New answers tagged poisson-distribution
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If I understand your question correctly, your constraint is implied when you take the product to get the complete likelihood. You could write it again explicitly if you like, which would be the best approach to avoid ambiguity. (But it is generally implied given we are talking about a Poisson likelihood.) It will be true that for any data point $k_i < 0$,...
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The two concepts are entirely different.
The quasi-Poisson is often used as a fudge when modelling count data that is overdispersed. quasi-poisson, refers to the conditional distribution of the response variable; instead of assuming that this conditional distribution is a Poisson distribution with mean $\lambda_i = \exp(\eta_i)$ (where $\eta_i$ is the value ...
answered Jan 13 at 16:49
Reinstate Monica - G. Simpson
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No, the sample mean of an independent sample from a Poisson distribution is NOT Poisson distributed. It is clear from the fact that a Poisson random variable can only have integer values, but the mean of such a sample does not need to be an integer.
But, the sum of the sample ($n$ times the mean) do have a Poisson distribution, and all wanted probabilities ...
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Your question is not very clear, I will try to formulate it mathematically. Then you can verify if this formulation is what you intended. $N$ is a random number of accidents,
$$ N \mid \Lambda \sim \mathcal{Pois}(\Lambda), $$
where the Poisson mean $\Lambda$ has a Gamma distribution. Then marginally, $N$ has a negative binomial distribution.
Then for each ...
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So you assume a continuous mixture of Poisson distributions, say
$$ X \mid \Lambda=\lambda \sim \mathcal{Pois}(\lambda),$$
where $\Lambda$ then has some distribution on the positive line.
If we assume that $\Lambda$ has a gamma distribution, results that $X$ has a marginal negative binomial distribution. You asks what happens if $\Lambda$ has a normal ...
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If $Y(t) = \sum_{n=1}^\infty \mathsf 1_{(0,t]}(S_n)$ is the counting process associated with the renewal times $\{S_n\}$, then by definition $$\{Y(t) = n\} = \{S_n\leqslant Y(t) < S_{n+1}\}. $$ This is simply because $n$ renewals have occured by time $t$, and the $(n+1)^{\mathrm{th}}$ renewal has yet to occur. As for $\{S_{n+1}<t\}\subset\{S_n<t\}$, ...
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NO. The quasi-Poisson **is not a distribution* at all, it is an estimation method. There is no distribution model that leads to the quasi-Poisson estimating equations, but still it is found to be useful because it has good asymptotic properties, and is a way to get around the often unreasonable property of the Poisson distribution that the variance equals ...
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You have stated in your question that the response variable population_density (quantifying animals per square kilometer) is continuous. That is unsurprising, since that variable is the ratio of a count variable and an area value. Presumably then, it must have been based constructed from an underlying count of animals in an area, and the area_size. If you ...
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If you want to model with a Poisson distribution then your outcome data should be in counts, not converted to densities. Modeling with a Poisson in terms of counts has the simplifying advantage that the variance equals the mean, but if you convert the counts to densities then I suspect that the software wouldn't know how to proceed (although I have no ...
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As pointed out in comments, with a certain positive (maybe small, but always positive) probability, the expression $Y_1/(Y_1+Y_2)$ becomes $0/0$, so the random variable $Y_1/(Y_1+Y_2)$ is not unambiguously defined. You must clarify your question, how do you define your random variable in that case?
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Each individual barrier is likely to be differently frequent, and each person to have a different propensity to endorse barriers, so your response is going to be a sum of binaries with different p(1). If your model captures the variation in p's you may be able to model the individual answers as Bernoulli in a mixed model, but even so, the sum will be Poisson ...
answered Dec 22 '19 at 3:03
Glen_b -Reinstate Monica
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