Percentage with negative numbers
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I have several competitors in an economic market, each one with a score of a strength indicator. Scores can vary in [-inf,inf]. For a subset of competitors, I need to calculate their relative importance.
With all positive scores (like in CASE A), I've been calculating relative importance using the following formula (like in percentages):
CASE A
Score 1 = 1 -- > r = 0.1111
Score 2 = 2 --> r = 0.2222
Score 3 = 6 --> r = 0.6667
How can I replicate my intent in CASE B, where I have some negative scores? Here, the above-mentioned formula does not reach my goal anymore.
CASE B
Score 1 = 1 -- > r = ?
Score 2 = 2 --> r = ?
Score 3 = 6 --> r = ?
Score 4 = -100 --> r = ?
percentages ratio
asked Nov 19 at 8:03
Forinstance
1011
|
show 1 more comment
up vote
0
down vote
favorite
I have several competitors in an economic market, each one with a score of a strength indicator. Scores can vary in [-inf,inf]. For a subset of competitors, I need to calculate their relative importance.
With all positive scores (like in CASE A), I've been calculating relative importance using the following formula (like in percentages):
CASE A
Score 1 = 1 -- > r = 0.1111
Score 2 = 2 --> r = 0.2222
Score 3 = 6 --> r = 0.6667
How can I replicate my intent in CASE B, where I have some negative scores? Here, the above-mentioned formula does not reach my goal anymore.
CASE B
Score 1 = 1 -- > r = ?
Score 2 = 2 --> r = ?
Score 3 = 6 --> r = ?
Score 4 = -100 --> r = ?
percentages ratio
asked Nov 19 at 8:03
Forinstance
1011
You should specify what is the meaning of a negative score. Do you expect that a competitor with negative score has a negative relative importance? It would be helpful if you could provide an example of the situation that you want to model.
– Dmitry
Nov 19 at 8:14
Thank you Dmitry. Negative numbers come out from the fact that the original scores are standardized with respect to all the market players and not only to the competitors in my subset. Standardization has been achieved subtracting the mean and dividing by the standard deviation. So it means below the average, but not in every case I can access the original average. Not sure if I explained myself clearly.
– Forinstance
Nov 19 at 8:18
why wouldn't you just shift all the scores by the lowest value? Say, $tilde{r}_i=r_i - min_{j}(r_j)$. For your example this will give: $r_1=101,dots,r_4=0$
– Dmitry
Nov 19 at 8:24
Because this would give me a problem, I think: I would have 101,102,106 and 0, in case B. Doing the percentage afterwards, I would get: 0.3269, 0.3301, 0.3430 and 0. In orginal scores score 2 was the double of score 1 (2 vs 1), which is however not represeted in these final numbers (0.3269 vs 0.3301). On the other hand, this is represented in case A (0.11 vs 0.22).
– Forinstance
Nov 19 at 8:35
1
OK, I'll try to come up with some plausible expression later.
– Dmitry
Nov 19 at 8:46
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have several competitors in an economic market, each one with a score of a strength indicator. Scores can vary in [-inf,inf]. For a subset of competitors, I need to calculate their relative importance.
With all positive scores (like in CASE A), I've been calculating relative importance using the following formula (like in percentages):
CASE A
Score 1 = 1 -- > r = 0.1111
Score 2 = 2 --> r = 0.2222
Score 3 = 6 --> r = 0.6667
How can I replicate my intent in CASE B, where I have some negative scores? Here, the above-mentioned formula does not reach my goal anymore.
CASE B
Score 1 = 1 -- > r = ?
Score 2 = 2 --> r = ?
Score 3 = 6 --> r = ?
Score 4 = -100 --> r = ?
percentages ratio
asked Nov 19 at 8:03
Forinstance
1011
I have several competitors in an economic market, each one with a score of a strength indicator. Scores can vary in [-inf,inf]. For a subset of competitors, I need to calculate their relative importance.
With all positive scores (like in CASE A), I've been calculating relative importance using the following formula (like in percentages):
CASE A
Score 1 = 1 -- > r = 0.1111
Score 2 = 2 --> r = 0.2222
Score 3 = 6 --> r = 0.6667
How can I replicate my intent in CASE B, where I have some negative scores? Here, the above-mentioned formula does not reach my goal anymore.
CASE B
Score 1 = 1 -- > r = ?
Score 2 = 2 --> r = ?
