|
Size: 919
Comment:
|
Size: 994
Comment:
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 6: | Line 6: |
with average score per rating $$\frac{\sum_{k=1}^R (k=1)k}{R}$$. |
What is the expected total discrepancy score in a R choice task?
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
$$\sum_{k=1}^R (k=1)k $$
with
average score per rating
$$\frac{\sum_{k=1}^R (k=1)k}{R}$$.
For example
R = 4
k=1 |
k=2 |
k=3 |
k=4 |
True Rank |
|||
0 |
1 |
2 |
3 |
1 |
|||
1 |
0 |
1 |
2 |
2 |
|||
2 |
1 |
0 |
1 |
3 |
|||
3 |
2 |
1 |
0 |
4 |
|||
Expected total score assuming random guesses at true rank = 2(1+2+3)+2(1+1+2)= 20 = 1x2 + 2x3 + 3x4 = $$\sum_{k=1}^4 (k=1)k $$.