3
$\begingroup$

The standard 2PL IRT model takes the following form $$P(X = 1\ \big|\ \theta, a, b) = \frac{e^{a(\theta - b)}}{1 + e^{a(\theta - b)}}$$ where $b$ represents the "item difficulty" and informs the latent trait ability, and $a$ represents the "item discriminability" and is a reflection of the ability of the item to discriminate between people or treatments with different ability levels.

For some reason, I find this formulation intuitively a bit more difficult to swallow as compared to a formulation which looks something more like $$P(X = 1\ \big|\ \theta, m, n) = \frac{e^{m\theta - n}}{1 + e^{m\theta - n}}$$ where $n$ still serves as a difficulty parameter, but I feel is more easily interpreted as a recentering of your data per item; and $m$ represents the scale of relative ability for the people or treatments you are trying to compare within that item.

Is there a reason why the first form is the standard as opposed to the suggestion parameterized with $(m, n)$? If I'm not mistaken, these would give very similar results if not identical - or am I mistaken?

$\endgroup$

1 Answer 1

3
$\begingroup$

You’re correct that the two would lead to the same distribution. The standard version gives $a$ and $b$ a more natural meaning. Holding the difficulty $b$ constant, you can adjust the slope of the sigmoid curve (the item discrimination) by adjusting $a$. You can then define reasonable prior distributions over discrimination $a$ and difficulty $b$. By contrast, what would the quantity $n=ab$ mean?

The other benefit is the connection to 1PL. The standard version of 2PL builds on 1PL while keeping the new part ($a$) separate. If you begin with some difficulty and ability estimates, you can use these to initialize a 2PL model and learn the $a$ values.

That being said, your version has the same number of free parameters. Without some prior distribution, there’s no difference in parameter tying between the two either. So again—same model, different interpretability.

$\endgroup$
4
  • $\begingroup$ If I wanted to interpret the variability in outcome for any particular item, the resultant distribution is centered around $n$ which I find cleaner than the product of two parameters $ab$. From the perspective of building on 1PL, this is actually cleaner as the introduction of the $a$ scalar actually impacts the result of $b$, where as you anticipate $n$ in $\theta - n$ and $m\theta - n$ to hold approx the same value $\endgroup$ Commented May 24, 2021 at 19:37
  • $\begingroup$ It may just be a case of preference $\endgroup$ Commented May 24, 2021 at 19:38
  • $\begingroup$ I don’t see it as “actually cleaner”. In your reparameterization, introducing $m$ for a given $\theta$ and $n$ alters both the location and the scale of the sigmoid. In the standard version, $a$ alters the scale and $b$ alters the location, independently of each other. It’s commonplace to parameterize distributions in terms of location and scale. (Which distribution is “centered around $n$“?) $\endgroup$ Commented May 24, 2021 at 19:45
  • $\begingroup$ I'm actually thinking about it from the perspective of a bayesian framework. Where if $n$ defines the center of responses per item, it's simpler to contextualize and tailor priors. You naturally would define a prior for $a$ such that a typical value is near 1, but it's odd as the center of the distribution of responses is at $ab$ $\endgroup$ Commented May 24, 2021 at 20:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.