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I have a variable $X$ which I am modelling with a mixture model: $$\begin{aligned} (X|A) &\sim \mathbb{1}_{0 \leq x < w \cdot m} \cdot \frac{\text{Gamma}(\alpha,0,\beta / m)}{k_1} \\ (X|B) &\sim \mathbb{1}_{x \geq w \cdot m - a} \cdot \frac{\text{Gamma}(\alpha,-a,\beta / m)}{k_2} \\ (X|C) &\sim \mathbb{1}_{- w\cdot (1-m) < x < 0} \cdot \frac{\text{NGamma}(\alpha,0,\beta/(1-m))}{k_3} \\ (X|D) &\sim \delta_a \end{aligned}$$ Where $\text{Gamma}(\alpha,\delta,\beta)$ is the gamma distribution (with pdf $\left(\Gamma(\alpha)\right)^{-1}\beta^\alpha (x-\delta)^{\alpha-1} e^{-(x-\delta)\beta}$), the $k_i$ are normalisation constants, $\text{NGamma}$ is a "negative gamma distribution", mirrored around the $y$ axis and with support in $(-\infty,0]$, $\delta_a$ is the degenerate distribution giving probability $1$ to the event $X=a$ and $0$ everywhere else, and $P(A)+P(B)+P(C)+P(D)=1$ so that all four cases are exhaustive.

I also have a variable $Y=\sum_i^N X_i$ where $\forall i:X_i \sim X$. I would like to find the distribution of $Y$.

I heard good things about finding the characteristic function of $X$ and then taking that to the $N^{th}$ degree and finding the distribution from there, but I do not know what the characteristic functions of the truncated gammas is. Does anyone have any pointers?

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  • $\begingroup$ This is a mixture model, not a "mixed" model (which is altogether different). I changed your post to reflect that and to fix some evident typographical errors. Please review the formulas to make sure I did not make some unwarranted assumptions about your intentions. $\endgroup$ – whuber Sep 15 '16 at 17:40
  • $\begingroup$ The only correction that was unnecessary was that from $k_1$ to $k_3$ on the third formula - both have the same normalisation constant - but that's not a problem. Thank you for the corrections! $\endgroup$ – Pedro Carvalho Sep 15 '16 at 17:47
  • $\begingroup$ It was not evident that they should have the same normalization constant, because although they have the same shape, their scales differ. But I see now that the truncation point has been adjusted to guarantee $k_1=k_3$. $k_2$ apparently has a simple predictable relationship with $k_1$, too. $\endgroup$ – whuber Sep 15 '16 at 17:56
  • $\begingroup$ Yes, $k_2 = 1 - k_1$, I could've just used a single $k$ normalisation constant in all three cases. $\endgroup$ – Pedro Carvalho Sep 15 '16 at 18:07
  • $\begingroup$ @whuber while not a "mixed model" as you say, it is a mixed distribution since it has both discrete and continuous components. I suspect that may be the original intent. $\endgroup$ – Glen_b -Reinstate Monica Sep 15 '16 at 18:33

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