Recent content by katxt

There is a trap that you might not have thought of. With least squares it is possible that the simple answer gives you a theta which is can't be true. For instance if the X's are 10, 10, 10, and 20, then the average is 12.5 and theta would be 10 + 2x2.5 = 15. But 20 is impossible in U(10, 15)...

Try drawing the distribution of U(10, 10+theta) and draw a line at the mean. If it makes things easier to see, draw U(10, 10+8) i.e. U(10, 18) and find the mean.

Can you find the mean of U(10, 10+theta)?

Make a new row between X and P, labelled Z. Fill in the Z row. Ignore the X row and continue.

I'm not suggesting that OLS is a better model, just that the results from OLS are easily interpreted and should give an indication of what you can expect from the Poisson after they have been manipulated to give real world meanings.

Just to give you an idea of the sort of values you are likely to get when you finish you interpretation, you could try the usual OLS regression where the coefficients often have real world meanings which should be close to your final Poisson interpretations.

X_1 and X_1 are not independent so E(X_1**2) does not = E(X_1)*E(X_1)

You could start with an online difference in proportions calculator. https://www.medcalc.org/calc/comparison_of_proportions.php

Interesting problem, though.

A little, but I have no intuitive feel for this foreign and statistics intense game. All I can suggest is that you try both ways and see which has more success.

You're right - the jackknife should strictly use the means of all the subsamples. However, if there are too many to enumerate, then you will get a good approximation by taking a random sample of the subsample means. If, for instance, you have a sample of 50 and you consider all the subsamples of...

I think we are talking at cross purposes here. The short answer, in my opinion, is that you can decrease the numbers to 85% and the variability will automatically decrease to 85% without you doing anything, or you can use the untouched numbers and decrease the mean and SD at the end. If this...

Yes Do an experiment in Excel. Type some random numbers. Find the mean and SD. Multiply the numbers by 0.85. Find the mean and SD. Compare the means and SDs

I wouldn't dismiss the jackknife too quickly. The delete-d jackknife method has a factor ( n-d)/d which is just 1 in your case. The only real difference is that you are taking a random sample of possible deletions instead of listing them all.

Is this something to do with the jackknife?