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#2 Help Me ! » residuse integral » 2008-03-04 02:04:54

jianingli
Replies: 1

what is the result of

#5 Re: Help Me ! » a problem about chi-square distribution » 2008-02-25 20:44:34

\begin{array}{align}{c}
\bar P_{out}  &= \delta \lambda \int_0^1 {\frac{1}{{(n!)^{2K} }}(\psi \phi )^{nK} \delta ^{2nK} t^{2nK} dt}  \\
  &= \frac{{\delta ^{2nK + 1} \lambda \prod\limits_{k = 1}^K {(\psi _k \phi _k )^n } }}{{(n!)^{2K} (2nK + 1)}} \\
  &= \frac{{\lambda \prod\limits_{k = 1}^K {(\psi _k \phi _k )^n } }}{{(n!)^{2K} (2nK + 1)}}\left( {\frac{{2^{2R}  - 1}}{\gamma }} \right)^{2nK + 1}  \\
\end{array}{align}

#7 Help Me ! » a problem about chi-square distribution » 2008-02-24 18:15:13

jianingli
Replies: 4

both

and
is
random variables with degree of freedom
.
So how can I get

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