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Math Help - A couple of questions

  1. #1
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    A couple of questions

    Hi i was working through my Mathematical Statistics book and i came across a couple of examples that have got me stumped.


    A Die is cast independently N times until a six appears on the up side of the die.
    a) Find probability p(N) and prove it is a discrete distribution
    b) Find p(N=2,4,6,8,...)
    c) Find p(N=3,6,9,12,...)
    d) Find the cdf for random variable N

    I know that the pmf is p(N) = ( {5} / {6})^{N-1}  ({1} /{6})
    And it gives the example of p(N=1,3,5,7,...) = ({6}/{11})


    Edit:
    For a) is this proof that it is a discrete distribution?
    p(N) = \sum_{x=1}^{\infty} ({5} / {6})^{N-1}  ({1} /{6}) = {1}

    Second Question:
    Let probability density f(x)=x/2 for 0<x<2 and 0 elsewhere. Compute E(1/X), compute cumulative distribution function and probability density function of Y=1/X, find the expected value and variance of Y.

    Heres what i did so far:

    f(x) = \frac{x}{2}
    E(\frac{1}{x}) = \int_{-\infty}^{\infty} (\frac{1}{x}) f(x) dx =  \int_{0}^{2} (\frac{1}{x}) (\frac{x}{2}) dx = {1}

    As for Y=1/X i have no idea how to get it. Is it a transformation? My book is terrible and i am lost.

    cdf =
    {0 if x < 0}
    {x if 0 \leq x < 1}
    {1 if x \geq 1}

    pdf = (i have no idea)
    Last edited by Xan21; September 30th 2008 at 03:20 PM.
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  2. #2
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    I am not sure about your question. But here is the example part .
    P\left( {\left\{ {1,3,5, \cdots } \right\}} \right) = \sum\limits_{k = 0}^\infty  {\left( {\frac{5}<br />
{6}} \right)^{2k} \left( {\frac{1}<br />
{6}} \right) = \frac{{\frac{1}<br />
{6}}}<br />
{{1 - \left( {\frac{5}<br />
{6}} \right)^2 }} = \frac{6}<br />
{{11}}}
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  3. #3
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    Thanks. However i dont understand how you came to get
    \frac{\frac{1}{6}}{1-({\frac{5}{6}})^2}

    but I got this for b)
    p(2,4,6,8,...) = \sum_{x=1}^{\infty} ({5} / {6})^{2x-1} ({1} /{6}) = ({5}/{11})

    and c)
    p(3,6,9,12,...) = \sum_{x=1}^{\infty} ({5}/{6})^{3x} ({1}/{6}) = ({125}/{546})

    as for d) i have no idea how to do a cdf if there is no limit...
    Last edited by Xan21; September 30th 2008 at 02:36 PM.
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  4. #4
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    Quote Originally Posted by Xan21 View Post
    p(2,4,6,8,...) = \sum_{x=1}^{\infty} ({5} / {6})^{2x} ({1} /{6}) = ({5}/{11})
    While ({5}/{11}) is correct for part a, \sum_{x=1}^{\infty} ({5} / {6})^{2x} ({1} /{6}) = \frac {25}{66} \not= \frac{5}{11}).
    How did you get that sum?

    The correct sum is p(2,4,6,8,...) = \sum_{x=1}^{\infty} ({5} / {6})^{2x-1} ({1} /{6}) = ({5}/{11})
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  5. #5
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    Oh i did put the "-1" in there.... just missed it when i was typing it up.
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  6. #6
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    The cdf for random variable N is simply equal to P(1,2,3,...N), which is \sum _{k=1} ^N (1/6)(5/6)^{k-1} = (1/6)\cdot \frac{1 - (5/6)^N}{1 - 5/6} = 1 - (5/6)^N.
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  7. #7
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    Quote Originally Posted by icemanfan View Post
    The cdf for random variable N is simply equal to P(1,2,3,...N), which is \sum _{k=1} ^N (1/6)(5/6)^{k-1} = (1/6)\cdot \frac{1 - (5/6)^N}{1 - 5/6} = 1 - (5/6)^N.
    ok i see... thanks. Added another question to the top!
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  8. #8
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    Regarding the added question: I'm not sure about this, but I think E(1/X) = \frac{1}{E(X)}.
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  9. #9
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    Quote Originally Posted by icemanfan View Post
    Regarding the added question: I'm not sure about this, but I think E(1/X) = \frac{1}{E(X)}.
    Think about this: \sum\limits_{k = 1}^\infty  {x^k }  = \frac{x}<br />
{{1 - x}}\quad  \Rightarrow \quad \sum\limits_{k = 1}^\infty  {kx^{k - 1} }  = \frac{1}<br />
{{\left( {1 - x} \right)^2 }}<br />
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  10. #10
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    Quote Originally Posted by Plato View Post
    Think about this: \sum\limits_{k = 1}^\infty  {x^k }  = \frac{x}<br />
{{1 - x}}\quad  \Rightarrow \quad \sum\limits_{k = 1}^\infty  {kx^{k - 1} }  = \frac{1}<br />
{{\left( {1 - x} \right)^2 }}<br />
    Is that to me or him? Cuz i thought i did the E(1/X) part right... thats how they do it in the book....
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