#1




Question on HW2.1
I was, out of curiosity, trying to derive an exact theoretical expression for the expected value of v_min, and this is what I got (apologies for the poor typesetting):
E( #heads_min ) = sum from i=0 to i=10 ( P(#heads_min = i) * i ) P(#heads_min = i) = 1000 * P( one gets i intersect the others get >= i) for which *1000 means that out of the 1000 trials, I choose one trial to be the minimum head trial. P( one gets i intersect the others get >= i) = P( one gets i ) * P(others get >= i) = p(i, 10, 0.5) * [1  F(i1, 10, 0.5)] ^ 999 for which p is the probability mass function and F is the cumulative mass function, with parameters (number successes, total number, prob. of success). I used i1 for F since >= i includes i itself. and this expression yielded the result E( #heads_min ) = 3.6814, and expected v_min is thus 0.36814, which is incorrect. Would anyone please point to me where my mistake is? I would really really appreciate it! Here is the R code I used for the calculation: phmin = function(i) { u = 0.5 # probability of head N = 10 # number of flips per trial m = 1000 # number of trials one_min = dbinom(i, N, u) others_more = pbinom(i1, N, u, lower.tail = FALSE) ^ (m1) return(1000 * one_min * others_more) } ivals = (0:10) expect_hmin = sum(phmin(ivals) %*% ivals) expect_hmin / 10 
#2




Re: Question on HW2.1
This formula assumes that the events "one gets i intersect the others get >= i" (with the "one" fixed at one coin at a time) are disjoint, otherwise this would be an overcount. These events are in fact overlapping. For instance, the case where all the coins get exactly i (which is admittedly an extremely unlikely event but chosen here as a simple illustration) would be counted in all 1000 cases, although it happens only "once."
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#3




Re: Question on HW2.1
Thank you so much, this clears it up exactly and I changed it and got the correct answer! I'll keep working.

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