Weak convergence with equilities instead of inequilities

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manpreet Tuteehub forum best answer Best Answer 2 years ago

There is a n">question in a syllabus called "Mathematische statistiek":

"If every XnXn and XX possess a discrete distribution supported on a finite set of integers, show that XnXnconverges in distribution to XX iff P(Xn=x)P(X=x)P(Xn=x)→P(X=x) for every xx."

This is what I tried: First assume the convergence holds. Let k1<k2<...<knk1n be all the integers XXcan reach. For iN{1}i∈N−{1} we have

P(Xnki)P(X
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manpreet 2 years ago

 

P(Xn=ki)=P(Xnki+0.1)P(Xnki0.1)P(Xki+0.1)P(Xki0.1)=P(X=ki)P(Xn=ki)=P(Xn≤ki+0.1)−P(Xn≤ki−0.1)→→P(X≤ki+0.1)−P(X≤ki−0.1)=P(X=ki)
since the unique integer value kiki belongs to the interval (ki0.1,ki+0.1](ki−0.1,ki+0.1]. The convergence takes place since CDF of XX is continuous at the points 
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