Wikipedia:Reference desk/Archives/Mathematics/2013 February 12

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February 12 edit

Cosets and normal subgroup equality edit

Hello. I recently came across an exercise, and would be very grateful if someone could help.

Given a normal subgroup H in G, with g in G, if  , then is  ?

Neuroxic (talk) 11:31, 12 February 2013 (UTC)[reply]

Any subgroup H contains the identity. Think what happens with that. Dmcq (talk) 12:11, 12 February 2013 (UTC)[reply]
That question is a special case of the following question, which may actually be easier (intuitively) to solve: if  , then is  ? Basically you're just asking "Is   for some subgroup  ?" and it is a very well known fact that the answer is yes, and the proof is also very well known and straightforward:   --AnalysisAlgebra (talk) 19:21, 12 February 2013 (UTC)[reply]
There's no need for   to be normal, by the way. --AnalysisAlgebra (talk) 19:23, 12 February 2013 (UTC)[reply]

Estimating parameters of a distribution from censored data edit

Suppose you hypothesize that your data are drawn from, say, a lognormal distribution or a gamma distribution. Your data are right-censored at the value x*. That is, you have exact data whenever xx*, but whenever x>x* all you know is that x>x*.

(1) How do you estimate the parameters of the hypothesized distribution? I know it would be by maximum likelihood, but how do you handle the censored data?

(2) Is there a command in, say, SAS that will do this?

(3) How do you test the hypothesis that the data did in fact come from that distribution? Duoduoduo (talk) 18:37, 12 February 2013 (UTC)[reply]

(1) If your data consists of uncensored points   and n censored points, and your pdf with parameters   is  , then the likelihood of   is  . There's ostensibly an inconsistency in that you use a density for the contribution to likelihood of some points and a probability for others, but since likelihood functions are only meaningful up to a constant factor it doesn't matter. -- Meni Rosenfeld (talk) 19:49, 12 February 2013 (UTC)[reply]
Thanks, Meni! Duoduoduo (talk) 14:19, 14 February 2013 (UTC)[reply]