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fixed mislabeled variable
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Jonathan Wooldridge
Jonathan Wooldridge
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Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

Which expands to the polynomial: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$$x_{j}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

Which expands to the polynomial: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

Which expands to the polynomial: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{j}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

edited for clarity
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Jonathan Wooldridge
Jonathan Wooldridge
  • 11
  • 2

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

The distribution of outcomes can be modeled asWhich expands to the polynomial expansion: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

The distribution of outcomes can be modeled as the polynomial expansion: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

Which expands to the polynomial: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

link to plot
Source Link
Jonathan Wooldridge
Jonathan Wooldridge
  • 11
  • 2

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

The distribution of outcomes can be modeled as the polynomial expansion: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

The distribution of outcomes can be modeled as the polynomial expansion: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

Not sure if this directly answers your question, but it might help in further analysis.

Given a set of probability coefficients $\sum _ {i = 0}^n a_i z^i$

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$

The distribution of outcomes can be modeled as the polynomial expansion: $\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3}$

If we were to introduce multiple variables $x_{i}$ then the resulting distribution is the product of those polynomials

variable $P(z^{0})$ $P(z^{1})$ $P(z^{2})$ $P(z^{3})$
$x_{1}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{8}$ $\frac{1}{4}$
$x_{2}$ $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{9}$ $\frac{16}{45}$

$x_{1} \times x_{2} = (\frac{1}{4} z^{0} + \frac{3}{8} z^{1} + \frac{1}{8} z^{2} + \frac{1}{4} z^{3})\times(\frac{1}{3} z^{0} + \frac{1}{5} z^{1} + \frac{1}{9} z^{2} + \frac{16}{45} z^{3})\\ = \frac{4 x^6}{45}+\frac{13 x^5}{180}+\frac{71 x^4}{360}+\frac{43 x^3}{180}+\frac{13 x^2}{90}+\frac{7 x}{40}+\frac{1}{12}$

link to plot

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Jonathan Wooldridge
Jonathan Wooldridge
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Jonathan Wooldridge
Jonathan Wooldridge
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added 6 characters in body
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Jonathan Wooldridge
Jonathan Wooldridge
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deleted 7 characters in body
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Jonathan Wooldridge
Jonathan Wooldridge
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added 3 characters in body
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Jonathan Wooldridge
Jonathan Wooldridge
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Jonathan Wooldridge
Jonathan Wooldridge
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