Apr
2
comment A test for a binomial r.v. vs. sum of two binomial r.v.'s?
I've been thinking about the link that you posted, and I am now pretty convinced that a test that compares the sample value $x$ to some threshold $t$, rejecting null if $x>t$, is UMP. This is because size $\alpha$ of this test doesn't depend on $q$ (and $M$). This also agrees with my plots of the likelihood function in Mathematica (though still I haven't been able to formally prove that it's increasing)... Thoughts?
Mar
31
comment A test for a binomial r.v. vs. sum of two binomial r.v.'s?
Thanks for a helpful link. My variables are in fact being added (though a convolution can be thought of as a mixture of random variables). I think if I can show that the likelihood ratio is non-decreasing monotonically over the support $1,2,\ldots,N$, then I can claim that LRT is an UMP test (per Lehmann&Romano Ch. 3). However, I am having trouble showing that the likelihood ratio is non-decreasing even though plotting it in Wolfram Mathematica suggests it is. I posted a question about that to math.SE: math.stackexchange.com/questions/734453/…
Mar
31
asked Is this function monotonically non-decreasing?
Mar
31
revised A test for a binomial r.v. vs. sum of two binomial r.v.'s?
slight change to notation again
Mar
31
revised A test for a binomial r.v. vs. sum of two binomial r.v.'s?
changed some notation
Mar
31
comment A test for a binomial r.v. vs. sum of two binomial r.v.'s?
Interesting. I only know of the monotone and non-decreasing condition on the likelihood ratio for UMP, but had trouble proving that (since one of the likelihood functions is a convolution). What do you mean by the "same form of a test"? Perhaps you could provide a reference to a relevant lemma?
Mar
31
asked A test for a binomial r.v. vs. sum of two binomial r.v.'s?
Mar
29
awarded Supporter
Mar
28
revised Integral involving a confluent hypergeometric function
Improved readability of the answer by putting the mathematical symbols into LaTeX.
Mar
28
accepted Integral involving a confluent hypergeometric function
Mar
28
comment Integral involving a confluent hypergeometric function
I was just thinking about this question the other day, and, as you said, the trick here is the summation representation of ${}_1F_1$ (which I didn't think about when I posted the question -- it came up as a step in a long series of calculations in Mathematica). Anyway, thanks for your answer (I edited tiny bit of formatting for mathematical symbols).
Mar
27
accepted How to numerically evaluate the CDF of this random variable?
Mar
27
comment How to numerically evaluate the CDF of this random variable?
Ahh, I see -- there is $x-t$ in the binomial. I must say, this is a fiendishly clever answer. I never though of using generating functions for this (and, as the matter of fact, until now I didn't know what generating functions were good for other than exercises in a graduate probability class...) Thanks!
Mar
27
comment How to numerically evaluate the CDF of this random variable?
Thanks for the edit! I still have questions. First, what is $A_x$? It is summation over $x$ from $x=0$ to $x=m$? I summed it that way (using Mathematica) and it spat out an expression involving ${}_2F_1$...
Mar
26
comment How to numerically evaluate the CDF of this random variable?
Thanks for the answer. I understand the normal approximation for the binomial distribution, but am stuck on your first sentence. Writing out the sum $Y+Z$ as a convolution, I obtain the following: $$\sum_{t=0}^x\binom{n}{t}(p+q-pq)^t(1-p-q+pq)^{n-t}\binom{m-n}{x-t}q^{x-t}(1-q)^{m-n-x+t}.$$ Since $1-p-q+pq=(1-p)(1-q)$, I can simplify the convolution as follows: $$\sum_{t=0}^x\binom{n}{t}(p+q-pq)^t(1-p)^{n-t}\binom{m-n}{x-t}q^{x-t}(1-q)^{m-x}$$ but I can't get the expression in (1). Could please elaborate how you get there?
Mar
26
revised How to numerically evaluate the CDF of this random variable?
fixed typo
Mar
26
asked How to numerically evaluate the CDF of this random variable?
Mar
24
accepted Estimating the variance of error in empirical approximation to a distribution
Mar
24
comment Estimating the variance of error in empirical approximation to a distribution
Aha -- the key word is "Bernoulli"! I think I understand how this works now. All this is new to me, and I appreciate your help.
Mar
24
comment What is known about the distribution of the errors in empirical approximation of a CDF?
Follow-up question posted: mathoverflow.net/questions/161282/…
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