3 ms·
Not quite; the probability of n/2 successes in n trials is given as Binomial(n,p) not p^n. p^n is correct for a single sequence but there are many possible sequ
by b10nic 3y ago
Not quite; the probability of n/2 successes in n trials is given as Binomial(n,p) not p^n. p^n is correct for a single sequence but there are many possible sequences that result half heads, half tails and so you have a factor of "N choose X" or the so called "Binomial Coefficient".
> (0.4)^20 × (0.6)^20
and I think you mean (0.4)^10 × (0.6)^10 or more generally p^x*(1-p)^n-x.
- l_e_o_n 3y agoI'm talking about the whole sequence; you're talking about the number of heads (or) tails in the sequence. The number of heads is a sufficient statistic, so we'll get the same likelihood ratios out, but the likelihood values themselves will be larger. You could make a similar point about the original CrossValidated Normal(0, 1)^N example by summarizing the data with the mean and sum of squares. This doesn't work if the data were Cauchy(0, 1)^N instead.