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← Revision 10 as of 2013-03-08 10:17:10 ⇥
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| = What is the expected total rank in a R choice task? = | = What is the expected total discrepancy score in a R choice task? = |
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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 of the absolute value of discrepancies equals | Suppose we have R possible choices and each of these is equally likely to be the true one. |
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| $$\sum_{k=1}^R (k=1)k $$ | If we consider a discrepancy as the difference between the true choice and the one given by a subject then |
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| For example | The expected total discrepancy of the ''absolute value'' of discrepancies equals |
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| R = 4 | $$\sum_{k=1}^R (k=1)k $$, $$1 \leq k \leq R $$ 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 the table below gives all the abs(discrepancies) for the case where R = 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 = $$\sum_{k=1}^4 (k=1)k $$. | 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 the table below gives all the abs(discrepancies) for the case where R = 4.
k=1 |
k=2 |
k=3 |
k=4 |
True Rank |
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0 |
1 |
2 |
3 |
1 |
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1 |
0 |
1 |
2 |
2 |
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2 |
1 |
0 |
1 |
3 |
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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.