# Tag Info

5

Consider the sum $T =\sum_{i=1}^{10} X_i,$ where $X_i \stackrel{iid}{\sim} \mathsf{Pois}(\lambda = 5).$ Then $T \sim \mathsf{Pois}(\lambda_{10}=50),$ and $P(T \le 36) + P(T \ge 65) \approx 0.048.$ In R: ppois(36, 50)  0.02375891 1-ppois(64, 50)  0.02360321 Thus an exact test of $H_0: \lambda = 5$ vs. $H_a: \lambda \ne 5$ at about the 4.8% level is to ...

2

One attractive class of models is obtained by discretizing any non-negative random variable $X.$ Let $X$ be in a distribution family $\mathcal{F}.$ Let $\alpha\gt 0$ be a number and set $$Y = \lfloor X/\alpha \rfloor.$$ This is the process of "binning" by assigning $X$ to one of the intervals $[0,\alpha),$ $[\alpha,2\alpha),$ and so on, ...

2

(2) The waiting time for the next event is $W \sim \mathsf{Exp}(\mathrm{rate\,}=\lambda = 6.)$ You seek $P(W \le 3/4) = 0.9889.$ One expects six events per hour (one every 10 min.) on average, so it is very likely the next event occurs within 45 min. In R, where pexp denotes an exponential CDF, and using R as a calculator: pexp(3/4, 6)  0.988891 1 - exp(-...

1

The support of the random vector $(X_j, Y)$ is the set $\mathcal A = \left \{ (p,q) \in \mathbb N^2 \mid p \leq q \right \}$. The distribution of $(X_j, Y)$ is given by the probabilities this vector takes for each elements of $\mathcal A$. Thus let $p \leq q$, then: \begin{align*} \mathbb P\left (X_j = p, Y= q \right ) &= \mathbb P \left (Y=q \mid X_j = ...

1

You should try to fit a Tweedie distribution, as mentionned by Tylerr, it is adapted to zero inflated count. You could find several examples on the web, like this one: https://towardsdatascience.com/insurance-risk-pricing-tweedie-approach-1d71207268fc

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