New answers tagged

3

There are alternatives to Poisson regression, which are typically more general. The Poisson distribution, after all, has only a single parameter, and is equidispersed, which is often a very questionable constraint, so most generalizations allow for overdispersion. One of the more common generalizations is the Negative Binomial distribution, where typically ...


0

After some research and the help of a comment by @Glen_b I will answer my question. The correct keyword is prediction interval; Wikipedia has this article about the subject. A recent reference is Improved Closed-Form Prediction Intervals for Binomial and Poisson Distributions (Krishnamoorthy & Peng, 2011) [PDF link].


0

You are suffering from class imbalance, because you have many instances to learn from for the first class but very few for the last. To reduce the class imbalance for a discretised continuous target variable like this, one would typically set the bin bounds such that the number of instances in each class is roughly equal. In your case, your bin bounds would ...


4

Note first that those $X_i = 0$ contribute nothing to the sum. We can exclude them by remembering that if we generate a Poisson variate $x$ with mean $\lambda$ and then a Binomial $z$ with probability parameter $p$ and size parameter equal to $x$, $z \sim \text{Poisson}(p\lambda)$; consequently the number of nonzero elements of our sum, label it $M$, $\sim \...


2

I have no experience on this field, but what about method of moments? We could work it out: \begin{align*} \mathbb{E}\left(Z\right)&=\mathbb{E}\left[\mathbb{E}\left(Z\mid N\right)\right]=\mathbb{E}\left[N\cdot\left(p_2-1+p_1+p_2\right)\right]=\lambda \left(2p_2+p_1-1\right)\\ \mathbb{E}\left(Z^2\right)&=\mathbb{E}\left[\mathbb{E}\left(Z^2\mid N\...


3

You are interested in the probability distribution associated with the distribution of the sum of $n_1$ and $n_2$ independent Poisson random variates with means $\lambda_1$ and $\lambda_2$ respectively. The sum $k$ of two (or more) independent Poisson variates is distributed Poisson with mean equal to the sum of the individual means: $$P(k|n_1, \lambda_1, ...


0

In the census document that @milos linked to we see: Of all women aged 15 to 50, 17.2 percent had one child, 23.1 percent had two children, and 18.5 percent had three or more children. and therefore 41.2% had no children. In your Poisson estimate the numbers are indeed quite different. That estimate has far too few with 0 children and too many with 3 ...


2

You can use proper scoring rules that look at the whole distribution of probabilities for your count variable. The choice of the most appropriate (proper) scoring rule in your case will depend on your objectives.


0

It depends on the distribution of the count data. Consider a Poisson regression, or a Negative Binomial Regression, depending on how your counts are distributed. That said, sometimes you can have OLs with count data: see OLS regression with count data


2

Glen's answer notes that you can check the code for this function, but I'm not sure if you know how to do this, so I'll augment his answer by showing you how. To check the code, just load the relevant library and type in the function name without any arguments: library(stats) poisson.test function (x, T = 1, r = 1, alternative = c("two.sided", "less", ...


7

When you say "another Poisson rate" ... if that other Poisson rate is derived from data then you are comparing data with data. I'll assume you mean against some prespecified/theoretical rate (i.e. that you're performing a one-sample test). You didn't state whether you were doing a one-tailed or two-tailed test. I'll discuss both What it's doing is using ...


Top 50 recent answers are included