Can I express this random variable in terms of known distributions?

Question

By computing the Laplace transform of the total length of a random tree (the nested Kingman coalescent tree with coalescence rates $\gamma$ for the individuals and $\gamma'$ for the species), we would like to express it in terms of known distributions.

• When we start with one species and $n$ individuals, it is the classic Kingman $n$-coalescent and the result is known. The Laplace transform of the total length of the tree is $$\mathbb{E}\left[e^{-\sigma L_{1,n}}\right]=\prod_{j=1}^{n-1}\frac{\frac{\gamma}{2}j}{\sigma+\frac{\gamma}{2}j}=\mathbb{E}\left[e^{-\sigma\sum_{j=2}^n\mathcal{E}_j}\right],$$ where $(\mathcal{E}_j)_{2\leq j\leq n}$ are independent and $\mathcal{E}_j$ is exponentially distributed with parameter $\frac{\gamma(j-1)}{2}$.

• Now we would like to extend this result when we have more than one species. If we start with two species, $n$ individuals in the first species and 1 in the second one, the Laplace transform is $$\mathbb{E}\left[e^{-\sigma L_{2,(n,1)}}\right]=\frac{\frac{\gamma'}{2}}{\sigma+\frac{\gamma'}{2}}\prod_{j=1}^{n-1}\frac{\frac{\gamma}{2}j}{\sigma+\frac{\gamma}{2}j}=\mathbb{E}\left[e^{-\sigma\left(\mathcal{E}'_2+\sum_{j=2}^n\mathcal{E}_j\right)}\right],$$ where $\mathcal{E}'_2$ is exponentially distributed with parameter $\frac{\gamma'}{2}$ and independent from $(\mathcal{E}_j)_{2\leq j\leq n}$.

• But as soon as there is more than one individuals in the second species, we were not able to express the length in terms of known distribution. Here is the Laplace transform for 2 species and 2 individuals in each species: \begin{align}\mathbb{E}\left[e^{-\sigma L_{2,(2,2)}}\right] & = \frac{1}{4\sigma+2\gamma+\gamma'}\frac{\frac{\gamma}{2}}{\sigma+\frac{\gamma}{2}}\frac{\gamma}{\sigma+\gamma}\left(2\gamma\frac{\frac{\gamma'}{2}}{\sigma+\frac{\gamma'}{2}}+\gamma'\frac{\frac{3\gamma}{2}}{\sigma+\frac{3\gamma}{2}}\right)\\ & = \frac{\frac{\gamma}{2}}{\sigma+\frac{\gamma}{2}}\frac{\gamma}{\sigma+\gamma}\frac{\frac{3\gamma}{2}}{\sigma+\frac{3\gamma}{2}}\frac{\frac{\gamma'}{2}}{\sigma+\frac{\gamma'}{2}}\frac{4\sigma+3\gamma+\gamma'}{4\sigma+2\gamma+\gamma'}. \end{align} We recognize again some exponential distribution from the first terms, but what about the last one. Can we say something about it? Maybe in terms of some conditional Laplace transform (if it makes sense)?

Answer 1

Your last expression for the Laplace transform is incorrect, because its value at $\sigma=0$ is not $1$.

Your penultimate expression for this Laplace transform can be rewritten as follows: \begin{align}& \frac{1}{4\sigma+2\gamma+\gamma'}\frac{\frac{\gamma}{2}}{\sigma+\frac{\gamma}{2}}\frac{\gamma}{\sigma+\gamma}\left(2\gamma\frac{\frac{\gamma'}{2}}{\sigma+\frac{\gamma'}{2}}+\gamma'\frac{\frac{3\gamma}{2}}{\sigma+\frac{3\gamma}{2}}\right)\\ &=(1-p)\frac{2\gamma+\gamma'}{4\sigma+2\gamma+\gamma'}\frac{\frac{\gamma}{2}}{\sigma+\frac{\gamma}{2}}\frac{\gamma}{\sigma+\gamma}\frac{\frac{\gamma'}{2}}{\sigma+\frac{\gamma'}{2}}\\ &+p\frac{2\gamma+\gamma'}{4\sigma+2\gamma+\gamma'}\frac{\frac{\gamma}{2}}{\sigma+\frac{\gamma}{2}}\frac{\gamma}{\sigma+\gamma}\frac{\frac{3\gamma}{2}}{\sigma+\frac{3\gamma}{2}}, \end{align} where $$p:=\frac{\gamma'}{2\gamma+\gamma'}\in(0,1).$$ So, your distribution is a $(1-p,p)$-mixture of two convolutions of exponential distributions with different rates; it is also the convolution of a $(1-p,p)$-mixture of two exponential distributions with a convolution of exponential distributions.