 1$. Is it true, that $P(exists lim_n to infty X_n) = 1$?
I know, that as $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$, and, by Jensen inequality for conditional expectations $P(E[X_n+1^2|X_1, … X_n] geq E^2[X_n+1|X_1, … X_n]) = 1$, we have that this question can be solved by proving one of the following statements:
$$P(exists lim_n to infty E[X_n+1|X_1, … X_n]) = 1$$
$$P(exists lim_n to infty E[X_n+1^2|X_1, … X_n]) = 1$$
However, they do not seem to be any easier to prove.
probability probability-theory convergence stochastic-processes martingales
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
$endgroup$
Suppose $X_n_n = 1^infty$ is a discrete-time submartingale (a sequence of random variables, such that $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$), such, that $forall n in mathbbN EX_n^2 < infty text and P(X_n > 0) = 1$. Is it true, that $P(exists lim_n to infty X_n) = 1$?
I know, that as $P(E[X_n+1|X_1, … X_n] geq X_n) = 1$, and, by Jensen inequality for conditional expectations $P(E[X_n+1^2|X_1, … X_n] geq E^2[X_n+1|X_1, … X_n]) = 1$, we have that this question can be solved by proving one of the following statements:
$$P(exists lim_n to infty E[X_n+1|X_1, … X_n]) = 1$$
$$P(exists lim_n to infty E[X_n+1^2|X_1, … X_n]) = 1$$
However, they do not seem to be any easier to prove.
probability probability-theory convergence stochastic-processes martingales
probability probability-theory convergence stochastic-processes martingales
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
edited Mar 16 at 9:27
Yanior Weg
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
asked Mar 16 at 8:42
Yanior WegYanior Weg
2,65211346
2,65211346
add a comment |
add a comment |
1 Answer
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$begingroup$
A simple counter example is $X_n=n$ for all $n$.
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
add a comment |
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$begingroup$
A simple counter example is $X_n=n$ for all $n$.
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
add a comment |
$begingroup$
A simple counter example is $X_n=n$ for all $n$.
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
add a comment |
$begingroup$
A simple counter example is $X_n=n$ for all $n$.
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
A simple counter example is $X_n=n$ for all $n$.
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
answered Mar 16 at 12:32
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
69.4k53170
add a comment |
add a comment |
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