# Standard errors of parameters in IRT

DeMars's Item Response Theory makes the following claims about parameter estimation in 3PL models:

1. The standard error of $$b$$ (difficulty) will be smaller when there are more examinees with $$\theta$$ (trait level) values near $$b$$.
2. The $$a$$ (discrimination) parameter will be estimated best when there is a range of values $$\theta$$ among the examinees.

Can anyone offer an intuition or mathematical argument for these? Is there a closed form for these parameters' SE where they are made evident?

• Are there associated page numbers with these quotes? There may be important context behind these claims. I don't believe 1) is true in general, and 2) requires some qualification and may not necessarily be true either. – philchalmers Oct 7 '19 at 16:55
• The claims start at the last paragraph of page 32, 2010 edition. No relevant context is given as far as I can see, but I also don't think the authors intended these claims as theorems, but rather as practical advice – Bananin Oct 7 '19 at 17:00

For a given question and a given examinee, the probability of a correct answer is $$logit(a(\theta - b))$$. Most precise estimations of a probability parameter come when it is near to 0.5, so its logit should be near 0, so here $$\theta$$ should be near to $$b$$.

Of course, as $$\theta$$ is assumed to distribute as a standard normal hidden variable, estimations of $$b$$ are most precise when $$b$$ is close to 0; any value far from 0 (they come from questions with a lot of imbalance between correct and incorrect answeres) generally has a bigger standard error.

I want to point out that when $$b$$ is close to 0 also estimates of $$a$$ get better, to put it in other words: better balance between correct and incorrect answers brings to better estimation of association between questions. On the other hand, with this formulation of the model you can see by yourself that $$a$$ heavily affects estimated value of $$b$$.

In contrast to that, I don't quite agree with the second claim. If $$\theta$$ presents a big range, this means that it has heavy tails, or outliers, points with great leverage that have the power to invalid the model. On the other hand, since variance of $$\theta$$ is fixed, when it has more outliers, its normal values are shrinked, so, in my understanding any good effect on $$\hat a$$ should be annulled. Let's see if someone brings up some good explanation for that claim.

Standard errors of $$a$$ and $$b$$ is given by observed Fisher information, no closed form exists for it.

• Thanks for the answer, it's great but Ido have one question still. How come you say $theta$ is assumed as a standard normal? I thought both $theta$ and the item parameters were both estimated (by JMLE or MMLE), not fixed. – Bananin Oct 17 '19 at 19:11
• $\theta$ is a latent variable which is not observed in any way, so you don't know its position (mean) nor its scale (variance). that's why, since some assumption about it is necessary, it is fixed to have mean 0 and variance 1. Normal distribution is very convenient in general and it's also the maximum entropy distribution given these constraints, but assuming it doesn't bring any further practical consequence – carlo Oct 17 '19 at 20:31
• This whole speech was about unconditioned distribution of $\theta$, after estimating the model you can infer each $\theta_i$ knowing $y_i$. Of course assumptions on its unconditional distribution are still needed, otherwise you woudn't even know what scale to use. – carlo Oct 17 '19 at 20:33