Score 3 = 6 --> r = ?
Score 4 = -100 --> r = ?
percentages ratio
percentages ratio
asked Nov 19 at 8:03
Forinstance
1011
asked Nov 19 at 8:03
Forinstance
1011
asked Nov 19 at 8:03
Forinstance
1011
asked Nov 19 at 8:03
Forinstance
1011
asked Nov 19 at 8:03
Forinstance
1011
1011
You should specify what is the meaning of a negative score. Do you expect that a competitor with negative score has a negative relative importance? It would be helpful if you could provide an example of the situation that you want to model.
– Dmitry
Nov 19 at 8:14
Thank you Dmitry. Negative numbers come out from the fact that the original scores are standardized with respect to all the market players and not only to the competitors in my subset. Standardization has been achieved subtracting the mean and dividing by the standard deviation. So it means below the average, but not in every case I can access the original average. Not sure if I explained myself clearly.
– Forinstance
Nov 19 at 8:18
why wouldn't you just shift all the scores by the lowest value? Say, $tilde{r}_i=r_i - min_{j}(r_j)$. For your example this will give: $r_1=101,dots,r_4=0$
– Dmitry
Nov 19 at 8:24
Because this would give me a problem, I think: I would have 101,102,106 and 0, in case B. Doing the percentage afterwards, I would get: 0.3269, 0.3301, 0.3430 and 0. In orginal scores score 2 was the double of score 1 (2 vs 1), which is however not represeted in these final numbers (0.3269 vs 0.3301). On the other hand, this is represented in case A (0.11 vs 0.22).
– Forinstance
Nov 19 at 8:35
1
OK, I'll try to come up with some plausible expression later.
– Dmitry
Nov 19 at 8:46
|
show 1 more comment
You should specify what is the meaning of a negative score. Do you expect that a competitor with negative score has a negative relative importance? It would be helpful if you could provide an example of the situation that you want to model.
– Dmitry
Nov 19 at 8:14
Thank you Dmitry. Negative numbers come out from the fact that the original scores are standardized with respect to all the market players and not only to the competitors in my subset. Standardization has been achieved subtracting the mean and dividing by the standard deviation. So it means below the average, but not in every case I can access the original average. Not sure if I explained myself clearly.
– Forinstance
Nov 19 at 8:18
why wouldn't you just shift all the scores by the lowest value? Say, $tilde{r}_i=r_i - min_{j}(r_j)$. For your example this will give: $r_1=101,dots,r_4=0$
– Dmitry
Nov 19 at 8:24
Because this would give me a problem, I think: I would have 101,102,106 and 0, in case B. Doing the percentage afterwards, I would get: 0.3269, 0.3301, 0.3430 and 0. In orginal scores score 2 was the double of score 1 (2 vs 1), which is however not represeted in these final numbers (0.3269 vs 0.3301). On the other hand, this is represented in case A (0.11 vs 0.22).
– Forinstance
Nov 19 at 8:35
1
OK, I'll try to come up with some plausible expression later.
– Dmitry
Nov 19 at 8:46
You should specify what is the meaning of a negative score. Do you expect that a competitor with negative score has a negative relative importance? It would be helpful if you could provide an example of the situation that you want to model.
– Dmitry
Nov 19 at 8:14
You should specify what is the meaning of a negative score. Do you expect that a competitor with negative score has a negative relative importance? It would be helpful if you could provide an example of the situation that you want to model.
– Dmitry
Nov 19 at 8:14
Thank you Dmitry. Negative numbers come out from the fact that the original scores are standardized with respect to all the market players and not only to the competitors in my subset. Standardization has been achieved subtracting the mean and dividing by the standard deviation. So it means below the average, but not in every case I can access the original average. Not sure if I explained myself clearly.
– Forinstance
Nov 19 at 8:18
Thank you Dmitry. Negative numbers come out from the fact that the original scores are standardized with respect to all the market players and not only to the competitors in my subset. Standardization has been achieved subtracting the mean and dividing by the standard deviation. So it means below the average, but not in every case I can access the original average. Not sure if I explained myself clearly.
