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| = What is the expected total discrepancy score in a R choice task? = | |
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| = What is the expected total rank in a R choice task? = | Suppose we have R possible choices and each of these is equally likely to be the true one. |
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| Suppose we have R choices from 1 to k and each of these is equally likely to be the true rank. The expected total rank equals | If we consider a discrepancy as the difference between the true choice and the one given by a subject then |
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| $$\sum_{k=1}^R (k=1)k $$ | The expected total discrepancy of the ''absolute value'' of discrepancies equals |
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| For example | $$\sum_{k=1}^R (k=1)k $$, $$1 \leq k \leq R $$ with the average sum of the absolute values of discrepancies per rating equal to $$\frac{\sum_{k=1}^R (k=1)k}{R}$$. For example |
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| 0 1 2 3 1 | ||||||||<style="TEXT-ALIGN: center"> '''k=1''' || '''k=2''' || '''k=3''' || '''k=4''' || '''True Rank''' || ||||||||<style="TEXT-ALIGN: center"> '''0''' || '''1''' || '''2''' || '''3''' || '''1''' || ||||||||<style="TEXT-ALIGN: center"> '''1''' || '''0''' || '''1''' || '''2''' || '''2''' || ||||||||<style="TEXT-ALIGN: center"> '''2''' || '''1''' || '''0''' || '''1''' || '''3''' || ||||||||<style="TEXT-ALIGN: center"> '''3''' || '''2''' || '''1''' || '''0''' || '''4''' || |
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Expected total score assuming random guesses at true rank = 2(1+2+3)+2(1+1+2)= 20 = 1x2 + 2x3 + 3x4 = $$\sum_{k=1}^4 (k=1)k $$. The average sum of abs(discrepancies) per rating = 20/4 = 5. |
What is the expected total discrepancy score in a R choice task?
Suppose we have R possible choices and each of these is equally likely to be the true one.
If we consider a discrepancy as the difference between the true choice and the one given by a subject then
The expected total discrepancy of the absolute value of discrepancies equals
$$\sum_{k=1}^R (k=1)k $$, $$1 \leq k \leq R $$
with the average sum of the absolute values of discrepancies per rating equal to
$$\frac{\sum_{k=1}^R (k=1)k}{R}$$.
For example
R = 4
k=1 |
k=2 |
k=3 |
k=4 |
True Rank |
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Expected total score assuming random guesses at true rank = 2(1+2+3)+2(1+1+2)= 20 = 1x2 + 2x3 + 3x4 = $$\sum_{k=1}^4 (k=1)k $$.
The average sum of abs(discrepancies) per rating = 20/4 = 5.