An Easy Strong Law - Weierstrass approximation theorem

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An Easy Strong Law - Weierstrass approximation theorem

Postby colman » Fri Dec 07, 2012 7:29 pm

Chapter 7 7.4
page 74

Could the following part of the proof be made clearer. In my case I don't follow the final manipulating of the equations (3rd to 4th say)

|B_{n}(p) - f(p)| <= E(Y_{n})
= E(Y_{n};Z_{n} <= \delta ) + E(Y_{n} ; Z_{n} > \delta)
<= 1/2 \epsilon P (Z_{n} <= \delta ) + 2KP(Z_{n} > \delta)
<= 1/2 \epsilon + 2K/(4n \delta^{2} ) :: where does the the 2K/(4n \delta^{2} ) come from?!
"Earlier, we chose a fixed \delta at (a). We now choose n so that
2K/(4n \delta^{2} ) < 1/2 \epsilon

...."
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Re: An Easy Strong Law - Weierstrass approximation theorem

Postby mj » Mon Dec 10, 2012 2:39 am

isn't this just the last line of section 7.3?
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Re: An Easy Strong Law - Weierstrass approximation theorem

Postby colman » Wed Dec 12, 2012 1:48 pm

Well yes, it does come about from the workings... and that refers to Thm 7.1, however its in thm 7.2 (Strong law) I see the first signs of the, lets say of a 4 appearing "... note that it imposes a 'finite fourth moment' condition..". I suspect it is from here that it unravels itself to produce:
np(1 - p) <= n/4

In this case my question would become, how does it happen that E(X^{4}) <= K, what so particular about X^4 v's X^3?
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Re: An Easy Strong Law - Weierstrass approximation theorem

Postby mj » Mon Dec 17, 2012 5:34 am

i think the issue here is that we get a bound on variances of variances.
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Re: An Easy Strong Law - Weierstrass approximation theorem

Postby colman » Tue Dec 18, 2012 1:06 pm

Alright, thanks for the reply, I'll take it from the top again keeping these words in mind.
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Re: An Easy Strong Law - Weierstrass approximation theorem

Postby maliha » Tue Jan 07, 2014 6:56 am

I had a question. If random variables X and Y are independent of sigma field \Omega, does that imply that X-Y is also independent of \Omega. More generally is f(X,Y) independent of sigma field \Omega if X and Y are both independent of \Omega?
Intuitively I think it should be the case, but I am unable to prove it.
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Re: An Easy Strong Law - Weierstrass approximation theorem

Postby colman » Mon May 05, 2014 11:04 am

Your asking if a new set S:= f(X,Y) stays independent.
As long as the conditions on 'f', preserve the sigma algebra requirments, the new set should also be indepentent.
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Re: An Easy Strong Law - Weierstrass approximation theorem

Postby Minicar » Thu Apr 19, 2018 2:40 am

As a good comment, I am glad to have come to this forum.
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