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I just want to understand why tightness condition implies that the measure sequence would admit a converging subsequence. I wonder if someone can explain to me how Prokhorov's theorem is proved intuitively without too much analysis formalism. Why not being able to escape to infinity warrants convergence?

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    $\begingroup$ Kindly state the version of Prokhorov's theorem that you know, so that others can begin their answer without confusion of notation etc. Also, it helps (but is not necessary) if you've seen a formal proof of Prokhorov's theorem. $\endgroup$ – Teresa Lisbon Sep 17 at 3:58
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I always thought the example given in Billingsley 's book was illustrative as to why tightness is needed for this result. Consider the sequence $F_n$ of distribution functions (or the associated probability measures on $\mathbb{R}$) corresponding to $Unif(-n,n)$ random variables. In this case every subsequence converges to the limiting "extended distribution function" $F_\infty(x)=1/2$. Notice that this CDF does not correspond to a probability measure on $\mathbb{R}$, and intuitively why this is occurring is that the mass of the distributions of $F_n$ is escaping to infinity. In general it can be shown that every sequence of CDF's must have a subsequence converging to some sort of "extended distribution function" like the example above (this is usually called Helly's selection theorem). Tightness it turns out is a necessary and sufficient condition to imply that this extended distribution function corresponds to a probability measure. See the below slides regarding Helly's theorem.

https://faculty.math.illinois.edu/~psdey/MATH561SP19/week10.pdf

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