Introduce a latent <Bernoulli distribution> indicator $A_i$, with $A_i=1$ denoting a structural zero and $\Pr(A_i=1)=\pi$. Conditional on $A_i=0$, retain the <negative binomial regression> with $\lambda_i=e^{x_i^T\beta}$ and $x_i=(1,d_i)^T$. This gives a <zero-inflated negative binomial model>:
$$
\boxed{\Pr(Y_i=0)=\pi+(1-\pi)f_{\rm NB}(0;r,\lambda_i),\quad \Pr(Y_i=y)=(1-\pi)f_{\rm NB}(y;r,\lambda_i)\quad(y>0).}
$$
A participant who fishes may still catch zero, so an observed zero does not reveal the latent class. This distinguishes <zero inflation> from a <hurdle model>, which truncates the count component at zero. The marginal <expectation> becomes $(1-\pi)\lambda_i$, and the <law of total variance> gives
$$
\operatorname{Var}(Y_i)=(1-\pi)\left(\lambda_i+\frac{\lambda_i^2}{r}\right)+\pi(1-\pi)\lambda_i^2.
$$
A constant $\pi$ is a parsimonious starting model. If there is evidence that nonparticipation changes with day or other recorded predictors, use <logistic regression> for $\pi_i$ instead. Conditional <independence> of visitors remains an assumption; extra zeros alone do not establish its validity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 206 / 1 / d / Solution

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A hurdle model separately models whether a count is zero and, conditional on being positive, uses a zero-truncated count distribution. Unlike <zero inflation>, the positive-count component cannot generate zeros.

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 Incoming links: Hurdle model