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    With an unbiased epilator.

    Now I've got your attention, have a question.



    We define skewness as

    



\gamma = \dfrac{E[(X-\mu)^{3}]}{\sigma^3}

    Show that for the Poisson distribution,

    



\gamma = \mu^{-0.5}
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    (Original post by Krollo)
    With an unbiased epilator.

    Now I've got your attention, have a question.



    We define skewness as

    



\gamma = \dfrac{E[(X-\mu)^{3}]}{\sigma^3}

    Show that for the Poisson distribution,

    



\gamma = \mu^{-0.5}
    There may be a smoother way to do this, but I've never really dealt with skewness before. . .
    For a Poisson, you know that: \sigma^2=\mu.
    From here, you can expand \mathbb{E}[(X-\mu)^3], and rearrange a little, you'll need to compute the third moment of the Poisson r.v., which I get to be \mu^3+3\mu^2+\mu.
    Terms cancel, and the result follows.

    EDIT: Just realised, you're not after help are you? I'll go back to sleep.
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    (Original post by Krollo)
    ...
    Probably not the intended method, but:

    \lambda = E(X^2) - \lambda^2 \Rightarrow E(X^2) = \lambda + \lambda^2

    \displaystyle E(X^3) = e^{-\mu} \sum_{n=0}^{\infty} \frac{\mu^{x}x^3}{x!} = e^{-\mu} \sum_{n=0}^{\infty} x^2 \frac{\mu^x}{(x-1)!}

    Write: x^2 = x^2 - 1 + 1 = (x-1)(x+1) + 1 to get:

    \displaystyle \frac{E(X^3)}{e^{-\mu}} = \sum_{n=0}^{\infty} \frac{(x+1)\mu^x}{(x-2)!} + \sum_{n=0}^{\infty} \frac{\mu^x}{(x-1)!}

    Focusing on the first term and re-writing x + 1 = (x-2) + 3

    \displaystyle \frac{x+1}{(x-2)!} = \frac{1}{(x-3)!} + \frac{3}{(x-2)!}

    So E(X^3) = \mu^3 + 3 \mu^2 + \mu

    Hence: E((X-\mu)^3) = \mu^3 + 3 \mu^2 + \mu -3\mu(\mu + \mu^2) + 3\mu^3   -\mu^3 = \mu

    So: \displaystyle \frac{\mu}{\mu^{1.5}} = \frac{1}{\mu^0.5} = \mu^{-1/2} as required.
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    (Original post by Krollo)
    With an unbiased epilator.
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    (Original post by Zacken)
    Probably not the intended method, but:

    \lambda = E(X^2) - \lambda^2 \Rightarrow E(X^2) = \lambda + \lambda^2

    \displaystyle E(X^3) = e^{-\mu} \sum_{n=0}^{\infty} \frac{\mu^{x}x^3}{x!} = e^{-\mu} \sum_{n=0}^{\infty} x^2 \frac{\mu^x}{(x-1)!}

    Write: x^2 = x^2 - 1 + 1 = (x-1)(x+1) + 1 to get:

    \displaystyle \frac{E(X^3)}{e^{-\mu}} = \sum_{n=0}^{\infty} \frac{(x+1)\mu^x}{(x-2)!} + \sum_{n=0}^{\infty} \frac{\mu^x}{(x-1)!}

    Focusing on the first term and re-writing x + 1 = (x-2) + 3

    \displaystyle \frac{x+1}{(x-2)!} = \frac{1}{(x-3)!} + \frac{3}{(x-2)!}

    So E(X^3) = \mu^3 + 3 \mu^2 + \mu

    Hence: E((X-\mu)^3) = \mu^3 + 3 \mu^2 + \mu -3\mu(\mu + \mu^2) + 3\mu^3   -\mu^3 = \cdots

    Okay, what have I done wrong?
    Read the definition again...
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    (Original post by Krollo)
    Read the definition again...
    Awesome. Now that I've gotten over being illiterate, s'all good. Lovely question.
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    Now I need to find some more interesting stats questions.

    In the meantime, a silly little thing (in the sense of utterly removed from reality) for you to work on:

    The standard normal variates X, Y and Z are such that X+Y=Z. Given that they have integer standard deviations a, b and c respectively, show that an odd number of a, b and c is even.

    Posted from TSR Mobile
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    (Original post by Krollo)
    Now I need to find some more interesting stats questions.

    In the meantime, a silly little thing (in the sense of utterly removed from reality) for you to work on:

    The standard normal variates X, Y and Z are such that X+Y=Z. Given that they have integer standard deviations a, b and c respectively, show that an odd number of a, b and c is even.

