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Can someone hint me on solving the following equation to find $P^*$

\begin{equation} P^* =\sum^\infty_{n=m}\{P(m).P(R_{n\mid m} \} \end{equation}

Where, \begin{equation} P(m) = e^{-\lambda t}(\lambda t)^m / m! \end{equation}

\begin{equation} P(R_{n\mid m}) = \sum^{n}_{j=0}(-1)^jC^j_n(1 - j.\phi)^m \end{equation}

The final solution is as follows:

\begin{equation} P^* = (1 - e^{-\phi\lambda t})^n \end{equation}

NOTE: It is not a part of any exam or assignment, I am reading a resarch paper. and I have never solved equations where I have Summation of such peculiar terms...If someone doesnot want to give the whole solution,any hint would still be highly appreciated.

SJa
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  • What is $\lambda_{1,a}$? Is it real or complex? – probably_someone Jun 28 '17 at 04:43
  • It is a real value. It is a parameter of the exponential distribution. I have edited it to make it simplified form – SJa Jun 28 '17 at 04:45
  • You are simply supposed to note that $$P^=\sum_{m=0}^\infty e^{-\lambda t}\frac{(\lambda t)^m}{m!}\sum_{j=0}^n(-1)^j{n\choose j}(1-j\phi)^m$$ is $$P^=e^{-\lambda t}\sum_{j=0}^n(-1)^j{n\choose j}\sum_{m=0}^\infty\frac{(\lambda t)^m}{m!}(1-j\phi)^m=e^{-\lambda t}\sum_{j=0}^n(-1)^j{n\choose j}e^{\lambda t(1-j\phi)}$$ that is, $$P^=\sum_{j=0}^n(-1)^j{n\choose j}e^{-j\lambda t\phi}=(1-e^{-\lambda t\phi})^n$$ Note the serious misprint in your formula for $P^$, which should involve $$\sum_{m=0}^\infty$$ instead of $$\sum_{n=m}^\infty$$ – Did Jun 28 '17 at 05:34
  • Thanks Did for the solution, Can you please give me some reference or know how the way you simplified the respective terms to $e^{\lambda t (1-j\phi)}$. and the other term to $(1 - e^{-\lambda t n})^n$.

    Thanks. Also, although I am pretty sure after reading the paper that it should start from $n=m$ and author also does the same equation as I have quoted. I will have a look at it. Thanks again

     – SJa Jun 28 '17 at 06:05

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