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Math Help - Show that E(X)=np

  1. #1
    Member oldguynewstudent's Avatar
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    Show that E(X)=np

    Show that E(X)=np when X is a binomial random variable.

    E(X)=\sum_{x=1}^{n}xp(1-p)^{x-1}

    We need to factor out np and change the limits to n-1

    I get np\sum_{x=1}^{n-1}x(1-p)^{x-1}

    Now let y=x-1 so we get np\sum_{y=0}^{n-1}(y+1)(1-p)^{y}

    Here's where I get stuck. Could anyone show me how to arrange this into a sum of the geometric series? All hints are welcome.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    E(X)=\sum_{x=1}^{n}xp(1-p)^{x-1}
    Where did this come from? I think it should be

    \displaystyle E(X)=\sum_{k=1}^{n}k\binom{n}{k}p^k(1-p)^{n-k}

    Edit: Actually though I think the answer can be gotten in the manner of this post. I think you'll just end up with \displaystyle\sum_{k=1}^n p
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  3. #3
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    If X is binomial, then the summation is wrong; your pmf has the kernal of a geometric random variable, but if X is really geometric then you need to be summing over the positive integers.

    I'll assume you actually want to find EX for geometric X; what you want to calculate is

     E(X) = \sum_{x = 1} ^ \infty x p (1 - p) ^ {x-1}

    In which case the gimmick is to notice that <br />
x(1-p)^{x - 1} = -\frac{d}{dp}(1-p)^x .<br />

    The idea behind this is to interchange the infinite sum and the taking of the derivative, which is justified, though you would have to take a course in real analysis to see why.

    If you really do want to calculate the expected value of a binomial, your pmf is totally wrong, and it would probably help if you used the right one, i.e. what undefined posted.
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  4. #4
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by theodds View Post
    If X is binomial, then the summation is wrong; your pmf has the kernal of a geometric random variable, but if X is really geometric then you need to be summing over the positive integers.

    I'll assume you actually want to find EX for geometric X; what you want to calculate is

     E(X) = \sum_{x = 1} ^ \infty x p (1 - p) ^ {x-1}

    In which case the gimmick is to notice that <br />
x(1-p)^{x - 1} = -\frac{d}{dp}(1-p)^x .<br />

    The idea behind this is to interchange the infinite sum and the taking of the derivative, which is justified, though you would have to take a course in real analysis to see why.

    If you really do want to calculate the expected value of a binomial, your pmf is totally wrong, and it would probably help if you used the right one, i.e. what undefined posted.
    Thank you for the quick response. Yes I did have it stated incorrectly. The professor went through the whole chapter in one lecture and it was scheduled for 2 lectures in the syllabus so I am a little behind on this. Attached is the definition in the book and the problem.

    With this in mind, can you give me a hint on how to factor out np ( p is already factored out) and get the limits to y=0 to y=n-1? That is where I am having trouble. Thanks.
    Attached Thumbnails Attached Thumbnails Show that E(X)=np-3_64b.jpg   Show that E(X)=np-3_64a.jpg  
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  5. #5
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    That isn't the definition of the binomial pmf, it's the definition of the geometric pmf. The summation you want to calculate is the one posted by undefined.
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  6. #6
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by theodds View Post
    That isn't the definition of the binomial pmf, it's the definition of the geometric pmf. The summation you want to calculate is the one posted by undefined.
    Thank you so much for clearing this up. I must really concentrate on my combinatorics class now, since those problems are due tomorrow and this isn't due until next Wed. I will try to get back to this later tonight. I really appreciate all the help I get from the wonderful people here.
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  7. #7
    MHF Contributor matheagle's Avatar
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    Why not use the MGF instead here.
    Have you covered that yet?
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