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Math Help - Independant Bernoulli Random Varible

  1. #1
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    Independant Bernoulli Random Varible

    SupposeY1; Y2; Y3; : : : are independent Bernoulli random variables with

    P(Yi = 0) = 1 =(1/i!)
    P(Yi = 1) = 1=i!

    Defining X=Y1+Y2+Y3+...
    Show that E[x]=e-1

    Not to sure how to do this queston, so wondering anyone can help me?

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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Independant Bernoulli Random Varible

    Note that the expected value E has linear properties which we can use here. Unfortunately I'm confused with your notations. Can you explain what you mean with 1=\frac{1}{i!}? We know that Y_1,Y_2, \ldots are discrete randiom variables, hence E[Y_i] = \sum_{x} x p_x. Finally, E[X] = \sum_{i=1}^{\infty} E[Y_i]
    Last edited by Siron; February 21st 2013 at 09:02 AM.
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  3. #3
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    Re: Independant Bernoulli Random Varible

    Sorry there has been a typing error, it should read as
    P(Yi = 0) = 1 -(1/i!)
    P(Yi = 1) = 1/i!
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    MHF Contributor Siron's Avatar
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    Re: Independant Bernoulli Random Varible

    Quote Originally Posted by wrexlive1 View Post
    Sorry there has been a typing error, it should read as
    P(Yi = 0) = 1 -(1/i!)
    P(Yi = 1) = 1/i!
    Okay. We can compute the expected value of Y_i as follows E[Y_i] = 0\left(1-\frac{1}{i!}\right)+1\left(\frac{1}{i!}\right) =\frac{1}{i!}.
    Therefore, E[X]=\sum_{i=1}^{\infty} E[Y_i]=\sum_{i=1}^{\infty} \frac{1}{i!} = \sum_{i=0}^{\infty} \frac{1}{i!}-1. Using the fact that e^x =  \sum_{i=0}^{\infty} \frac{x^i}{i!} we have \sum_{i=0}^{\infty} \frac{1}{i!}=e. Hence, E[X]=e-1.
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