Conditions for a.s convergence of a gamma series

Question

Let $(X_n)$ be a sequence of independent random variable, $X_n$ having the gamma distribution, $$f_{X_n}(x)=\frac{\alpha_n^{p_n}}{\Gamma(p_n)}e^{-x\alpha_n}x^{p_n-1}1_{[0,+\infty[}(x).$$

Find necessary and sufficient conditions on $(\alpha_n)_n,(p_n)_n,$ so that $\sum_nX_n$ converges a.s.

Supposing that $\sum_n\frac{p_n}{\alpha_n}<+\infty$ and $\sum_n\frac{p_n}{\alpha_n^2}<+\infty,$ and since $\sum_n{Var(X_n-E[X_n]})<+\infty$ and $\sum_nX_n=\sum_n(X_n-E[X_n])+\sum_nE[X_n],$ this means that $\sum_nX_n$ converges a.s.

For the converse, I think the easiest way to do it is to use the 3 series theorem, which means we have to find conditions, so that the series $$\sum_n\frac{\alpha_n^{p_n}}{\Gamma(p_n)}\int_{1}^{+\infty}e^{-x\alpha_n}x^{p_n-1}dx=\sum_n1-\frac{\alpha_n^{p_n}}{\Gamma(p_n)}\int_0^1e^{-x\alpha_n}x^{p_n-1}dx,$$ $$\sum_{n}\frac{\alpha_n^{p_n}}{\Gamma(p_n)}\int_{0}^1e^{-x\alpha_n}x^{p_n}dx=\sum_n\frac{\alpha_n^{p_n-1}}{\Gamma(p_n)}(p_n\int_{0}^{1}e^{-x\alpha_n}x^{p_n-1}dx-e^{-\alpha_n})$$ must converge, and here I am stuck, how to continue, using this facts.

Answer 1

First of all, the converge of both series $ \sum_n\frac{ p_n}{\alpha_n}, \sum_n \frac{ p_n}{\alpha_n^2}$ are not necessary. For example, $$ p_n=n^{-3},\quad \alpha_n=n^{-1}. $$ Then $$ \sum_n\mathsf{E}[X_n]=\sum_n\frac{p_n}{\alpha_n}=\sum_n\frac1{n^2}<\infty, \qquad \sum_n\frac{p_n}{\alpha^2_n}=\sum_n\frac1{n}=\infty. $$ Hence $ \sum_n X_n <\infty $ converge a.s. from $ X_n\ge 0 $ and $ \sum_n\mathsf{E}[X_n]<\infty $.

Now we suppose $ p_n\ge \delta>0 $ and prove that if $ \sum_nX_n<\infty $, then $ \sum_n\frac{ p_n}{\alpha_n}<\infty, \sum_n \frac{p_n}{\alpha_n^2}<\infty$.

Since $ X_n $ is $ \Gamma(p_n,\alpha_n) $-distributed, then the character function of $ X_n $ is $$ \phi_n(t)=\mathsf{E}[e^{itX_n}]=\frac1{(1-it/\alpha_n)^{p_n}}, \quad \log\phi_n(t)=-p_n\log\Bigl(1-\frac{it}{\alpha_n}\Bigr). $$ Now from $ \sum_n X_n <\infty $ a.s., it is easy to deduce the follows $$ \sum_{n\ge 1}p_n\log\Bigl(1+\frac{t^2}{\alpha^2_n}\Bigr)=-\sum_{n\ge 1}\log|\phi_n(t)|^2<\infty. \; (|t|<a) \tag{1} $$ From (1) we have $ p_n\log\Bigl(1+\frac{t^2}{\alpha_n^2}\Bigr)\to0$. Using $ p_n\ge \delta>0 $, we have $\alpha_n\to\infty$ and $$\frac{p_n\log(1+t^2/\alpha_n^2)}{p_n/\alpha_n^2}\to t^2, $$ Combining (1) and above limit we get $$\sum_n\frac{p_n}{\alpha_n^2}<\infty. \tag{2} $$ Meanwihile, \begin{align*} \Bigl|\log\phi_n(t)-\frac{itp_n}{\alpha_n}\Bigr| &=\Bigl|-p_n\log\Bigl(1-\frac{it}{\alpha_n}\Bigr)-\frac{itp_n}{\alpha_n}\Bigr|\\ &\le \frac{p_n}{\alpha_n^2}t^2,\quad\text{as } \frac{|t|}{\alpha_n}<\frac12. \end{align*} Hence from (1),(2) we may get $$\Bigl|\sum_{n\ge n_t} \log\phi_n(t)-it\sum_{n\ge n_t}\frac{p_n}{\alpha_n}\Bigr| \le \sum_{n\ge n_t} \Bigl|\log\phi_n(t)-\frac{itp_n}{\alpha_n}\Bigr| \le t^2\sum_{n\ge n_t}\frac{p_n}{\alpha_n^2}. $$ and $$ \sum_n\frac{p_n}{\alpha_n}<\infty. $$ At last, the sufficient and necessary condition for $\sum_nX_n<\infty $ a.s. is $$ \sum_np_n\Bigl[\mathrm{log}\Bigl(1+\dfrac1{\alpha_n^2}\Bigr)+ \mathrm{arctg\dfrac1{\alpha_n}}\Bigr]<\infty.$$ The proof is similar to above.