    Posted from TSR Mobile
    We have: a^2 + b^2 = c^2 - which is a Pythagorean triple. Since squares are congruent to 0, 1 mod 2 we have:

    Case 1: a^2 + b^2 \equiv 0 \pmod{2} and since a^2, b^2 are squares then they are either both even or both odd. In both cases, c is even so that's an odd number of even.

    Case 2: a^2 + b^2 \equiv 1 \pmod{2} and since a^2, b^2 are squares, then one is even and the other is odd, so in both cases, c is odd so that's an odd number of even.

    I hope I haven't mucked this up somehow, I'm generally horrid at this sort of stuff. :ahee:
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    Does anyone have S1 and C1 edexcel textbooks pdf??
    Thanks in advaance )
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    (Original post by AyeshaSohail101)
    Does anyone have S1 and C1 edexcel textbooks pdf??
    Thanks in advaance )
    Please don't go off topic in someone else's thread. You can create another thread and ask.
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    (Original post by Student403)
    Please don't go off topic in someone else's thread. You can create another thread and ask.
    Oh sorry to bother you :/ actually im new here and i dont know how to create a thread yet.
    Sorry again it wont happen again
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    (Original post by AyeshaSohail101)
    Oh sorry to bother you :/ actually im new here and i dont know how to create a thread yet.
    Sorry again it wont happen again
    Oh don't worry about it I'll show you by posting a message on your profile in a sec!
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    (Original post by Student403)
    Please don't go off topic in someone else's thread. You can create another thread and ask.
    When you try and stop someone's conversation but they ignore you...#FEELSBADMAN



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    Jk, Not srs BibleThump
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    (Original post by XxKingSniprxX)
    When you try and stop someone's conversation but they ignore you...#FEELSBADMAN

    xD sh
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    (Original post by Zacken)
    I hope I haven't mucked this up somehow, I'm generally horrid at this sort of stuff. :ahee:
    It looks fine to me... but we are assuming X and Y are independent, I think

    Alsop, I thought the phrase "standard normal variates" suggests standard deviation of 1?
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    (Original post by Mathemagicien)
    It looks fine to me... but we are assuming X and Y are independent, I think

    Alsop, I thought the phrase "standard normal variates" suggests standard deviation of 1?
    Yep, we're assuming that they're independent. I don't know how to do it if don't assume independence - I've only done up to S3. :dontknow:

    Good spot, probably just a slip in Krollo's phrasing.
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    (Original post by Zacken)
    Yep, we're assuming that they're independent. I don't know how to do it if don't assume independence - I've only done up to S3. :dontknow:

    Good spot, probably just a slip in Krollo's phrasing.
    Its rather neat for dependant variables: Var(A+B)=Var(A)+2Covar(A,B)+Var( B), similar to the identity (a+b)^2 = A^2 + 2ab + b^2

    And its easy to prove using the definitions of Variance and Covariance: Var(A)=E([A-E{A}]^2), Covar(A,B)=E([A-E{A}][B-E{B}])

    You enjoying Stats then? Are you going to be doing the full 3 Alevels in Maths too?
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    (Original post by Mathemagicien)
    Its rather neat for dependant variables: Var(A+B)=Var(A)+2Covar(A,B)+Var( B), similar to the identity (a+b)^2 = A^2 + 2ab + b^2

    And its easy to prove using the definitions of Variance and Covariance: Var(A)=E([A-E{A}]^2), Covar(A,B)=E([A-E{A}][B-E{B}])

    You enjoying Stats then? Are you going to be doing the full 3 Alevels in Maths too?
    Ah, yes. Covariance. I've got no clue what that is though, just seen it in the formula booklets. Thanks for that. :yep:

    Yeah, I'm enjoying it quite a bit more than mechanics at the moment. I'm only doing it as an insurance module, no way I'll be doing AFM (even if I could, I wouldn't. My exam board doesn't offer it). You doing AFM?
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    (Original post by Zacken)
    Ah, yes. Covariance. I've got no clue what that is though, just seen it in the formula booklets. Thanks for that. :yep:

    Yeah, I'm enjoying it quite a bit more than mechanics at the moment. I'm only doing it as an insurance module, no way I'll be doing AFM (even if I could, I wouldn't. My exam board doesn't offer it). You doing AFM?
    Wait what? Edezcel IAL doesn't have AFM?
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    (Original post by Student403)
    Wait what? Edezcel IAL doesn't have AFM?
    Nah, they don't even offer modules beyond S3 and M3.
 
 
 
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