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Math Help - Finding the expected value of x^2 in binomial distribution

  1. #1
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    Finding the expected value of x^2 in binomial distribution

    Here is the summation (from i=0 to n) of i^2 * (n!/(n-i)!(i!)) * p^i * q^(n-i)

    I know that the answer is equal to np[(n-1)p+1].

    I am not really sure how to get what is in the bracket of the answer.

    Please help me out.

    Thanks.
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  2. #2
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    There are easier ways to do this than evaluating that sum

    You know that
    E(X^2) = Var(X) + E^2(X)


    You also know that by definition the binomial variable X is the sum of n independant bernoulli trials. Its not difficult to compute the mean and variance of the sum of n independant bernoulli trials, then you can use the result above.
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  3. #3
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    THank you for your quick reply.

    However, I would actually like to know how to get that answer, which is np[(n-1)p+1] using the summation and binomial distribution properties.

    Could you please show me the steps?
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  4. #4
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    Instead of recognising the left hand sum as an expectation, you can use the algebra in the following link to show explicitly that it is (n-1)p.
    Mean or Expectation Value

    Spoiler:

    latex in case i need it later
    \begin{eqnarray*}
    E(X^2) &=& \sum_{i=0}^n \left( i^2 \frac{n!}{i!(n-i)!} p^{i} (1-p)^{n-i}\right)\\
    && \text{note first term in sum is 0 and rewrite limit}\\
    &=& \sum_{i=1}^n \left( i^2 \frac{n!}{i!(n-i)!} p^{i} (1-p)^{n-i}\right)\\
    &=& np\sum_{i=1}^n \left( i \frac{(n-1)!}{(i-1)!(n-i)!} p^{i-1} (1-p)^{n-i}\right)\\
    && \text{let } j=i-1 \text{ , so }i=j+1\\
    &=& np\sum_{j=0}^{n-1} \left( i \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)\\
    &=& np\sum_{j=0}^{n-1} \left( (j+1) \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)\\
    &=& np\sum_{j=0}^{n-1} \left( (j) \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)+ np\sum_{j=0}^{n-1} \left( \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)\\
    && \text{recognise right hand sum as PDF of Bi(n-1,p)}\\
    &=& np\sum_{j=0}^{n-1} \left( (j) \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)+ np \times 1\\
    && \text{recognise left hand sum as Expectation of Bi(n-1,p)}\\
    &=& np\left((n-1)p \right)+ np \times 1\\
    &=& np\left((n-1)p + 1 \right)


    \end{eqnarray*}
    Last edited by SpringFan25; May 8th 2011 at 05:27 AM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    It's easier to find E(X(X-1)) first and then use E(X(X-1))= E(X^2)-np
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