The expected value of Beta Function












0












$begingroup$


Estimate the probability of success



Suppose I send 10 tasks to my machine. 6 out of 10 tasks success, and 4 failed.
These outcomes is summarized by $X$ as a binary variable, 1 is task success, and 0 if task fail. We know that $X$ is continuous random variable



The expected value of a continuous random variable is dependent on the probability density function used to model the probability that the variable
will have a certain value. Therefor, I exploit Beta distribution to estimate the probability of success for next tasks. I will ${alpha}$ as input of the number past success tasks
and ${beta}$ as the number of past fail tasks



Expected value



begin{equation}
E(x) = frac{alpha+1}{alpha+beta+2}
end{equation}



In my example, $alpha = 6$ and $beta = 4$. Thus, the $E(x)$ = 0.58.



Does every think looks good?










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$endgroup$

















    0












    $begingroup$


    Estimate the probability of success



    Suppose I send 10 tasks to my machine. 6 out of 10 tasks success, and 4 failed.
    These outcomes is summarized by $X$ as a binary variable, 1 is task success, and 0 if task fail. We know that $X$ is continuous random variable



    The expected value of a continuous random variable is dependent on the probability density function used to model the probability that the variable
    will have a certain value. Therefor, I exploit Beta distribution to estimate the probability of success for next tasks. I will ${alpha}$ as input of the number past success tasks
    and ${beta}$ as the number of past fail tasks



    Expected value



    begin{equation}
    E(x) = frac{alpha+1}{alpha+beta+2}
    end{equation}



    In my example, $alpha = 6$ and $beta = 4$. Thus, the $E(x)$ = 0.58.



    Does every think looks good?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Estimate the probability of success



      Suppose I send 10 tasks to my machine. 6 out of 10 tasks success, and 4 failed.
      These outcomes is summarized by $X$ as a binary variable, 1 is task success, and 0 if task fail. We know that $X$ is continuous random variable



      The expected value of a continuous random variable is dependent on the probability density function used to model the probability that the variable
      will have a certain value. Therefor, I exploit Beta distribution to estimate the probability of success for next tasks. I will ${alpha}$ as input of the number past success tasks
      and ${beta}$ as the number of past fail tasks



      Expected value



      begin{equation}
      E(x) = frac{alpha+1}{alpha+beta+2}
      end{equation}



      In my example, $alpha = 6$ and $beta = 4$. Thus, the $E(x)$ = 0.58.



      Does every think looks good?










      share|cite|improve this question









      $endgroup$




      Estimate the probability of success



      Suppose I send 10 tasks to my machine. 6 out of 10 tasks success, and 4 failed.
      These outcomes is summarized by $X$ as a binary variable, 1 is task success, and 0 if task fail. We know that $X$ is continuous random variable



      The expected value of a continuous random variable is dependent on the probability density function used to model the probability that the variable
      will have a certain value. Therefor, I exploit Beta distribution to estimate the probability of success for next tasks. I will ${alpha}$ as input of the number past success tasks
      and ${beta}$ as the number of past fail tasks



      Expected value



      begin{equation}
      E(x) = frac{alpha+1}{alpha+beta+2}
      end{equation}



      In my example, $alpha = 6$ and $beta = 4$. Thus, the $E(x)$ = 0.58.



      Does every think looks good?







      probability expected-value beta-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 21:35









      joujou

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