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Convergence in Probability 
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#1
Jan2613, 09:50 AM

P: 247

I was a bit confused with the pages that I attached...
1) "An intuitive estimate of [itex]\theta[/itex] is the maximum of the sample". But we are only taking random samples, so even the maximum might be far from [itex]\theta[/itex], right? 2) I don't understand how [itex]E(Y_n) = (n/(n+1))\theta[/itex]. I thought that [itex]E(Y_n) = (Y_n)*pdf = (Y_n)(\frac{nt^n1}{\theta^n})[/itex]. 3) "Further, based on the cdf of Y_n, it is easily seen that [itex]Y_n \rightarrow \theta[/itex]". Does that mean that E(Y_n) converges to theta, so Y_n must also converge to theta? Thank you in advance 


#2
Jan2613, 10:48 AM

Sci Advisor
P: 3,297

Some of those questions are explained in this thread:
http://www.physicsforums.com/showthread.php?t=380389 


#3
Jan2613, 01:35 PM

P: 247




#4
Jan2613, 03:25 PM

Sci Advisor
P: 3,297

Convergence in Probability



#5
Jan2613, 03:26 PM

Sci Advisor
P: 6,076

For you third question, define Y_{n}. 


#6
Jan2613, 03:31 PM

P: 247




#7
Jan2813, 02:43 PM

Sci Advisor
P: 6,076

Since E(Y_{n}) > θ and θ is the maximum of the distribution, the probability that lim Y_{n} is < θ must be 0.
The proof is straightforward. Let A be the event that the limit is < θ, then: E(Y_{n}) = E(Y_{n}A)P(A) + E(Y_{n}A')P(A') > θ only if P(A) = 0. 


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