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Math Help - Expected value of probability distribution

  1. #1
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    Expected value of probability distribution

    The number X of particles emitted as the result of an experiment is a random variable with probability distribution:

    P(X=k)=\left(\frac{1}{2}\right)^{k+1}\ for\ k \geq 0

    What is the expected number of particles emitted during one experiment

    This is how I've gone so far:

    E(X)=\sum_{k=0}^{\infty}kP(X=k)

    =\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\rig  ht)^3+ 3\left(\frac{1}{2}\right)^4+...

    I'm confused as to how to evaluate this sum
    Last edited by acevipa; June 4th 2010 at 10:57 PM.
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  2. #2
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    Quote Originally Posted by acevipa View Post
    The number X of particles emitted as the result of an experiment is a random variable with probability distribution:

    P(X=k)=\left(\frac{1}{2}\right)^{k+1}\ for\ k \geq 0

    What is the expected number of particles emitted during one experiment

    This is how I've gone so far:

    E(X)=\sum_{k=0}^{\infty}kP(X=k)

    =\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\rig  ht)^3+ 3\left(\frac{1}{2}\right)^4+...

    I'm confused as to how to evaluate this sum
    It is well known that the sum of an infinite geometric series is \sum_{k=0}^{+\infty} r^k = \frac{1}{1-r} for |r| < 1. Therefore:

    \frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = ....

    \Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = ....

    \Rightarrow \sum_{k=0}^{+\infty}k r^{k-1} = ....

    \Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = r^2(....)
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  3. #3
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    im pretty rusty on infinite sums and it would be interesting to know if there is a closed form summation of that.

    Nevertheless, you dont need to do anything difficult to solve this problem. You can rewrite as follows:

    P(x=k) = 0.5^k \times 0.5

    This is the PF of a geometric distribution with p=0.5. The mean of a geometric distribution is a standard result: \frac{1}{p}


    edit too slow
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  4. #4
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    oops i had the wrong form of the geometric distribution.

    its \frac{1-p}{p}
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    It is well known that the sum of an infinite geometric series is \sum_{k=0}^{+\infty} r^k = \frac{1}{1-r} for |r| < 1. Therefore:

    \frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = ....

    \Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = ....

    \Rightarrow \sum_{k=0}^{+\infty}k r^{k-1} = ....

    \Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = r^2(....)
    Yes, but for my case,

    \left(\sum_{k=0}^{+\infty} r^k\right) = 2

    So if you differentiate that, it just becomes 0
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  6. #6
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    Quote Originally Posted by acevipa View Post
    Yes, but for my case,

    \left(\sum_{k=0}^{+\infty} r^k\right) = 2

    So if you differentiate that, it just becomes 0
    No, the starting point is the first line of my previous reply. Yuo are trying to find a general formula. The, after you find it, you substitute r = 1/2.
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  7. #7
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    Ok, so is this correct:

    \frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = \frac{d}{dr} \left(\frac{1}{1-r}\right) for |r|<1.

    =\frac{1}{(1-r)^2}

    \Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = <br />
\sum_{k=0}^{+\infty}k r^{k-1}

    \Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = \frac{r^2}{(1-r)^2} where r=\frac{1}{2}

    =\frac{\left(\frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)^2}

    =\frac{\left(\frac{1}{2}\right)^2}{\left(\frac{1}{  2}\right)^2}

    =1
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  8. #8
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    Quote Originally Posted by acevipa View Post
    Ok, so is this correct:

    \frac{d}{dr} \left(\sum_{k=0}^{+\infty} r^k\right) = \frac{d}{dr} \left(\frac{1}{1-r}\right) for |r|<1.

    =\frac{1}{(1-r)^2}

    \Rightarrow \sum_{k=0}^{+\infty}\frac{d}{dr} \left( r^k \right) = <br />
\sum_{k=0}^{+\infty}k r^{k-1}

    \Rightarrow \sum_{k=0}^{+\infty}k r^{k+1} = \frac{r^2}{(1-r)^2} where r=\frac{1}{2}

    =\frac{\left(\frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)^2}

    =\frac{\left(\frac{1}{2}\right)^2}{\left(\frac{1}{  2}\right)^2}

    =1
    Yes. (Verification is simple using the well known formula given here: Geometric distribution - Wikipedia, the free encyclopedia).
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Yes. (Verification is simple using the well known formula given here: Geometric distribution - Wikipedia, the free encyclopedia).
    Thanks for that.
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