zeeshas901
New member
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- Feb 14, 2020
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Let [MATH]{u}=[u~~1-u][/MATH] and [MATH]\boldsymbol{v}=[v~~1-v][/MATH] be two known row vectors of proportions. Furthermore, the matrix
[MATH]\boldsymbol{P}= \begin{bmatrix} p_{1} & 1-p_{1}\\ 1-p_{2} & p_{2} \end{bmatrix}. [/MATH]is a right stochastic matrix having determinant, [MATH]\det({\boldsymbol{P}})=p_{1}+p_{2}-1[/MATH]. Is there exist a matrix [MATH]\boldsymbol{D}[/MATH] for which
[MATH]f(\boldsymbol{uD},\boldsymbol{vD})=f(\boldsymbol{u},\boldsymbol{v})=\sum\limits_{i=1}^{2}|\frac{v_{i} - u_{i}}{v_{i} + u_{i}}|[/MATH]?
I have defined [MATH]\boldsymbol{D}[/MATH] as [MATH]\boldsymbol{D}=(\det({\boldsymbol{P}}))^{-1}\boldsymbol{P}\boldsymbol{R}=\text{diag}(\boldsymbol{P})[/MATH] where the matrix [MATH]\boldsymbol{R}[/MATH] is of the form
[MATH]\boldsymbol{R}= \begin{bmatrix} p_{1}p_{2} & -p_{2}(1-p_{1})\\ -p_{1}(1-p_{2}) & p_{1}p_{2} \end{bmatrix}. [/MATH]Thus, the relationship
[MATH]f(\boldsymbol{uD},\boldsymbol{vD})=f(\boldsymbol{u},\boldsymbol{v})=\sum\limits_{i=1}^{2}|\frac{v_{i} - u_{i}}{v_{i} + u_{i}}|[/MATH] holds.
Example
Consider [MATH]\boldsymbol{u}=[5/10~~5/10][/MATH], [MATH]\boldsymbol{v}=[7/10~~3/10][/MATH] and
[MATH]\boldsymbol{P}= \begin{bmatrix} 5/10 & 5/10\\ 2/10 & 8/10\\ \end{bmatrix}.[/MATH]
Thus,
[MATH]\boldsymbol{R}= \begin{bmatrix} 40/100 & -40/100\\ -10/100 & 40/100\\ \end{bmatrix}[/MATH]
and
[MATH]\boldsymbol{D}= \begin{bmatrix} 5/10 & 0\\ 0 & 8/10\\ \end{bmatrix} [/MATH]
The proposed measure of distance is [MATH]f(\boldsymbol{uD},\boldsymbol{vD})=0.417=f(\boldsymbol{u},\boldsymbol{v})[/MATH].
Question
I have two questions:
(i) Is the above method makes any sense? That is, the way I have defined [MATH]\boldsymbol{D}[/MATH] and [MATH]\boldsymbol{R}[/MATH]?
(ii) How can I generalise [MATH]\boldsymbol{R}[/MATH] for any dimension? I am stuck here.
[MATH]\boldsymbol{P}= \begin{bmatrix} p_{1} & 1-p_{1}\\ 1-p_{2} & p_{2} \end{bmatrix}. [/MATH]is a right stochastic matrix having determinant, [MATH]\det({\boldsymbol{P}})=p_{1}+p_{2}-1[/MATH]. Is there exist a matrix [MATH]\boldsymbol{D}[/MATH] for which
[MATH]f(\boldsymbol{uD},\boldsymbol{vD})=f(\boldsymbol{u},\boldsymbol{v})=\sum\limits_{i=1}^{2}|\frac{v_{i} - u_{i}}{v_{i} + u_{i}}|[/MATH]?
I have defined [MATH]\boldsymbol{D}[/MATH] as [MATH]\boldsymbol{D}=(\det({\boldsymbol{P}}))^{-1}\boldsymbol{P}\boldsymbol{R}=\text{diag}(\boldsymbol{P})[/MATH] where the matrix [MATH]\boldsymbol{R}[/MATH] is of the form
[MATH]\boldsymbol{R}= \begin{bmatrix} p_{1}p_{2} & -p_{2}(1-p_{1})\\ -p_{1}(1-p_{2}) & p_{1}p_{2} \end{bmatrix}. [/MATH]Thus, the relationship
[MATH]f(\boldsymbol{uD},\boldsymbol{vD})=f(\boldsymbol{u},\boldsymbol{v})=\sum\limits_{i=1}^{2}|\frac{v_{i} - u_{i}}{v_{i} + u_{i}}|[/MATH] holds.
Example
Consider [MATH]\boldsymbol{u}=[5/10~~5/10][/MATH], [MATH]\boldsymbol{v}=[7/10~~3/10][/MATH] and
[MATH]\boldsymbol{P}= \begin{bmatrix} 5/10 & 5/10\\ 2/10 & 8/10\\ \end{bmatrix}.[/MATH]
Thus,
[MATH]\boldsymbol{R}= \begin{bmatrix} 40/100 & -40/100\\ -10/100 & 40/100\\ \end{bmatrix}[/MATH]
and
[MATH]\boldsymbol{D}= \begin{bmatrix} 5/10 & 0\\ 0 & 8/10\\ \end{bmatrix} [/MATH]
The proposed measure of distance is [MATH]f(\boldsymbol{uD},\boldsymbol{vD})=0.417=f(\boldsymbol{u},\boldsymbol{v})[/MATH].
Question
I have two questions:
(i) Is the above method makes any sense? That is, the way I have defined [MATH]\boldsymbol{D}[/MATH] and [MATH]\boldsymbol{R}[/MATH]?
(ii) How can I generalise [MATH]\boldsymbol{R}[/MATH] for any dimension? I am stuck here.