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rfaenger

09:45

Hello. How would you choose $\mu$ and $\lambda$ for a normal distribution to become similar to a poisson distribution with average number of events $\lambda$? Is there a analytical way to calculate this?

Indra

10:15

Hi, Anyone knows the answer to this:
https://stats.stackexchange.com/questions/388884/how-to-convert-irt-theta-score-to-a-percentage-score

Q: How to convert IRT theta score to a percentage score

I am trying to implement an adaptive test using 3PL IRT model. We need to screen the candidates and label their expertise as (beginner, intermediate, or expert). We also need a percentage score for each examinee. When the examinee is given a sufficient number of items, the initial estimate...

4 hours later…

whuber

14:15

@rfaenger It depends on what you mean by "similar" and how small lambda might be. Assuming lambda is large enough for the approximation to be useful, usually one matches the first two moments: the normal mean and variance are both equal to lambda.

Here's an R example: plot(0:100, dpois(0:100, 50)); curve(dnorm(x, 50, sqrt(50)), add=TRUE)

whuber

14:32

If you're willing to work a tiny bit harder, a variance-stabilizing transformation does a good job and might apply to values of lambda as small as 3 or so.

Here it is as an `R` function `f`: `
f <- function(x, lambda) {
pnorm(sqrt(x+1/2), sqrt(lambda), 1/2) - pnorm(sqrt(pmax(0, x-1/2)), sqrt(lambda), 1/2)
};

lambda <- 3;
x <- 0:(3*lambda);
y <- dpois(x, lambda);
i <- y > 1e-6;

plot(x[i], y[i], pch=21, bg="#e02020");
lines(x[i], f(x[i], lambda), type="h", add=TRUE)`

gung

15:03

@rfaenger, please ask that on the main site. There we have better facilities for asking
& answering questions (e.g. formatting options will work) and the information
will be available for people with the same question. That isn't a chat item.

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