expectation

Aug 2009
16
0
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.
 

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Nov 2009
517
130
Big Red, NY
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.
\(\displaystyle \sum x[\frac{1}{2}\frac{e^{-\lambda}\lambda^x}{x!} + \frac{1}{2}\frac{e^{-\mu}\mu^x}{x!}] \)

\(\displaystyle = \sum [x\frac{1}{2}\frac{e^{-\lambda}\lambda^x}{x!} + x\frac{1}{2}\frac{e^{-\mu}\mu^x}{x!}] \)

\(\displaystyle = \sum x\frac{1}{2}\frac{e^{-\lambda}\lambda^x}{x!} + \sum x\frac{1}{2}\frac{e^{-\mu}\mu^x}{x!} \)

\(\displaystyle = \frac{e^{-\lambda}}{2}\sum \frac{x\lambda^x}{x!} + \frac{e^{-\mu}}{2}\sum\frac{x\mu^x}{x!} \)
 
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Aug 2009
16
0
Many thanks! that was very clear.

May I also ask how to get from line 2 to line 3? I know that it is partly due to the taylors series expansion but I'm a bit confused as to where the x went in the third line.
 
Nov 2009
517
130
Big Red, NY
Many thanks! that was very clear.

May I also ask how to get from line 2 to line 3? I know that it is partly due to the taylors series expansion but I'm a bit confused as to where the x went in the third line.
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?
 
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Aug 2009
16
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Yes, it is, thank you very much, this expansion has been very useful.