– Forinstance
Nov 19 at 8:18
why wouldn't you just shift all the scores by the lowest value? Say, $tilde{r}_i=r_i - min_{j}(r_j)$. For your example this will give: $r_1=101,dots,r_4=0$
– Dmitry
Nov 19 at 8:24
why wouldn't you just shift all the scores by the lowest value? Say, $tilde{r}_i=r_i - min_{j}(r_j)$. For your example this will give: $r_1=101,dots,r_4=0$
– Dmitry
Nov 19 at 8:24
Because this would give me a problem, I think: I would have 101,102,106 and 0, in case B. Doing the percentage afterwards, I would get: 0.3269, 0.3301, 0.3430 and 0. In orginal scores score 2 was the double of score 1 (2 vs 1), which is however not represeted in these final numbers (0.3269 vs 0.3301). On the other hand, this is represented in case A (0.11 vs 0.22).
– Forinstance
Nov 19 at 8:35
Because this would give me a problem, I think: I would have 101,102,106 and 0, in case B. Doing the percentage afterwards, I would get: 0.3269, 0.3301, 0.3430 and 0. In orginal scores score 2 was the double of score 1 (2 vs 1), which is however not represeted in these final numbers (0.3269 vs 0.3301). On the other hand, this is represented in case A (0.11 vs 0.22).
– Forinstance
Nov 19 at 8:35
1
1
OK, I'll try to come up with some plausible expression later.
– Dmitry
Nov 19 at 8:46
OK, I'll try to come up with some plausible expression later.
– Dmitry
Nov 19 at 8:46
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
Let's stick to your notation and define the score of the $i$th competitor by $x_i$ and the relative importance by $r_i$.
The definition $r_i=frac{x_i}{sum_{j=1}^n |x_j|}$ has the following drawbacks:
- The relative importance can be negative if $x_i<0$. This is not easy to interpret. Let's say you have two competitors with relative importance $r_i=0.1$ and $r_j=-0.1$. How would you compare these numbers?
- Large negative score strongly influences the resulting values of the relative importance which is counterintuitive.
One possible option would be to define the relative importance using an exponential function:
$$r_i=frac{alpha^{x_i}}{sum_{j=1}^n alpha^{x_j}},$$
with the base $alpha>1$. This formulation has an advantage of penalizing very small (negative) scores. However, the relative importance grows nonlinearly with $x_i$, i.e., $x_i=2x_j$ does not imply $r_i=2r_j$.
answered Nov 21 at 13:29
Dmitry
604517
Thanks a lot! Is there any other solution that would keep the relationship linear? I.e. keeping the proportion xi=2xj?
– Forinstance
Dec 4 at 15:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let's stick to your notation and define the score of the $i$th competitor by $x_i$ and the relative importance by $r_i$.
The definition $r_i=frac{x_i}{sum_{j=1}^n |x_j|}$ has the following drawbacks:
- The relative importance can be negative if $x_i<0$. This is not easy to interpret. Let's say you have two competitors with relative importance $r_i=0.1$ and $r_j=-0.1$. How would you compare these numbers?
- Large negative score strongly influences the resulting values of the relative importance which is counterintuitive.
One possible option would be to define the relative importance using an exponential function:
$$r_i=frac{alpha^{x_i}}{sum_{j=1}^n alpha^{x_j}},$$
with the base $alpha>1$. This formulation has an advantage of penalizing very small (negative) scores. However, the relative importance grows nonlinearly with $x_i$, i.e., $x_i=2x_j$ does not imply $r_i=2r_j$.
answered Nov 21 at 13:29
Dmitry
604517
Thanks a lot! Is there any other solution that would keep the relationship linear? I.e. keeping the proportion xi=2xj?
– Forinstance
Dec 4 at 15:03
add a comment |
up vote
1
down vote
Let's stick to your notation and define the score of the $i$th competitor by $x_i$ and the relative importance by $r_i$.
The definition $r_i=frac{x_i}{sum_{j=1}^n |x_j|}$ has the following drawbacks:
- The relative importance can be negative if $x_i<0$. This is not easy to interpret. Let's say you have two competitors with relative importance $r_i=0.1$ and $r_j=-0.1$. How would you compare these numbers?
- Large negative score strongly influences the resulting values of the relative importance which is counterintuitive.
One possible option would be to define the relative importance using an exponential function:
$$r_i=frac{alpha^{x_i}}{sum_{j=1}^n alpha^{x_j}},$$
with the base $alpha>1$. This formulation has an advantage of penalizing very small (negative) scores. However, the relative importance grows nonlinearly with $x_i$, i.e., $x_i=2x_j$ does not imply $r_i=2r_j$.
answered Nov 21 at 13:29
Dmitry
604517
Thanks a lot! Is there any other solution that would keep the relationship linear? I.e. keeping the proportion xi=2xj?
