POTW Convergence in Probability

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If a sequence of real random variables {X_n} converges in expectation to a limit μ and the variance of X_n approaches zero, then X_n converges to μ in probability. The proof relies on the properties of variance and the definition of convergence in probability. As the variance decreases, the distribution of X_n becomes increasingly concentrated around the mean μ. This indicates that for any ε > 0, the probability that X_n deviates from μ by more than ε approaches zero. Thus, the conditions provided ensure convergence in probability.
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Prove that if ##\{X_n\}_{n = 1}^\infty## is a sequence of real random variables on probability space ##(\Omega, \mathscr{F},\mathbb{P})## such that ##\lim_n \mathbb{E}[X_n] = \mu## and ##\lim_n \operatorname{Var}[X_n] = 0##, then ##X_n## converges to ##\mu## in probability.
 
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We're going to use
https://en.m.wikipedia.org/wiki/Chebyshev's_inequality

For any ##m##, there exists ##N## such that for ##n>N##, we have ##P(|X_n-\mu | > 1/m) < 1/m^2##, applying Chebyshev's inequality with ##\sigma < 1/m^3## and ##k=m##, and ##N## picked such that ##var(X_n) < 1/m^3## for ##n>N##.

That's pretty much it, since ##m## it's arbitrary.

For any ##\epsilon >0##, for any ##m## such that ##1/m<\epsilon##, and for ##n## large enough, we have ##P(|X_n-\epsilon) |<1/m##. Since ##m## it's arbitrary if we make ##n## big enough, in the limit as n goes to infinity this goes to zero. Hence all the probability weight must be on ##\mu## as desired.