Laplace transform of the gamma pdf; $\mathcal L\{k+1\}(s)=\frac{\mathcal L\{k\}(s)}{1+θs}$

Question

The gamma distribution has pdf $$f(t,k,θ)=\frac{t^{k-1}e^{-t/θ}}{θ^k(k-1)!}$$ Show that if the pdf's Laplace transform is $\mathcal L\{k\}(s)$, then $$\mathcal L\{k+1\}(s)=\frac{\mathcal L\{k\}(s)}{1+θs}$$ I am wondering how to show this. Do I need to apply integration by parts? Thanks.

Answer 1

HINT, we have that:

$$\mathscr{L}_t\left[t^{\text{k}-1}\cdot\frac{e^{-\frac{t}{\theta}}}{\theta^\text{k}\cdot\left(\text{k}-1\right)!}\right]_{\left(\text{s}\right)}:=\int_0^\infty t^{\text{k}-1}\cdot\frac{e^{-\frac{t}{\theta}}}{\theta^\text{k}\cdot\left(\text{k}-1\right)!}\cdot e^{-\text{s}t}\space\text{d}t\tag1$$

Now, we have that:

$$t^{\text{k}-1}\cdot\frac{e^{-\frac{t}{\theta}}}{\theta^\text{k}\cdot\left(\text{k}-1\right)!}\cdot e^{-\text{s}t}=\frac{\theta^{-\text{k}}}{\Gamma\left(\text{k}\right)}\cdot t^{\text{k}-1}\cdot\exp\left\{-t\left(\text{s}+\frac{1}{\theta}\right)\right\}\tag2$$

Try to prove that:

$$\mathscr{L}_t\left[t^{\text{k}-1}\cdot\exp\left\{-t\left(\text{s}+\frac{1}{\theta}\right)\right\}\right]_{\left(\text{s}\right)}=\left(\frac{1}{\theta}+\text{s}\right)^{-\text{k}}\cdot\Gamma\left(\text{k}\right)\tag3$$

When $\Re\left(\frac{1}{\theta}+\text{s}\right)>0\space\wedge\space\Re\left(\text{k}\right)>0$

So, we end up with:

$$\mathscr{L}_t\left[t^{\text{k}-1}\cdot\frac{e^{-\frac{t}{\theta}}}{\theta^\text{k}\cdot\left(\text{k}-1\right)!}\right]_{\left(\text{s}\right)}=\left(\frac{1}{\theta}+\text{s}\right)^{-\text{k}}\cdot\theta^{-\text{k}}\tag4$$