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Thread: Steps showing (x^y)*((1-x)^(1-y))=((x/(1-x))^y)*(1-x)=(e^(zy))/(1+e^z)

  1. #1
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    Steps showing (x^y)*((1-x)^(1-y))=((x/(1-x))^y)*(1-x)=(e^(zy))/(1+e^z)

    Working on demonstration that the binomial distribution can be expressed in generic exponential form.
    But having trouble with intermediate algebra steps to show (where z=log(x/(1-x))
    (x^y)*((1-x)^(1-y))
    =((x/(1-x))^y)*(1-x)
    =(e^(zy))/(1+e^z)
    Can someone fill me in?

  2. #2
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    Hello, lumi!

    Working on demonstration that the binomial distribution can be expressed in generic exponential form.
    But having trouble with intermediate algebra steps, where .[tex]e^z \,=\,\dfrac{x}{1-x}[/tex]

    [tex]x^y(1-x)^{1-y} \;=\;\left(\dfrac{x}{1-x}\right)^y\cdot(1-x) \;=\;\dfrac{e^{zy}}{1+e^z}[/tex]

    Can someone fill me in?

    [tex]\displaystyle x^y(1-x)^{1-y} \;=\;x^y\cdot (1-x)^1
    \cdot(1-x)^{-y} \;=\; x^y\cdot(1-x)\cdot\dfrac{1}{(1-x)^y} [/tex]

    . . . . . . . . . . [tex]=\;\dfrac{x^y}{(1-x)^y}\cdot(1-x) \;=\;\left(\dfrac{x}{1-x}\right)^y\cdot(1-x) [/tex] .[1]


    Let [tex]\color{blue}{\dfrac{x}{1-x} \:=\:e^z} \quad\Rightarrow\quad x \:=\:e^z(1-x) \quad\Rightarrow \quad x \:=\:e^z - xe^z [/tex]

    . . [tex]x + xe^z \:=\:e^z \quad\Rightarrow\quad x(1+e^z) \:=\:e^z \quad\Rightarrow\quad x \:=\:\dfrac{e^z}{1+e^z} [/tex]

    . . [tex]1-x \:=\:1 - \dfrac{e^z}{1+e^z} \quad\Rightarrow\quad \color{blue}{1 - x \:=\:\dfrac{1}{1+e^z}}[/tex]


    Substitute into [1]: .[tex](e^z)^y\cdot\dfrac{1}{1+e^z} \;=\;\dfrac{e^{zy}}{1+e^z}
    [/tex]

  3. #3
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    Thx soroban!!


    Very clear, thank you very much. Goes to show there's an art to it!!
    Guess I was confused by z=log(x/(1-x)). Is that equivalent to z=ln(x/(1-x))?
    No, not numerically but doing by log10, one would just put a 10 instead of e.
    .
    Last edited by lumi; 10-02-2013 at 08:49 AM.

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