Anjum:

To find E[x|x>0] you need to integrate:

x * p(x | x>0) dx

which is:

x * p(x ^ x>0)/p(x>0)

since your integration is from 0 to +inf, the numerator reduces to

x * p(x) / p(x>0)

You're forgetting the denominator, which is the factor of 2 you're missing.

-Andrew

Anjum: Sorry for the confusion. You want to find another distribution where the SAMPLE you took has the same mean and variance. For instance, imagine getting a sample and it happens to have come from a chi-squared (just for illustration) distribution. If you assume Ho is true, you will do your stuff according to parts a and b to get a locations for c1,c2 and distribution of r(x). Compare that to what you would get if you had calculated locations for c1 and c2 (and therefore r(x)) assuming the same points had come from a chi-squared distribution. Is that helpful?

Edited by author 02-18-2004 01:37 PM
Michael: Thanks for your response to my message /m18. Just to clarify it again, so you are saying that I need to find another distribution with same Mean and Variance to compare the value of E(r(x)) to the one obtained under H*. So if I want to consider a normal distribution with N = 100, then under H* I will have to use x~N(50.5,833.25) to get the same variance and mean?

To respond to /m20. Noah, I did the exact same integration and came up with the same conflict between my theoretical and experimental results. As far the integration is concerned, I am pretty confident that it is right. I think we are missing some other factor or some other concept that is making our experimental results different. btw, in your integration expression you forgot to write a constant factor of (1/(sigma*(2*pi^0.5))).

I am trying to do the symbolic integration from 0 to
infinity of

xe^(x^2/sigma^2)dx

to find the average value of x on the right side of the
gaussian. I get sigma/(2*(2*pi)^.5) but my expermintal results give close to 1 minus that value, and I am confident they are correct. Much more confindent than in my
very rusty integration by substitution skills. Does anybody know how to integrate the above correctly?

Edited by author 02-18-2004 11:56 AM
Anjum: In both C and D, assume that you are working with the same set of points (i.e., the same mean and variance) as the sample for A and B. Find other non-Gaussian distributions such that the expected c1 and c2 locations under those distributions produce an r(x) less than (part C) and greater than (part D) the r(x) for the Guassian distribution [for the same observation].

A question on part C) and D) of problem 2:
It is asking us to find a distribution for which E(r(x)) will be less or more than the one under H* (Null hypothesis). But wouldn't E(r(x) depend on the parameters of a distibution? For example: E(r(x)) will be very big under H* if x ~ N(200,500) but will not be that big if x~N(0,1). So a uniform distribution from [0,1] may have smaller E(r(x)) than the later case of H* but may be larger than N(200,500) case.
Am I reading the problem wrong?
Thanks.

Anjum:

My bad. It is so obvious. SUM(sigma_i) is not expected to be 0. IT IS ZERO.

Sometimes you just get lost after you stare at it for too long. :)

Comment on /m15: Jay, Sum(sigma_i) is assumed to be 0, since it is Sum(O_i - e_i), which has expected value of zero. I think that is the assumption at work here.

When we derive Pearson Chi^2 test from LRT Chi^2, we have:
2*SUM[(e_i+sigma_i)*(sigma_i/e_i - sigma_i^2/2*e_i^2)]
= 2*SUM[sigma_i-(sigma^2/2*e_i^2)+(sigma^2/e_i^2)-...]
~SUM[(o_i-e_i)^2/e_i].

Somehow, we got rid of term sigma_i. It seems we assume SUM(sigma_i)=0. Why? Am I missing some assumption here?

Answer to /m12: Your reasoning about 4 degrees of freedom seems right to me. Have you repeated your experiment often enough to be sure that the empirical d.f. is 5?

Answer to /m11: Yes, your understanding is right.

Edited by author 02-16-2004 02:11 PM
Thanks for all your responses.
One more question regarding degrees of freedom --
i) I am generating a multinomial distribution using a fair dice. My null hypothesis is that dice is a fair dice, meaning I have one parameter to estimate with a true value of 1/6. Alterate hypothesis is that each number has different probability of coming up, so total of (6-1) parameters to estimate. From Prof. Weber's notes, the degree of freedom of the chi-squared statistic should be (k-1)-p that is (6-1)-1 and that is 4. But my statistic is following chi-squared distribution with d.f. 5. What is wrong? I am using expression (2) from Prof. Weber's notes to compute my statitics value.


btw: I came across the following paper, a readable and relavant paper. http://www.ecb.int/pub/wp/ecbwp083.pdf