Answer 2

The necessary and sufficient condition for $ \sum_nX_n<\infty $ a.s. is $$ \sum_n p_n\Big[\log\Big(1+\frac{1}{\alpha^2}\Big) + \arctan\frac1{\alpha_n}\Big]<\infty. $$

Outline of proof: Since $ X_n $ is $ \Gamma(p_n,\alpha_n) $-distributed, then $X_n\ge 0 $ and its characteristic funtion $ \phi_n(t)=\mathsf{E}[e^{itX_n}] $ has following expressions, $$ \phi_n(t)= \Big(1-\frac{it}{\alpha_n} \Big)^{-p_n} =r_n(t)e^{\theta_n(t)}, $$ and $$ -\log r_n(t)=\frac{p_n}2\log\Big(1+\frac{t^2}{\alpha^2_n}\Big)\ge 0, \quad \theta_n(t)=p_n\arctan\Big(\frac{t}{\alpha_n}\Big). $$ Now let $ S_n=\sum\limits_{j=1}^nX_j$, then \begin{align*} \psi_n(t)&\triangleq\mathsf{E}[e^{itS_n}]\\ &=\prod_{j=1}^n\phi_j(t) =\Big[\prod_{j=1}^nr_j(t)\Big]\exp\Big[\sum_{j=1}^n\theta_j(t)\Big]\\ &=\prod_{j=1}^n\Big[\Big(1+\dfrac{t^2}{\alpha_j^2}\Big)^{p_j/2}\Big] \exp\Big[i\sum_{j=1}^np_j\arctan\Big(\frac{t}{\alpha_j}\Big)\Big]. \end{align*} and \begin{align*} & \sum_nX_n<\infty \; \text{ a.s. }\\ &\iff \psi_n(t)\to \psi(t) \quad \forall t \text{ and } \lim\limits_{t\to0}\psi(t)=1.\\ &\iff\begin{cases} 0\le L(t) \triangleq -\sum\limits_n\log r_n^2(t)<\infty \; \forall t \text{ and } \lim\limits_{t\to0}L(t)=0\\ \Theta(t)\triangleq \sum\limits_n\theta_n(t) \text{ is convergent }\forall t \text{ and } \lim\limits_{t\to0}\Theta(t)=0 \end{cases}\\ &\iff\begin{cases} 0\le L(t) \triangleq \sum\limits_n p_n\log\Big(1+\dfrac{t^2}{\alpha^2_n}\Big)<\infty \;\forall t\text{ and } \lim\limits_{t\to0}L(t)=0,\\ \Theta(t)\triangleq \sum\limits_np_n\arctan\Big(\dfrac{t}{\alpha_n}\Big) \text{ is conv. }\forall t \text{ and } \lim\limits_{t\to0}\Theta(t)=0. \end{cases}\\ &\iff\begin{cases} 0\le\sum\limits_n p_n\log\Big(1+\dfrac1{\alpha^2_n}\Big)<\infty,\\ 0\le\sum\limits_np_n\arctan\Big(\dfrac1{\alpha_n}\Big)<\infty. \end{cases}\\ &\iff 0\le\sum\limits_n p_n\log\Big(1+\dfrac1{\alpha_n}\Big)<\infty. \end{align*}