Relating a Gamma Distribution to an Exponential one? Question related to Gamma and Exponential random variables.
Suppose I have a Gamma random variables with shape and scale parameter $m$ and $\theta$ i.e $$X\sim\Gamma(m,\theta)$$ respectively.
Can I re-write it as an exponential RV with certain rate parameter that $$X\sim \text{exp}(?)$$
If yes what would the rate parameter be?
 A: Assume that $X\sim\exp(\Lambda)$ where $\Lambda$ is almost surely positive with density $g$, then the density $f$ of $X$ is $$f(x)=\int_0^\infty \lambda \mathrm e^{-\lambda x}g(\lambda )\mathrm d\lambda ,$$ hence one is looking for some $g$ such that, for every positive $x$, $$\int_0^\infty \lambda \mathrm e^{-\lambda x}g(\lambda )\mathrm d\lambda =\frac{x^{m-1}\mathrm e^{-x/\theta}}{\Gamma(m)\theta^m}.$$ Equivalently, one asks that $$\int_0^\infty \theta\lambda \mathrm e^{-\lambda \theta x}g(\lambda )\mathrm d\lambda =\frac{x^{m-1}\mathrm e^{-x}}{\Gamma(m)}.$$ Consider the function $h$ defined by $$h(u)=g\left(\frac{u}\theta\right)\frac{u}\theta,$$ then $$\int_0^\infty \mathrm e^{-u x}h(u)\mathrm du =\frac{x^{m-1}\mathrm e^{-x}}{\Gamma(m)}.$$
To produce the factor $\mathrm e^{-x}$ on the RHS, assume that $h(u)=0$ for every $u\leqslant1$ and $h(u)=k(u-1)$ for every $u\gt1$, then one asks that $$\int_0^\infty \mathrm e^{-u x}k(u)\mathrm du =\frac{x^{m-1}}{\Gamma(m)}.$$ Note finally that, for every positive $a$,
$$\int_0^\infty \mathrm e^{-u x}u^{a-1}\mathrm du=\frac1{x^a}\int_0^\infty \mathrm e^{-u }u^{a-1}\mathrm du=\frac{\Gamma(a)}{x^a},$$ hence the identities above are solved for $a=1-m$, that is, $$k(u)=\frac1{\Gamma(m)\Gamma(1-m)u^m}=\frac{\sin(\pi m)}{\pi\,u^m},$$ and finally, $$g(\lambda)=\frac1\lambda h(\theta\lambda)=\frac{\sin(\pi m)}{\pi\,\lambda\,(\theta\lambda-1)^m}\mathbf 1_{\lambda\gt1/\theta}.$$ This yields a legitimate density for every $$0\lt m\lt1.$$ If $m=1$, $\Lambda$ has no density but one can choose $$\Lambda=\frac1\theta\ \text{almost surely}.$$ If $m\gt1$, there is no solution since the gamma density with parameters $(m,\theta)$ is increasing on an interval $x\lt x_*$ while every exponential density is decreasing on $x\gt0$.
A: The distribution is $\dfrac 1{\Gamma(m)}\cdot \left(\dfrac x\theta\right)^{m-1} e^{-x/\theta}\dfrac{dx}\theta$.
If $m=1$, then this is $e^{-x/\theta}\dfrac{dx}\theta$, and that is an exponential distribution.  The rate parameter is the reciprocal of the scale parameter, so the rate is $1/\theta$.
If $m>1$ then notice that the density (which is what is mulitplied by $dx$, thus $\dfrac 1{\Gamma(m)}\cdot \left(\dfrac x\theta\right)^{m-1} e^{-x/\theta}\dfrac1\theta$ approaches $0$ as $x\downarrow0$, and the density of an exponential distribution does not do that.  And if $m<1$ then the density approaches $\infty$ as $x\downarrow0$ and the density of the exponential distribution does not do that.
If the waiting time until the next "arrival" has a memoryless distribution with average value $\theta$, then the distribution of the waiting time until the $20$th arrival (for example) is not memoryless.  This is the case where $m=20$.  The reason it's not memoryless is that if you've waited a long time and the $20$th arrival hasn't happened yet (and you don't know how many of the first $19$ have happened) then it is more probable that the $20$th arrival will happen in the next short time interval than it would be if you had waited only a short time.  The reason is that if you've waited a long time, then it is probable that many of those first $19$ arrivals have already happened, and if you've waited only a short time, then it is not.  Since this distribution ($m=20$) is not memoryless, it is not an exponential distribution.
A: If your shape parameter m is a integer, then your Gamma-distributed variable X can be generated as a sum of m exponentially distributed variables Y each with rate parameter equal to the reciprocal of X's scale parameter. You can see this by comparing the moment-generating functions (MGFs) for both distributions and showing they are the same.
The MGF for gamma-distributed X is
$$
M_X(t)=(1-\theta t)^{-m}.
$$ The MGF for the exponentially-distributed Y's are each all
$$
M_Y(t) = (1-\theta t)^{-1}, \theta = 1/\lambda.
$$ The MGF for the sum of the m Y's would thus be
$$M_{\Sigma Y} (t) = [(1-\theta t)^{-1}]^m = (1-\theta t)^{-m} = M_X(t).$$
