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1

Let's first write down a density for a scale-shape parameterization for a Gamma and then shift it. Taking the density from Wikipedia (which has it correct), but making the variable $z$ rather than $x$: $$f(z;\alpha ,\theta )=\frac{z^{\alpha -1}e^{-z/\theta }}{\theta ^{\alpha }\Gamma (\alpha )}\quad {\text{ for }}z>0,\quad \alpha ,\theta >0$$ ...


0

The formula for the predicted mean value in your regression is $$ \textrm{total_oop} = \exp \left(\beta_0 + \beta_1 \cdot \textrm{PI} + \beta_2 \cdot\textrm{year} + \beta_3 \cdot \textrm{PI} \cdot\textrm{year} \right) $$ where PI is a dummy variable equal to 0 if someone doesn't have private insurance and 1 if they do. The intercept ($\beta_0$) is the ...


27

When you're considering simple parametric models for the conditional distribution of data (i.e. the distribution of each group, or the expected distribution for each combination of predictor variables), and you are dealing with a positive continuous distribution, the two common choices are Gamma and log-Normal. Besides satisfying the specification of the ...


2

One method is to find the distribution. Let $N$ be the number of messages, which (as you have determined) has a geometric distribution with $$\Pr(N=n) = 2^{-n},\ n=1, 2, 3, \ldots.$$ Conditional on $N$, you also have determined the distribution of the time $T$ to message $N:$ it is Gamma with parameter $N,$ given by $$\Pr(T \le t\mid N=n) = \frac{1}{\...


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