zeeshas901
New member
- Joined
- Feb 14, 2020
- Messages
- 10
Hello!
Can anyone assist regarding the following a scalar measure of the difference between to vectors, please?
Thank you!
Let [MATH]\boldsymbol{u}=[u_{1}~~u_{2}~~u_{3}][/MATH] and [MATH]\boldsymbol{v}=[v_{1}~~v_{2}~~v_{3}][/MATH] be two known row vectors of proportions. Suppose two row vectors [MATH]\boldsymbol{a}[/MATH] and [MATH]\boldsymbol{b}[/MATH] are defined as follows:
[MATH]\boldsymbol{a}=\left[\frac{u_{1}}{u_{1}+u_{2}}~~\frac{u_{2}}{u_{1}+u_{2}}\right]~\text{and}~\boldsymbol{b}=\left[\frac{v_{1}}{v_{1}+v_{2}}~~\frac{v_{2}}{v_{1}+v_{2}}\right][/MATH].
Does the following equation holds?
[MATH]f(\boldsymbol{a},\boldsymbol{b})\le f(\boldsymbol{u},\boldsymbol{v}),[/MATH]where the function for [MATH]\boldsymbol{x}[/MATH] and [MATH]\boldsymbol{y}[/MATH] is defined as:
[MATH]f(\boldsymbol{x},\boldsymbol{y})=\frac{1}{2}\sum\limits_{i=1}^{3}|y_{i}-x_{i}|.[/MATH]
I have approached the question in two ways.
Solution 1: Numerical Example
Firstly, I assumed some values for [MATH]\boldsymbol{u}=[5/10~~3/10~~2/10][/MATH] and [MATH]\boldsymbol{v}=[7/10~~2/10~~1/10][/MATH]. Then, I computed [MATH]f(\boldsymbol{u},\boldsymbol{v})=1/10[/MATH] and [MATH]f(\boldsymbol{a},\boldsymbol{b})=11/72.[/MATH] Here, clearly [MATH]f(\boldsymbol{a},\boldsymbol{b}) < f(\boldsymbol{u},\boldsymbol{v})[/MATH] holds.
Solution 2: General Case
Secondly, I tried to prove this for the general case as follows.
Consider,
[MATH]f(\boldsymbol{a},\boldsymbol{b})\le f(\boldsymbol{u},\boldsymbol{v})[/MATH]
[MATH]\frac{1}{2}\sum\limits_{i=1}^{2}|\frac{v_{i}}{v_{1}+v_{2}}-\frac{u_{i}}{u_{1}+u_{2}}|\le\frac{1}{2}\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{(v_{1}+v_{2})(u_{1}+u_{2})}\sum\limits_{i=1}^{2}|v_{i}-u_{i}|\le\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{(v_{1}+v_{2})(u_{1}+u_{2})}\le|v_{3}-u_{3}|[/MATH]
[MATH] 1 \le |v_{3}-u_{3}| (v_{1}+v_{2})(u_{1}+u_{2})[/MATH]
This last express does not look "correct" to me as [MATH]u_{i}[/MATH] and [MATH]v_{i}[/MATH] are proportions so the product of the LHR in the last expression "cannot" be greater than "1" (Isn't it?). Also note [MATH]u_{1}+u_{2}+u_{3}=1[/MATH] and [MATH]v_{1}+v_{2}+v_{3}=1[/MATH].
I am wondering where have I make a mistake in the second or first method? If both are correct, then why I am getting a different answer? Moreover, how may I correct if there is any mistake it?
Thank you!
Can anyone assist regarding the following a scalar measure of the difference between to vectors, please?
Thank you!
Let [MATH]\boldsymbol{u}=[u_{1}~~u_{2}~~u_{3}][/MATH] and [MATH]\boldsymbol{v}=[v_{1}~~v_{2}~~v_{3}][/MATH] be two known row vectors of proportions. Suppose two row vectors [MATH]\boldsymbol{a}[/MATH] and [MATH]\boldsymbol{b}[/MATH] are defined as follows:
[MATH]\boldsymbol{a}=\left[\frac{u_{1}}{u_{1}+u_{2}}~~\frac{u_{2}}{u_{1}+u_{2}}\right]~\text{and}~\boldsymbol{b}=\left[\frac{v_{1}}{v_{1}+v_{2}}~~\frac{v_{2}}{v_{1}+v_{2}}\right][/MATH].
Does the following equation holds?
[MATH]f(\boldsymbol{a},\boldsymbol{b})\le f(\boldsymbol{u},\boldsymbol{v}),[/MATH]where the function for [MATH]\boldsymbol{x}[/MATH] and [MATH]\boldsymbol{y}[/MATH] is defined as:
[MATH]f(\boldsymbol{x},\boldsymbol{y})=\frac{1}{2}\sum\limits_{i=1}^{3}|y_{i}-x_{i}|.[/MATH]
I have approached the question in two ways.
Solution 1: Numerical Example
Firstly, I assumed some values for [MATH]\boldsymbol{u}=[5/10~~3/10~~2/10][/MATH] and [MATH]\boldsymbol{v}=[7/10~~2/10~~1/10][/MATH]. Then, I computed [MATH]f(\boldsymbol{u},\boldsymbol{v})=1/10[/MATH] and [MATH]f(\boldsymbol{a},\boldsymbol{b})=11/72.[/MATH] Here, clearly [MATH]f(\boldsymbol{a},\boldsymbol{b}) < f(\boldsymbol{u},\boldsymbol{v})[/MATH] holds.
Solution 2: General Case
Secondly, I tried to prove this for the general case as follows.
Consider,
[MATH]f(\boldsymbol{a},\boldsymbol{b})\le f(\boldsymbol{u},\boldsymbol{v})[/MATH]
[MATH]\frac{1}{2}\sum\limits_{i=1}^{2}|\frac{v_{i}}{v_{1}+v_{2}}-\frac{u_{i}}{u_{1}+u_{2}}|\le\frac{1}{2}\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{(v_{1}+v_{2})(u_{1}+u_{2})}\sum\limits_{i=1}^{2}|v_{i}-u_{i}|\le\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{(v_{1}+v_{2})(u_{1}+u_{2})}\le|v_{3}-u_{3}|[/MATH]
[MATH] 1 \le |v_{3}-u_{3}| (v_{1}+v_{2})(u_{1}+u_{2})[/MATH]
This last express does not look "correct" to me as [MATH]u_{i}[/MATH] and [MATH]v_{i}[/MATH] are proportions so the product of the LHR in the last expression "cannot" be greater than "1" (Isn't it?). Also note [MATH]u_{1}+u_{2}+u_{3}=1[/MATH] and [MATH]v_{1}+v_{2}+v_{3}=1[/MATH].
I am wondering where have I make a mistake in the second or first method? If both are correct, then why I am getting a different answer? Moreover, how may I correct if there is any mistake it?
Thank you!