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LetZn, n1be a sequence of random variables andca constant such that for each

>0,P|Znc|>0asnq. Show that for any bounded continuous functiong,

E[g(Zn)]g(c)asnq.

Short Answer

Expert verified

SplitEgZn into two parts: wherex is nearly close toc and where xis far fromc and use assumptions to obtain the required.

Step by step solution

01

Given Information.

Zn,n1,be a sequence of random variables and ca constant such that for each

>0,P|Znc|>0asnq.

02

Explanation.

We are given thatgis a continuous function, which means that for every cand>0there exists role="math" localid="1649851182951" >0such that

|x-c||g(x)-g(c)|

Secondly, we are given thatgis a bounded function which means that there existsB>0such that

|g(x)|B

for everyx. Using the theorem about the expected value of a function of a random variable, we have that

EgZn=g(x)dPZn=|x-c|g(x)dPZn+|x-c|>g(x)dPZn

for some fixedc. Now, for xsuch that |x-c|we have that g(x)g(c)+and forxsuch that|x-c|>we have thatg(x)B. Therefore

EgZn(g(c)+)|x-c|dPZn+B|x-c|>dPZn=(g(c)+)PZn-c+BPZn-c>

Now, apply lim supnto both sides and use the assumption that PZn-c>0asn. It yields that PZn-c1asn. Therefore, we end up with

limsupnEgZng(c)+

03

Explanation.

On the other hand, we can make lower boundary onEgZn. Observe that for that same we have that for|x-c|holds g(x)g(c)-Also, for every xwe havelocalid="1649852924731" -Bg(x). Therefore, similarly to previous, we have that

EgZn(g(c)-)|x-c|dPZn-B|x-c|>dPZn

=(g(c)-)PZn-c-BPZn-c>

Applying limit inferior to both sides, we have that

liminfnEgZng(c)-

because ofPZn-c>0asn. Finally, we have that

g(c)-liminfnEgZnlimsupnEgZng(c)+

and since>0was arbitrary, we can get that 0to finally obtain that

limnEgZn=g(c)

so we have proved the claim.

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