Rasch Application of the Poisson Distribution
On page 18 of the text, Rasch has defined the misreadings in two tests as avi1 and avi2, and the sum of the two as av. He has also equated the observed number of misreadings in test i with λvi, the expected number. And from equation 20 in my previous blog:
| λvi | = | τi/ζv | (20) |
So if:
| avi | = | λvi |
and:
| av | = | avi1 + avi2 |
then:
| λv | = | (τi1 + τvi2)/ζv | (24) |
Rasch calls this the "additivity of impediments" (op. cit. page 16). Leaving aside the fact that additivity is not listed in the (MS) dictionary, I can't see the point in this nomenclature just yet. He also rearranges the equation, and suddenly replaces his bold equality with an approximation:
| ζv | ≈ | ( τi1 + τvi2)/av | (25) |
I shall not discuss the remaining text on that page, because it relates solely to reading tests, and is not of general interest.
The following section refers back to the Poisson estimation, which I have restated as equation 11:
| p{a|n} | ≈ | λae-l /a! | (11) |
And having spent the previous section adding the misreadings in two reading test, he now treats them as a sequence of events, and multiplies the Poisson estimation of the probability of the first event occurring with that of the second. So he substitutes equation 20 into equation 11 for both tests and multiplies the two together.
I can't do subscripts of subscripts or subscripts of superscripts, so where needed I shall write the subscript in the same font as the parameter it qualifies. So for the first test:
| p{avi1|n} | ≈ | (τi1/ζv)avi1e-(τi1/ζv) /avi1! |
And for the second test:
| p{avi2|n} | ≈ | ( τi2/ζv)avi2e-(τi2/ζv) /avi2! |
Putting the two together:
| p{avi1,avi2} | ≈ | ( τi1/ζv)avi1( τi2/ζv)avi2e-(( τi1+ τi2)/ζv) /avi1!avi2! | ||
| = | τi1avi1τi2avi2e-((τi1+τi2)/ζv) /ζvavavi1!avi2! | (26) |
How dull is that? Rasch then cites his "additivity theorem" (op. cit. page 19), turning dull into pretentious, in my humble opinion. The formula representing this theorem is my equation 24. And when I substitute 24 into 26 I get:
| p{avi1,avi1} | = | τi1avi1 τi2avi2e-λv /ζvavavi1!avi2! | (27) |
But that doesn't look very like his equation 5.2 (op. cit. page 19). Yet it is from here that Rasch divides his equivalent of 27 into his equivalent of 26 to eliminate the personal factor and end up with an an equation containing only difficulty or impediment factors.
Without wasting time on the detail of the algebra, there are two things I don't like about all this.
The first is the reliance on approximations, which frankly don't hold true except for very small numbers. In my July 28 blog I tinkered around with some numbers and you really need to be looking at probabilities of less than 5%. This may work for an easy reading test, but it really won't work for many other tests, especially when the difficulty of the test items has been deliberately tuned to match the ability of the candidate, as happens in many modern interactive tests.
The second is the messy algebra. I'm sure there must be a simpler way of eliminating one or other parameter with consecutive sittings of the same test by different candidates or different tests by the same candidate, although I can't think of one just now.
I shall therefore omit the remainder of the chapter, in which Rasch congratulates himself on what he has done and produces a lot more messy equations. I don't mind messy equations if they produce exact answers, but if they are build on approximations, they seem to me like trying to build a house with wonky bricks. And as for my resolve to address every equation, I'll restrict that resolve to the more interesting ones.
Comments