Following Problem is from probability theory:Define $G(n),P(n)\ge 0,n\in\mathbb{N}$,and such $$\begin{cases}G(n)=e^{-\lambda}\cdot\dfrac{\lambda^n}{n!},\lambda>0\\ \displaystyle\sum_{j=0}^{n}G(j)P(n-j)+(n+1)[P(n+1)-P(n)]=0,\forall n\in \mathbb{N} ,\\ \displaystyle \sum_{n=0}^{\infty}P_{n}=1,\text{ i.e. the measure of entire sample space i equal to one} \end{cases}$$
Compute the sum( the expectation)$\displaystyle\sum_{n=0}^{\infty}nP_{n}$
The equations provided lead to the following: \begin{align} \sum_{n=0}^{\infty} G_{n} \, t^{n} &= e^{- \lambda} \, \sum_{n=0}^{\infty} \frac{(\lambda \, t)^{n}}{n!} = e^{-\lambda \, (1 - t)}. \end{align} Now, \begin{align} 0 &= \sum_{n=0}^{\infty} (n+1) \left(P_{n+1} - P_{n} \right) \, t^{n} + \sum_{n=0}^{\infty} \, \sum_{k=0}^{n} G_{k} \, P_{n-k} \, t^{n} \\ &= \sum_{n=0}^{\infty} (n+1) \left(P_{n+1} - P_{n} \right) \, t^{n} + \sum_{n=0}^{\infty} \, \sum_{k=0}^{\infty} G_{k} \, P_{n} \, t^{n+k} \\ &= \frac{1}{t} \, \sum_{n=0}^{\infty} n P_{n} \, t^{n} - \sum_{n=0}^{\infty} n \, P_{n} \, t^{n} - P(t) + e^{-\lambda (1-t)} \, P(t) \\ &= \frac{1-t}{t} \, \sum_{n=0}^{\infty} n \, P_{n} \, t^{n} - (1 - e^{-\lambda(1-t)}) \, P(t) \end{align} this becomes \begin{align} \frac{\sum_{n=0}^{\infty} n \, P_{n} \, t^{n}}{P(t)} = \frac{t \, \left( 1 - e^{-\lambda (1-t)}\right)}{1-t} \end{align} where $P(t) = \sum_{n=0}^{\infty} P_{n} \, t^{n}$. By taking the limit as $t \to 1$ the result \begin{align} \lim_{t \to 1} \left\{ \frac{\sum_{n=0}^{\infty} n \, P_{n} \, t^{n}}{\sum_{n=0}^{\infty} P_{n} \, t^{n}} \right\} = \lim_{t \to 1} \left\{ \frac{t \, \left( 1 - e^{-\lambda (1-t)}\right)}{1-t} \right\} \end{align} leads to \begin{align} \sum_{n=0}^{\infty} n \, P_{n} = \lambda \end{align} which is the expected result.
\textenvironment. – Pedro Jul 21 '15 at 16:51