zeeshas901
New member
- Joined
- Feb 14, 2020
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- 10
Let [MATH] \mathbf{x}_{1}~\text{and}~\mathbf{x}_{2} [/MATH] be two row-vectors. The distance between [MATH] \mathbf{x}_{1}~\text{and}~\mathbf{x}_{2} [/MATH] is defined by
[MATH]d(\mathbf{x}_{1},\mathbf{x}_{2})=\left[\sum\limits_{i=1}^{2}\sum\limits_{j=1}^{D} \left( \frac{x_{ij}\hat{\alpha}_{i}}{\hat{\pi}_{j}} -1 \right)^{2}\right]^{1/2},[/MATH]
where
[MATH] \hat{\pi}_{j}=\frac{(x_{1j}\hat{\alpha}_{1})^{2}+(x_{2j}\hat{\alpha}_{2})^{2}}{x_{1j}\hat{\alpha}_{1}+x_{1j}\hat{\alpha}_{2}} \qquad (j=1,2,\cdots,D), [/MATH]and
[MATH]\hat{\alpha}_{i}=\frac{\sum_{j=1}^{D}x_{ij}/\hat{\pi}_{j}}{\sum_{j=1}^{D}\left(x_{ij}/\hat{\pi}_{j}\right)^{2}} \qquad (i=1,2) [/MATH]are vaues that minimize [MATH] d(\mathbf{x}_{1},\mathbf{x}_{2};\boldsymbol{\alpha}, \boldsymbol{\pi}) [/MATH].
I need to prove that [MATH]d[/MATH] is a true metric. I have proved that [MATH]d\ge0[/MATH] and [MATH]d(\mathbf{x}_{1},\mathbf{x}_{2})=d(\mathbf{x}_{2},\mathbf{x}_{1})[/MATH] (which is clear from the definition of d). But for triangular inequality:
[MATH]d(\mathbf{x}_{1},\mathbf{x}_{2})+d(\mathbf{x}_{2},\mathbf{x}_{3})\ge d(\mathbf{x}_{1},\mathbf{x}_{3})[/MATH]I am having trouble because for each [MATH]d(\mathbf{x}_{1},\mathbf{x}_{2}), d(\mathbf{x}_{2},\mathbf{x}_{3})~\text{and}~d(\mathbf{x}_{1},\mathbf{x}_{3})[/MATH] the parameter vectors [MATH]\boldsymbol{\alpha}[/MATH] and [MATH]\boldsymbol{\pi}[/MATH].
Can anyone assist or give an idea to prove the triangular inequality, please?
Thanks
[MATH]d(\mathbf{x}_{1},\mathbf{x}_{2})=\left[\sum\limits_{i=1}^{2}\sum\limits_{j=1}^{D} \left( \frac{x_{ij}\hat{\alpha}_{i}}{\hat{\pi}_{j}} -1 \right)^{2}\right]^{1/2},[/MATH]
where
[MATH] \hat{\pi}_{j}=\frac{(x_{1j}\hat{\alpha}_{1})^{2}+(x_{2j}\hat{\alpha}_{2})^{2}}{x_{1j}\hat{\alpha}_{1}+x_{1j}\hat{\alpha}_{2}} \qquad (j=1,2,\cdots,D), [/MATH]and
[MATH]\hat{\alpha}_{i}=\frac{\sum_{j=1}^{D}x_{ij}/\hat{\pi}_{j}}{\sum_{j=1}^{D}\left(x_{ij}/\hat{\pi}_{j}\right)^{2}} \qquad (i=1,2) [/MATH]are vaues that minimize [MATH] d(\mathbf{x}_{1},\mathbf{x}_{2};\boldsymbol{\alpha}, \boldsymbol{\pi}) [/MATH].
I need to prove that [MATH]d[/MATH] is a true metric. I have proved that [MATH]d\ge0[/MATH] and [MATH]d(\mathbf{x}_{1},\mathbf{x}_{2})=d(\mathbf{x}_{2},\mathbf{x}_{1})[/MATH] (which is clear from the definition of d). But for triangular inequality:
[MATH]d(\mathbf{x}_{1},\mathbf{x}_{2})+d(\mathbf{x}_{2},\mathbf{x}_{3})\ge d(\mathbf{x}_{1},\mathbf{x}_{3})[/MATH]I am having trouble because for each [MATH]d(\mathbf{x}_{1},\mathbf{x}_{2}), d(\mathbf{x}_{2},\mathbf{x}_{3})~\text{and}~d(\mathbf{x}_{1},\mathbf{x}_{3})[/MATH] the parameter vectors [MATH]\boldsymbol{\alpha}[/MATH] and [MATH]\boldsymbol{\pi}[/MATH].
Can anyone assist or give an idea to prove the triangular inequality, please?
Thanks