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

[lumi]

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?

 

[soroban]

Hello, lumi!

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

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

Can someone fill me in?

\(\displaystyle \displaystyle x^y(1-x)^{1-y} \;=\;x^y\cdot (1-x)^1
\cdot(1-x)^{-y} \;=\; x^y\cdot(1-x)\cdot\dfrac{1}{(1-x)^y} \)

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


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

. . \(\displaystyle x + xe^z \:=\:e^z \quad\Rightarrow\quad x(1+e^z) \:=\:e^z \quad\Rightarrow\quad x \:=\:\dfrac{e^z}{1+e^z} \)

. . \(\displaystyle 1-x \:=\:1 - \dfrac{e^z}{1+e^z} \quad\Rightarrow\quad \color{blue}{1 - x \:=\:\dfrac{1}{1+e^z}}\)


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

 

[lumi]

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.
.