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If $Y_n=E(X|mathcalF_n)$ for some $mathcalF_n subset mathcalF$ and if $Y_n to Y$ with probability one. Show $Y_n to Y$ in $L^1$.

Converging exponentially in probability implies convergence with probability one?Coniditional expectation for bounded random variablesShowing Uniform Integrability of Random variablesCharacterization of $n^-1 max_1 le k le N Y_k to 0$ in probability for $Y_n$ i.i.d.$X_n to X$ almost surely and $Y_n-X_n to 0$ in probability implies $Y_n to X$ in probability?Convergence in quadratic mean and in meanJensen's inequality and conditional expectationIf $mathcalL(X_n)=mathcalL(Y_n)$ and $X_n to X$ in probability then does there exists a Y such that $Y_n to Y$Necessity of uniform integrability in martingale convergence theoremShow that $Y_n := (prod_i=1^n X_i)^1/n$ converges with probability 1

0

1

$begingroup$

If $Y_n=E(X|mathcalF_n)$ for some $mathcalF_n subset mathcalF$ and if $Y_n to Y$ with probability one. Show $Y_n to Y$ in $L^1$.

I am trying to show that $Y_n$ is uniformly integrable (UI). If it is known that $E|X|<infty$, then there is a theorem that tells us that there exists a convex function $phi$ such that $phi(x)/x to_x infty$ and $E[phi(|X|)]<infty$. This should help us show that $Y_n$ are UI with Jensen's inequality. But how should I show $E|X|<infty$ or is there other ways?

probability-theory convergence conditional-expectation

share|cite|improve this question

edited Mar 29 at 7:11

saz

82.1k862131

asked Mar 29 at 2:05

Daniel LiDaniel Li

787414

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Conditional expectation is usually defined for integrable random variables.
  $endgroup$
  – d.k.o.
  Mar 29 at 3:27
  

  
  
  
  

add a comment | 

0

1

$begingroup$

If $Y_n=E(X|mathcalF_n)$ for some $mathcalF_n subset mathcalF$ and if $Y_n to Y$ with probability one. Show $Y_n to Y$ in $L^1$.

I am trying to show that $Y_n$ is uniformly integrable (UI). If it is known that $E|X|<infty$, then there is a theorem that tells us that there exists a convex function $phi$ such that $phi(x)/x to_x infty$ and $E[phi(|X|)]<infty$. This should help us show that $Y_n$ are UI with Jensen's inequality. But how should I show $E|X|<infty$ or is there other ways?

probability-theory convergence conditional-expectation

share|cite|improve this question

edited Mar 29 at 7:11

saz

82.1k862131

asked Mar 29 at 2:05

Daniel LiDaniel Li

787414

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Conditional expectation is usually defined for integrable random variables.
  $endgroup$
  – d.k.o.
  Mar 29 at 3:27
  

  
  
  
  

add a comment | 

$begingroup$

If $Y_n=E(X|mathcalF_n)$ for some $mathcalF_n subset mathcalF$ and if $Y_n to Y$ with probability one. Show $Y_n to Y$ in $L^1$.

I am trying to show that $Y_n$ is uniformly integrable (UI). If it is known that $E|X|<infty$, then there is a theorem that tells us that there exists a convex function $phi$ such that $phi(x)/x to_x infty$ and $E[phi(|X|)]<infty$. This should help us show that $Y_n$ are UI with Jensen's inequality. But how should I show $E|X|<infty$ or is there other ways?

probability-theory convergence conditional-expectation

share|cite|improve this question

edited Mar 29 at 7:11

saz

82.1k862131

asked Mar 29 at 2:05

Daniel LiDaniel Li

787414

$endgroup$

share|cite|improve this question

edited Mar 29 at 7:11

saz

82.1k862131

asked Mar 29 at 2:05

Daniel LiDaniel Li

787414

share|cite|improve this question

edited Mar 29 at 7:11

saz

82.1k862131

asked Mar 29 at 2:05

Daniel LiDaniel Li

787414

edited Mar 29 at 7:11

saz

82.1k862131

edited Mar 29 at 7:11

saz

82.1k862131

asked Mar 29 at 2:05

Daniel LiDaniel Li

787414

asked Mar 29 at 2:05

Daniel LiDaniel Li

787414