(New page: 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...) |
|||
| Line 1: | Line 1: | ||
| − | 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 | + | 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. |
Revision as of 13:06, 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.
