Let $\xi_1$ and $\xi_2$ be independent random variables:
$\xi_1 \simeq Geom(1/2), \xi_2 \simeq Geom(1/6)$.
How do you find the probability mass function of $\eta=\xi_1-\xi_2$ using convolution?
[Math] Difference between two independent geometric random variables
convolutionprobability
Related Solutions
The fact that this involves "convolutions" and sums of i.i.d. random variables makes me think of trying to deduce the distribution from Moment generating functions. Using the independence of the $\xi_i$, we have (for $t<\lambda$),
$$ \begin{eqnarray*} \mathbb{E}\left[e^{t\sum\limits_{n=1}^{\nu}\xi_{n}}\right]&{}={}&\sum\limits_{r=1}^{\infty}\int{\textbf{1}}_{\left\{\xi_{1}\leq 1,\,\ldots\,,\,\xi_{r-1}\leq 1\,,\,\xi_{r}>1 \right\}}e^{t\left(\sum\limits_{n=1}^{r}z_{n}\right)}f_{z_1}\ldots f_{z_r}dz_1\ldots dz_r\newline &{}={}&\sum\limits_{r=1}^{\infty}\left(\dfrac{\lambda}{\lambda{}-{}t}\right)^r e^{t-\lambda}\left(1{}-{}e^{t-\lambda}\right)^{r-1}\newline &{}={}&e^{t-\lambda}\dfrac{\lambda}{\lambda{}-{}t}\sum\limits_{r=1}^{\infty}\left(\dfrac{\lambda}{\lambda{}-{}t}\right)^{r-1} \left(1{}-{}e^{t-\lambda}\right)^{r-1}\newline &{}={}&\left(\dfrac{e^{t-\lambda}\dfrac{\lambda}{\lambda{}-{}t}}{1{}-{}\left(\dfrac{\lambda}{\lambda{}-{}t}\right) \left(1{}-{}e^{t-\lambda}\right)}\right)\,\newline &{}={}&\dfrac{1}{1{}-{}e^{\lambda{}-{}t}t/\lambda}\,. \end{eqnarray*} $$
This looks like $\sum\limits_{n=1}^{\nu}\xi_{n}$ is trying to be exponentially distributed with "rate" $\lambda/e^{\lambda{}-{}t}$, but I do not know this functional form by heart. Any ideas?
Edit: Not knowing the explicit inverse of the final generating function form above, I thought of examining each term in the equivalent series representation: perhaps the individual terms have nicer inverses. If this is the case, then a series representation might be sufficient. If we perform the substitution $u{}={}t/\lambda$, so that $u<1$, note that the moment generating function may be re-written as
$$ \sum\limits_{r=1}^{\infty}\left(\dfrac{1}{1-u}\right)^r\left(1-e^{\lambda\left(u-1\right)}\right)^{r-1}e^{\lambda\left(u-1\right)}{}={}\sum\limits_{r=1}^{\infty}\sum\limits_{k=0}^{r-1}{r-1\choose k}\left(\dfrac{1}{1-u}\right)^r(-)^ke^{(k+1)\lambda(u-1)}\,. $$
A series representation may be obtained, therefore, if we can invert the "atomic" moment generating functions
$$ \left(\dfrac{1}{1-u}\right)^r e^{(k+1)\lambda(u-1)}\,. $$
Heuristically, we wish to solve an integral of the kind
$$ \int\limits_{-\infty}^{1}\left(\dfrac{1}{1-u}\right)^r e^{(k+1)\lambda(u-1)-xu}\,\,\mbox{d}u\,. $$
For $x<\lambda(k+1)$, the integral's solution has the form $$ (-\lambda)^{r}e^{-x}\left(\dfrac{x^{r-1}}{(r-1)!}\log(\lambda(k+1)-x){}+{}f_{r-1}(k)\right) $$
where $f_{r-1}(k)$ is a rational function involving terms of, at most, degree "$r-1$" in $k$.
(Note: the explicit solution can be obtained by integrating the expression $\dfrac{(-\lambda)^re^{-x}}{\lambda(k+1)-x}$, "$r$"-times, w.r.t "$k$". A justification of this follows by differentiating the integral we wish to solve. Note, also, that our "$u$" substitution above was merely to make this presentation look nicer and puts this solution "off" by a factor of $\lambda$: the actual solution follows analogous operations using the $t$ variable, instead).
Best Answer
$$P(\xi_1 - \xi_2 = k) = \sum_{i=1}^{\infty} P(\xi_2 = i,\xi_1 = k+i) = \sum_{i=1}^{\infty}P(\xi_1=i)P(\xi_2 = k+i),$$ and you can calculate this sum easily using the mass functions of $\xi_1$ and $\xi_2$.