How to find moments/compute this integral?












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I have a steady state distribution which is of the form $$K[A+Bz]^{C}e^{D(1-z)}$$ where $A,ldots,D$ are constants. I want to find moments of $z$. I do not know how I might go about it so that I can get a closed form expression. My first instinct was to look at the Gamma function, but I have been unable to reduce this to a useful expression. The exact expressions for $A,ldots,D$ are as follows:



$A=4s$



$B=2c+h$



$C=frac{4Ns(c+h)}{(2c+h)^2}-1$



$D=frac{hN}{2(2c+h)}$



This seems to me a problem of manipulating constants right now, but I have been unable to get something I can work with with $A,B,D$. Any help is appreciated!



PS: $min[-1,1]$.










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    $begingroup$


    I have a steady state distribution which is of the form $$K[A+Bz]^{C}e^{D(1-z)}$$ where $A,ldots,D$ are constants. I want to find moments of $z$. I do not know how I might go about it so that I can get a closed form expression. My first instinct was to look at the Gamma function, but I have been unable to reduce this to a useful expression. The exact expressions for $A,ldots,D$ are as follows:



    $A=4s$



    $B=2c+h$



    $C=frac{4Ns(c+h)}{(2c+h)^2}-1$



    $D=frac{hN}{2(2c+h)}$



    This seems to me a problem of manipulating constants right now, but I have been unable to get something I can work with with $A,B,D$. Any help is appreciated!



    PS: $min[-1,1]$.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have a steady state distribution which is of the form $$K[A+Bz]^{C}e^{D(1-z)}$$ where $A,ldots,D$ are constants. I want to find moments of $z$. I do not know how I might go about it so that I can get a closed form expression. My first instinct was to look at the Gamma function, but I have been unable to reduce this to a useful expression. The exact expressions for $A,ldots,D$ are as follows:



      $A=4s$



      $B=2c+h$



      $C=frac{4Ns(c+h)}{(2c+h)^2}-1$



      $D=frac{hN}{2(2c+h)}$



      This seems to me a problem of manipulating constants right now, but I have been unable to get something I can work with with $A,B,D$. Any help is appreciated!



      PS: $min[-1,1]$.










      share|cite|improve this question









      $endgroup$




      I have a steady state distribution which is of the form $$K[A+Bz]^{C}e^{D(1-z)}$$ where $A,ldots,D$ are constants. I want to find moments of $z$. I do not know how I might go about it so that I can get a closed form expression. My first instinct was to look at the Gamma function, but I have been unable to reduce this to a useful expression. The exact expressions for $A,ldots,D$ are as follows:



      $A=4s$



      $B=2c+h$



      $C=frac{4Ns(c+h)}{(2c+h)^2}-1$



      $D=frac{hN}{2(2c+h)}$



      This seems to me a problem of manipulating constants right now, but I have been unable to get something I can work with with $A,B,D$. Any help is appreciated!



      PS: $min[-1,1]$.







      calculus probability integration moment-generating-functions






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      asked Jan 4 at 18:53









      BoshuBoshu

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