ku Posted August 14, 2005 Share Posted August 14, 2005 [math]X =^{d} R(0,\theta)[/math] and we obtain five independent observations on X: 1.2, 3.7, 2.1, 5.9, and 4.0. The median [math]\hat{M}[/math] is 3.7 and I'm told that [math]var(\hat{M})=\frac{\theta^2}{28}[/math]. How is this obtained? Do I use the formula [math]var(\hat{M})=\frac{1}{4nf(m)^2}[/math]? Link to comment Share on other sites More sharing options...

karljazzi Posted June 7, 2007 Share Posted June 7, 2007 [math]X =^{d} R(0,\theta)[/math] and we obtain five independent observations on X: 1.2, 3.7, 2.1, 5.9, and 4.0. The median [math]\hat{M}[/math] is 3.7 and I'm told that [math]var(\hat{M})=\frac{\theta^2}{28}[/math]. How is this obtained? Do I use the formula [math]var(\hat{M})=\frac{1}{4nf(m)^2}[/math]? I would like to know the answer too,I think the formula is only for large enough n,which is the asymptotic behavior,what about small sample size? Link to comment Share on other sites More sharing options...

cosine Posted June 16, 2007 Share Posted June 16, 2007 [math]X =^{d} R(0,\theta)[/math] and we obtain five independent observations on X: 1.2, 3.7, 2.1, 5.9, and 4.0. The median [math]\hat{M}[/math] is 3.7 and I'm told that [math]var(\hat{M})=\frac{\theta^2}{28}[/math]. How is this obtained? Do I use the formula [math]var(\hat{M})=\frac{1}{4nf(m)^2}[/math]? Could someone please explain what [math]X =^{d} R(0,\theta)[/math] means? What is an equals sign superscripted with a d mean? and is R some well-known probability distribution? Link to comment Share on other sites More sharing options...

Bignose Posted June 16, 2007 Share Posted June 16, 2007 The X =d normally means "X is distributed according to" and the R would be the probability distribution. But, I am unfamilar with what R would mean, too. Link to comment Share on other sites More sharing options...

cosine Posted June 16, 2007 Share Posted June 16, 2007 The X =d normally means "X is distributed according to" and the R would be the probability distribution. But, I am unfamilar with what R would mean, too. Hmm, I looked for examples of probability distributions on wikipedia denoted with an R, I found so far: http://en.wikipedia.org/wiki/Rayleigh_distribution http://en.wikipedia.org/wiki/Rice_distribution But I haven't any real idea what these are or are commonly used for Link to comment Share on other sites More sharing options...

Tartaglia Posted June 16, 2007 Share Posted June 16, 2007 R(0,theta) is almost certainly a rectangular uniform distribution from 0 to theta. The expectancy of the median is therefore theta/2. The expectancy of the median squared can be determined by considering the probability of picking a number x and having two others above and two below this number. ie a binomial. Then integrating x^2*f(x) between theta and 0 and then fiddling about with a substitution and factorials to convert the integration into a Beta distribution pdf. Then Var(M) = E(M^2)-E(M)^2 I am unable to decide whether theta is known or not and this would obviously have a significant bearing on the result Link to comment Share on other sites More sharing options...

Tartaglia Posted June 16, 2007 Share Posted June 16, 2007 I've just worked through this and get Variance of theta^2/28 The important points here are the pdf of the median m ia given by f(x) = 1/theta*(1-x/theta)^2*(x/theta)^2*5!/(2!2!1!) the first part 1/theta is the pdf of the rectangular uniform solution, the (1-x/theta)^2 is the probability of two others above x, the (x/theta)^2 is the probabilty of 2 below x and the 5!/(2!2!11) is the arrangements of x, two above and two below evaluating the integral of x*f(x) between theta and 0 gives E(M) = theta/2 as required evaluating the integral of x^2*f(x) between theta and 0 gives E(M^2) = 2*(theta^2)/7 Variance = theta^2/28 as required More generally for an odd sample size 2n-1 a general formula can be derived by substituting y = x/theta to convert to a Beta style pdf and then by manipulating the integral to equal 1. Link to comment Share on other sites More sharing options...

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