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  1. #1
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    expectation

    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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  2. #2
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    Quote Originally Posted by blueturkey View Post
    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.
    \sum x[\frac{1}{2}\frac{e^{-\lambda}\lambda^x}{x!} + \frac{1}{2}\frac{e^{-\mu}\mu^x}{x!}]

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

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

    = \frac{e^{-\lambda}}{2}\sum \frac{x\lambda^x}{x!} + \frac{e^{-\mu}}{2}\sum\frac{x\mu^x}{x!}
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  3. #3
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    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.
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  4. #4
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by blueturkey View Post
    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:

    \sum \frac{\lambda^x}{x!} = \underbrace{(1 + \lambda + \frac{\lambda^2}{2!} +...)} = e^{\lambda}

    Your sum is:

    \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?
    Last edited by Anonymous1; May 8th 2010 at 12:03 PM.
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  5. #5
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    Yes, it is, thank you very much, this expansion has been very useful.
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