– Forinstance
Dec 4 at 15:03
add a comment |
up vote
1
down vote
up vote
1
down vote
Let's stick to your notation and define the score of the $i$th competitor by $x_i$ and the relative importance by $r_i$.
The definition $r_i=frac{x_i}{sum_{j=1}^n |x_j|}$ has the following drawbacks:
- The relative importance can be negative if $x_i<0$. This is not easy to interpret. Let's say you have two competitors with relative importance $r_i=0.1$ and $r_j=-0.1$. How would you compare these numbers?
- Large negative score strongly influences the resulting values of the relative importance which is counterintuitive.
One possible option would be to define the relative importance using an exponential function:
$$r_i=frac{alpha^{x_i}}{sum_{j=1}^n alpha^{x_j}},$$
with the base $alpha>1$. This formulation has an advantage of penalizing very small (negative) scores. However, the relative importance grows nonlinearly with $x_i$, i.e., $x_i=2x_j$ does not imply $r_i=2r_j$.
answered Nov 21 at 13:29
Dmitry
604517
Let's stick to your notation and define the score of the $i$th competitor by $x_i$ and the relative importance by $r_i$.
The definition $r_i=frac{x_i}{sum_{j=1}^n |x_j|}$ has the following drawbacks:
- The relative importance can be negative if $x_i<0$. This is not easy to interpret. Let's say you have two competitors with relative importance $r_i=0.1$ and $r_j=-0.1$. How would you compare these numbers?
- Large negative score strongly influences the resulting values of the relative importance which is counterintuitive.
One possible option would be to define the relative importance using an exponential function:
$$r_i=frac{alpha^{x_i}}{sum_{j=1}^n alpha^{x_j}},$$
with the base $alpha>1$. This formulation has an advantage of penalizing very small (negative) scores. However, the relative importance grows nonlinearly with $x_i$, i.e., $x_i=2x_j$ does not imply $r_i=2r_j$.
answered Nov 21 at 13:29
Dmitry
604517
answered Nov 21 at 13:29
Dmitry
604517
answered Nov 21 at 13:29
Dmitry
604517
answered Nov 21 at 13:29
Dmitry
604517
604517
Thanks a lot! Is there any other solution that would keep the relationship linear? I.e. keeping the proportion xi=2xj?
– Forinstance
Dec 4 at 15:03
add a comment |
Thanks a lot! Is there any other solution that would keep the relationship linear? I.e. keeping the proportion xi=2xj?
– Forinstance
Dec 4 at 15:03
Thanks a lot! Is there any other solution that would keep the relationship linear? I.e. keeping the proportion xi=2xj?
– Forinstance
Dec 4 at 15:03
Thanks a lot! Is there any other solution that would keep the relationship linear? I.e. keeping the proportion xi=2xj?
– Forinstance
Dec 4 at 15:03
add a comment |
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You should specify what is the meaning of a negative score. Do you expect that a competitor with negative score has a negative relative importance? It would be helpful if you could provide an example of the situation that you want to model.
– Dmitry
Nov 19 at 8:14
Thank you Dmitry. Negative numbers come out from the fact that the original scores are standardized with respect to all the market players and not only to the competitors in my subset. Standardization has been achieved subtracting the mean and dividing by the standard deviation. So it means below the average, but not in every case I can access the original average. Not sure if I explained myself clearly.
– Forinstance
Nov 19 at 8:18
why wouldn't you just shift all the scores by the lowest value? Say, $tilde{r}_i=r_i - min_{j}(r_j)$. For your example this will give: $r_1=101,dots,r_4=0$
– Dmitry
Nov 19 at 8:24
Because this would give me a problem, I think: I would have 101,102,106 and 0, in case B. Doing the percentage afterwards, I would get: 0.3269, 0.3301, 0.3430 and 0. In orginal scores score 2 was the double of score 1 (2 vs 1), which is however not represeted in these final numbers (0.3269 vs 0.3301). On the other hand, this is represented in case A (0.11 vs 0.22).
– Forinstance
Nov 19 at 8:35
1
OK, I'll try to come up with some plausible expression later.
– Dmitry
Nov 19 at 8:46