Re: Cauchy PDF + Parameter Estimate
On 25 Feb 2002 07:56:56 -0800, [EMAIL PROTECTED] (kjetil halvorsen) wrote: It isstraightforward tlo write down the loglikelihood, and then whatever optimization routine (there must be one in Matlab) will help you! Just be careful when searching, because Cauchy likelihoods are frequently multi-modal. Duncan Murdoch = Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at http://jse.stat.ncsu.edu/ =
Re: Estimating priors for Bayesian analysis
On 4 May 2001 04:11:23 -0700, [EMAIL PROTECTED] (Will Hopkins) wrote: For example, I might believe that the individual's true score is 70 units, and that the likely range is +/- 10 units. So what describes likely? 90%, 95%, 99%...? Do Bayesians have any validated way to work that out? If they don't, then the whole Bayesian edifice might just come crashing down. I put this to a Bayesian who has been helping me, but I have received no reply from him since I sent the message, so I suspect the worst. It's hard to get the value very accurately, but the appropriate way is to look at betting behaviour. If you're claiming a 50% belief that the value is between 60 and 80, then you should be indifferent to accepting a bet at even odds of the value being inside or outside that interval. For other levels, it would be a bet with different odds: e.g. you'd be willing to offer or accept 9 to 1 odds that it is in your 90% interval. Of course, there are other psychological things affecting the decision to accept or reject such a bet (can I afford to lose? Is winning worth the trouble of thinking about it? Do I think gambling is immoral? etc.), but the idea of indifference to each alternative is the key idea. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: The meaning of the p value
On 2 Feb 2001 01:12:59 -0800, [EMAIL PROTECTED] (Will Hopkins) wrote: I've been involved in off-list discussion with Duncan Murdoch. At one stage there I was about to retire in disgrace. But sighs of relief... his objection is Bayesian. Just to clarify, I don't think this is a valid summary of what I said. What I said offline was just a longer version of what I said online in [EMAIL PROTECTED]. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: unequal n's: quadratic weights
Serve me right for not checking my work. Here's a correction: On Wed, 31 Jan 2001 12:37:13 GMT, [EMAIL PROTECTED] (Duncan Murdoch) wrote: To find it, do this. Suppose the quadratic curve is A x^2 + B x + C. Then you've got three equations: A (-1)^2 + B (-1) + C = 2.05 A (0)^2 + B (0) + C = 6.38 A (1)^2 + B (1) + C = 12.08 There's a unique solution to these equations; it is C = 6.38, A = (12.08+2.05)/2-C, B = (12.08-2.05)/2-C.Those are the least-squares parameter estimates. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: unequal n's: quadratic weights
On Tue, 30 Jan 2001 23:22:51 -0500, "K. Bloom" [EMAIL PROTECTED] wrote: If your answer IS the correct answer to my question, perhaps you would kindly explain it to me in a more concrete way suitable to my simple knowledge of the problem. My means are 2.05, 6.38, and 12.08 for the three groups respectively. In other words.. what does one calculate and how? I'm still not sure I understand your question completely, but what I think you're asking is this: You have 3 groups of observations. You'd graph them with x-coordinates equal to -1, 0 and +1. The y-coordinates would be the observed data. The means for those three groups are 2.05, 6.38, and 12.08, based on different numbers of observations in each group. You want the quadratic curve that provides the least-squares fit to your data. If that's the case, then the numbers of observations in each group doesn't matter. There's a quadratic curve that goes exactly through each group mean, and you can't find a better fit than that. To find it, do this. Suppose the quadratic curve is A x^2 + B x + C. Then you've got three equations: A (-1)^2 + B (-1) + C = 2.05 A (0)^2 + B (0) + C = 6.38 A (1)^2 + B (1) + C = 12.08 There's a unique solution to these equations; it is C = 6.38, A = (12.08+2.05)/2, B = (12.08-2.05)/2.Those are the least-squares parameter estimates. If you have more than three groups, then you won't be able to find an exact solution like this (you've only got three parameters to play with), and then the least-squares solution *does* depend on the group sizes. In general, you should solve linear least-squares problems using "multiple linear regression"; there are a ton of texts on that and I'd suggest you use one of those. In particular, with unequal group sizes you probably want to use "weighted least squares", with the weights equal to the group sizes. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: unequal n's: quadratic weights
On Tue, 30 Jan 2001 19:53:03 -0500, "K. Bloom" [EMAIL PROTECTED] wrote: Could someone please tell me how to calculate quadratic coefficients for unequal sample sizes (equal intervals, three groups)? The formula is not in the SPSS algorithm notes. Maybe I'm misunderstanding the question, but if you only have 3 groups, your estimates will interpolate the group means. Just solve the 3 linear equations in 3 unknowns, using whatever equation solving method you like. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Florida votes and statistical errors (fwd)
On 9 Dec 2000 17:28:28 -0800, [EMAIL PROTECTED] (Bob Hayden) wrote: It's not clear how you split 25 electors among 2 candidates when you don't know the popular vote, nor if you assume it was a tie. (You might then have a lawsuit over which elector will be split in two!-) There are lots of solutions that seem common-sensical if they guarantee a pre-determined outcome you find attractive. That's why I favor tossing a coin. Gore would win if you gave 13 of the votes to Bush, so he'd probably be gracious and give away the undetermined one. Would he still have a lead if the same procedure was used in every state? He got a lot of votes from California based on a very close popular vote. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Is the vote brouhaha due to statistical falicy?
Surely some voters would choose to leave the ballot blank. If I didn't like any of the presidential candidates, but I thought that some more local election was worth voting in, I might do that. In fact, in the local municipal election that we just had, I was allowed to vote for as many as 4 candidates, but only chose to vote for 2. I would hope that the recount is an attempt to correct counting errors, not to create votes where there were none. Duncan Murdoch On Sun, 19 Nov 2000 09:32:31 -0500, Bob Wheeler [EMAIL PROTECTED] wrote: The model is simplified, but I assume that (B) votes will be counted by "divining the voter's intent" through chance imperfections in the ballots. Thus the probability is 0.50. Duncan Murdoch wrote: On Sat, 18 Nov 2000 22:49:41 -0500, Bob Wheeler [EMAIL PROTECTED] wrote: There are two possibilities: (A) a vote was attempted; (B) no vote was attempted. Let us assume that for (A) the probability of a Gore vote is 0.62, and 0.50 for (B), If no vote was attempted, then surely the probability that a vote for Gore was attempted is 0, not 0.50. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: What is standard deviation exactly?
On Mon, 22 May 2000 13:24:25 +1000, "Glen Barnett"
[EMAIL PROTECTED] wrote:
I assume you're talking about sample standard deviations,
not population standard deviations (though interpretation
of what it represents is similar).
...
Note that the standard deviation can't exceed half the range
(largest value minus smallest value).
That's true for the n denominator ("population standard deviation"),
but not for n-1 ("sample standard deviation"). For example, if your
sample is just the two points 0 and 1, the sample standard deviation
is 0.71, and the range is 1.
Duncan Murdoch
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