Incomplete regularized Gamma function recursive relation
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I'm trying to found the detailed derivation of:
$Q(alpha + n, t) = Q(alpha, t) + t^alpha e^{-t} displaystylesum_{k=0}^{n-1} dfrac{t^k}{Gamma(alpha+k+1)}$
This is a relation necessary for the detailed proof of the Bayesian A/B test for countable type data. As you can see here:
Formulas for Bayesian A/B Testing
May you help me? My level is not so good in pure mathematics (and especially in functional analysis).
Thx in advance
functional-analysis gamma-function
$endgroup$
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$begingroup$
I'm trying to found the detailed derivation of:
$Q(alpha + n, t) = Q(alpha, t) + t^alpha e^{-t} displaystylesum_{k=0}^{n-1} dfrac{t^k}{Gamma(alpha+k+1)}$
This is a relation necessary for the detailed proof of the Bayesian A/B test for countable type data. As you can see here:
Formulas for Bayesian A/B Testing
May you help me? My level is not so good in pure mathematics (and especially in functional analysis).
Thx in advance
functional-analysis gamma-function
$endgroup$
add a comment |
$begingroup$
I'm trying to found the detailed derivation of:
$Q(alpha + n, t) = Q(alpha, t) + t^alpha e^{-t} displaystylesum_{k=0}^{n-1} dfrac{t^k}{Gamma(alpha+k+1)}$
This is a relation necessary for the detailed proof of the Bayesian A/B test for countable type data. As you can see here:
Formulas for Bayesian A/B Testing
May you help me? My level is not so good in pure mathematics (and especially in functional analysis).
Thx in advance
functional-analysis gamma-function
$endgroup$
I'm trying to found the detailed derivation of:
$Q(alpha + n, t) = Q(alpha, t) + t^alpha e^{-t} displaystylesum_{k=0}^{n-1} dfrac{t^k}{Gamma(alpha+k+1)}$
This is a relation necessary for the detailed proof of the Bayesian A/B test for countable type data. As you can see here:
Formulas for Bayesian A/B Testing
May you help me? My level is not so good in pure mathematics (and especially in functional analysis).
Thx in advance
functional-analysis gamma-function
functional-analysis gamma-function
edited Jan 22 at 17:39
Vincent ISOZ
asked Jan 22 at 7:13
Vincent ISOZVincent ISOZ
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63
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