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I am currently reading a book, where the author has shown that for a sequence of random variables $(Y_n)_{n \in \mathbb{N}}$ $(Y_n \geq 0 $ for all n) and $d<\infty$, it holds.

if: lim $sup_{n \in \mathbb{N}} \mathbb{E}[Y_n] \geq d$ and lim $inf_{n \in \mathbb{N}} Y_n \leq d$
then $Y_n$ converges in Probability to $d$

Unfortunetaly i have very little experience in proving convergence in probability. My first guess was to use Markov inequality, but so far it didnt yield any results.

StefanWK
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    Not sure if this is true as-stated. Take $d = 1$, $Y_n$ IID so that $\mathbb P(Y_n = 0) = \mathbb P(Y_n = 2) = .5$. We have $\mathbb E(Y_n) \equiv d$, while $\liminf Y_n = 0 \leq d$ almost surely, but $|Y_n - 1| \equiv 1$ for all $n$, and so there is no hope that $Y_n$ converges in probability to $1$ (or to any value for that matter). – A Blumenthal Dec 26 '17 at 18:16

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