Dear all.
I saw the following assertion.
“Suppose Xn (n>=1) is a sequence of non-negative random variables on a probability space (Ω,F,P) such that E[Xn]0 is a real constant .
Then the family (f’(Xn)*Xn*1{Xn>x0}) (n>=1) where 1{} is an index function is uniformly integrable.”
I tried to prove this assertion but failed.Please teach me how to prove.
1. Formulation of the problem requires additional assumption that $f'(x)$ exists at every $x\ge x_0$. Otherwise it may happen, that $X_n$ equals the "bad" value of $ x $ with positive probability so that $f'(X_n)$ would be undefined. Alternatively, one can assume that $f'(X_n) := f_-' (X_n)$, since the left derivatiwe for convex decreasing functions always exists and is finite except perhaps at the initial point. Then the sufficient assumption is that $f'(x_0)$ is finite.
2. Accordingly, below I am solving the problem with $f'(X_n)$ replaced by $g(X_n)$, where $ g$ is a positive decreasing function defined on $ [x_0, \infty)$, with $x_0\ge 0$ and $ \lim g(C) = 0$, as $C \to \infty$. Let us name
$ Y_n := g(X_n) X_n 1\{ X_n > x_0\} $
The integrals from the definition of uniform integrability of the sequence $Y_n$, $n=1,2,...$, can be estimated as follows, for C>x_0 :
$ \int_\{|Y_n \ge C\} \, Y_n \, dP
\le
\int_\{X_n \ge C/ g(x_0) \} \, g(C)\, X_n \, dP
\le
g(C) \, E\{ X_n} \to 0$ , as $ C \to \infty $
Hopefully, it helps.
Best regards, Joachim
1. Formulation of the problem requires additional assumption that $f'(x)$ exists at every $x\ge x_0$. Otherwise it may happen, that $X_n$ equals the "bad" value of $ x $ with positive probability so that $f'(X_n)$ would be undefined. Alternatively, one can assume that $f'(X_n) := f_-' (X_n)$, since the left derivatiwe for convex decreasing functions always exists and is finite except perhaps at the initial point. Then the sufficient assumption is that $f'(x_0)$ is finite.
2. Accordingly, below I am solving the problem with $f'(X_n)$ replaced by $g(X_n)$, where $ g$ is a positive decreasing function defined on $ [x_0, \infty)$, with $x_0\ge 0$ and $ \lim g(C) = 0$, as $C \to \infty$. Let us name
$ Y_n := g(X_n) X_n 1\{ X_n > x_0\} $
The integrals from the definition of uniform integrability of the sequence $Y_n$, $n=1,2,...$, can be estimated as follows, for C>x_0 :
$ \int_\{|Y_n \ge C\} \, Y_n \, dP
\le
\int_\{X_n \ge C/ g(x_0) \} \, g(C)\, X_n \, dP
\le
g(C) \, E\{ X_n} \to 0$ , as $ C \to \infty $
Hopefully, it helps.
Best regards, Joachim
Dear Prof. Domsta
Thanks a lot for your answer.
The assumptions of the original assertion include the continuous differentiability of the convex function as you pointed. It's my fault.
Now I understood your solution.
Best regards
Look at the work of Joe Diestel about Uniform Integrability. I attached the file.
Dear JorgeEliecer Hermandez Hermandez,
Lot of thanks for the file of Joe Diestel's paper., presenting so many facts related to the uniform integrability.
Best regards, Joachim Domsta
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