[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]?

[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?

[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?

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.

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

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

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.