# Standard 2PL IRT Formulation

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?

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.
• 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 Commented May 24, 2021 at 19:37
• 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 $n$“?) Commented May 24, 2021 at 19:45
• 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 $ab$ Commented May 24, 2021 at 20:44