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I followed the example given by Anand Gautam, and solved the equation :<math>\hat n_{ML} = \text{max}_n ( \binom{n}{1000000} p^{1000000} (1-p)^{n-1000000} )</math>. and the answer i got is n=(1e6)/p, which make sense to me. for example, u toss coin n times, got 5 heads,the probablity of getting head is 1/2, asking you to find n. it is quite obvious, n=5/0.5=10, in other words n=(# of heads)/p. | I followed the example given by Anand Gautam, and solved the equation :<math>\hat n_{ML} = \text{max}_n ( \binom{n}{1000000} p^{1000000} (1-p)^{n-1000000} )</math>. and the answer i got is n=(1e6)/p, which make sense to me. for example, u toss coin n times, got 5 heads,the probablity of getting head is 1/2, asking you to find n. it is quite obvious, n=5/0.5=10, in other words n=(# of heads)/p. | ||
| + | If you follow the example, and mimic it, you will end up getting | ||
| + | <math>0= p^{1000000} (1-p)^{n-1000000-1}(1m-np)\,\ </math>. | ||
| + | since P^1000000 is > or = 1 and (1-p)^(n-1000000-1) is > or 1. so we set 1m-np=0 and we will get n=1m/p | ||
Revision as of 12:17, 11 November 2008
I followed the example given by Anand Gautam, and solved the equation :$ \hat n_{ML} = \text{max}_n ( \binom{n}{1000000} p^{1000000} (1-p)^{n-1000000} ) $. and the answer i got is n=(1e6)/p, which make sense to me. for example, u toss coin n times, got 5 heads,the probablity of getting head is 1/2, asking you to find n. it is quite obvious, n=5/0.5=10, in other words n=(# of heads)/p. If you follow the example, and mimic it, you will end up getting $ 0= p^{1000000} (1-p)^{n-1000000-1}(1m-np)\,\ $. since P^1000000 is > or = 1 and (1-p)^(n-1000000-1) is > or 1. so we set 1m-np=0 and we will get n=1m/p
