Hi, I've got this problem with the solutions but I dont understand how to get from line 1 to line 2
I've attached the solution.
Im not sure why the sum to infinity of x.lamda^x results in lambda.e^lambda
Thanks in advance for any help clarifying.
Hi, I've got this problem with the solutions but I dont understand how to get from line 1 to line 2
I've attached the solution.
Im not sure why the sum to infinity of x.lamda^x results in lambda.e^lambda
Thanks in advance for any help clarifying.
$\displaystyle \sum x[\frac{1}{2}\frac{e^{-\lambda}\lambda^x}{x!} + \frac{1}{2}\frac{e^{-\mu}\mu^x}{x!}] $
$\displaystyle = \sum [x\frac{1}{2}\frac{e^{-\lambda}\lambda^x}{x!} + x\frac{1}{2}\frac{e^{-\mu}\mu^x}{x!}] $
$\displaystyle = \sum x\frac{1}{2}\frac{e^{-\lambda}\lambda^x}{x!} + \sum x\frac{1}{2}\frac{e^{-\mu}\mu^x}{x!} $
$\displaystyle = \frac{e^{-\lambda}}{2}\sum \frac{x\lambda^x}{x!} + \frac{e^{-\mu}}{2}\sum\frac{x\mu^x}{x!} $
The well known Taylor expansion is:
$\displaystyle \sum \frac{\lambda^x}{x!} = \underbrace{(1 + \lambda + \frac{\lambda^2}{2!} +...)} = e^{\lambda}$
Your sum is:
$\displaystyle \sum \frac{x\lambda^x}{x!} = 0 + \lambda + 2\frac{\lambda^2}{2!} + 3\frac{\lambda^3}{3!}+... = \lambda(\frac{\lambda}{\lambda} + 2\frac{\lambda^2}{\lambda 2!} +3\frac{\lambda^3}{\lambda 3!}+...) = \lambda\underbrace{(1 + \lambda + \frac{\lambda^2}{2!} +...)} = \lambda e^{\lambda}$
Is this clear?