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Gamma distribution has its pdf given by $f(t;k;\theta) = \frac{t^{k-1} e^{-t/\theta}}{\theta^k (k-1)!}$. Show that if the pdf's Laplace transform is $L_k (s)$, then $L_{k+1} (s) = \frac{L_k (s)}{1+\theta s}$.

My only success so far is writing $f(t;k+1;\theta) = \frac{t}{\theta k} f(t;k;\theta)$. I'm trying to use the definition of Laplace transform driectly but can't seem to be able to extract $L_k$ from the resulting (very complicated) equation to obtain the recurrence. Further, my notes say that for a function $g(t)$ with Laplace $G(s)$, the funtion $tg(t)$ will have its transform equal to $-dG/ds (s)$. This is pretty much the case here (bar the constants $1/\theta k$). Is there room for applying this? Thanks.

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Since $$L_{k+1}(x)=\int_0^{+\infty}\frac{t^ke^{-t/\theta}}{k!\theta^{k+1}}e^{-ts}\mathrm dt=\int_0^{+\infty}\frac{t^ke^{-t\left(s+\frac 1\theta\right)}}{k!\theta^{k+1}}\mathrm dt,$$ we obtain the wanted result after an integration by parts.

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  • $\begingroup$ Integration by parts is unnecessarily complicated. The change of variable $u=(s+1/\theta)t$ and the condition that $L(0)=1$ for every Laplace transform yield readily the expression of $L_{k+1}(s)$. But yes, to get the recursion in the question, it might be necessary to take the long road of integration by parts. $\endgroup$
    – Did
    Mar 23, 2014 at 18:10

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