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May 24th, 2015, 06:18 AM   #1
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Angry Seeking approximation method of a probability expression.

Dear all,

Given:

$P^R_n(i) = \beta ^ \left( i-m \right) ( 1 - \beta ) ^ \left( n-i+m \right) \binom{n-m}{i-m}$

and

$\phi$

Want:

$I^{min} _n = min \{ j: \sum_{i=j}^n P^R_n(i) < \phi \}$


I mean if there exists any simple ways of approximating or calculating the result of $\sum_{i=j}^n P^R_n(i)$.
Or approximating $\sum_{i=j}^n P^R_n(i)$ withgut multiplying one after one of the multipliers of the factorial.
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May 24th, 2015, 03:13 PM   #2
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Your statement is somewhat confusing. What role does m have in $\displaystyle P_n^R(i)$?
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May 24th, 2015, 07:06 PM   #3
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$m$ is also a given number less than $n$.
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May 24th, 2015, 07:31 PM   #4
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Angry

Quote:
Originally Posted by rubis View Post
Dear all,

Given:

$P^R_n(i) = \beta ^ \left( i-m \right) ( 1 - \beta ) ^ \left( n-i+m \right) \binom{n-m}{i-m}$

and

$\phi$

Want:

$I^{min} _n = min \{ j: \sum_{i=j}^n P^R_n(i) < \phi \}$


I mean if there exists any simple ways of approximating or calculating the result of $\sum_{i=j}^n P^R_n(i)$.
Or approximating $\sum_{i=j}^n P^R_n(i)$ withgut multiplying one after one of the multipliers of the factorial.

I mean if there exists any simple ways of approximating or calculating the result of $\sum_{i=j}^n P^R_n(i)$.
Or approximating $\binom{n-m}{i-m}$ without multiplying one after one multipliers of the factorial.
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May 24th, 2015, 08:42 PM   #5
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I think this website has been hacked, because I have found someone had modified my message!
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