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Is this claim true? The following is my attempt at the proof. I am unsure about the proof because I did not have to use the fact that $X_n\geq 0, \forall n\in \mathbb{N}$. Any feedback for confirmation will be greatly appreciated.

Claim: Suppose the sequence of nonnegative random variables, $X_n\geq 0$ satisfies $E[x_{n+1}]\leq E[X_n]r$, $\forall n\in \mathbb{N}$, where $0<r<1$.Then, $X_n$ converges to zero almost surely.

Proof: Given inequality implies $E[X_n]\leq E[X_0]r^n$. For any $\epsilon>0$, Markov inequality yields $P(\{X_n\geq \epsilon\})\leq \cfrac{E[X_n]}{\epsilon}\leq \cfrac{E[X_0]r^n}{\epsilon}$. Thus, $\sum_{n=0}^\infty P(\{X_n\geq \epsilon\})\leq \sum_{n=0}^\infty \cfrac{E[X_0]r^n}{\epsilon}=\cfrac{E[X_0]}{\epsilon}\cfrac{1}{1-r}<\infty$. Therefore by Borel Canteli Lemma, $X_n$ converges to zero almost surely.

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  • $\begingroup$ Where is your problem ? $\endgroup$
    – TheBridge
    Commented Mar 26 at 15:29
  • $\begingroup$ @TheBridge Would the claim still be true if I took away the assumption $X_n\geq 0$? My problem is mainly self-confidence, I guess. I am new to probability and I am teaching it myself. $\endgroup$
    – curiosity
    Commented Mar 26 at 15:32
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    $\begingroup$ You use the $X_n \ge 0$ condition when applying Markov's inequality. $\endgroup$
    – angryavian
    Commented Mar 26 at 15:38
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    $\begingroup$ Sorry I read your question too fast. To answer your question it is not true in general take the trivial example $X_n = -n $, then for any $0<r<1 $ your inequality is true as $-(n+1) < -n$ but $Lim X_n = -\infty$ so it's not 0. You need somehow to have a lower bound of some form on $X_n$ to prevent this. $\endgroup$
    – TheBridge
    Commented Mar 26 at 15:44
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    $\begingroup$ If you let $X$ be any nonnegative random variable with $E[X]=\infty$ and define $X_n=X$ for all $n$, then arguably this is a counter-example because $E[X_{n+1}]\leq E[X_n]r$ reduces to $\infty\leq \infty$. You just need to assume finite expectations, particularly for $E[X_1]$. $\endgroup$
    – Michael
    Commented Mar 26 at 18:24

1 Answer 1

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Assuming finiteness as mentioned in the comments $E[X_{1}]$. By the decreasing we get $\lim_{n}E[X_{n}]=0$. By Fatou's lemma for $X:=\liminf_{n}X_{n}\geq 0$ we get

$$E[X]\leq \liminf_{n}E[X_{n}]=0.$$

So we get $X=0$.

Your approach also works too because it shows that for every $\omega$ there exists $N(\omega)$ so that

$$0\leq X\leq \epsilon.$$

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  • $\begingroup$ Thank you so much. $\endgroup$
    – curiosity
    Commented Mar 28 at 17:01

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