expectation of Gamma distribution help

Question

If x∼Gamma(1,λ) how would i find the expected value E(e^bx) where b=aλ

I'm kinda stuck as to how to approach the question. Some help will be greatly appreciated

Thank you in advance

Answer 1

So you have $X\sim \mathcal{Gamma}(1,\lambda)$, which means that pdf of $X$ looks like $$w_X(x)=\lambda e^{-\lambda x}$$ And you need to find $\mathrm{E}\{e^{a\lambda X}\}$. Then $$\mathrm{E}\{e^{a\lambda X}\}=\int_0^\infty e^{a\lambda x}\lambda e^{-\lambda x} \;\mathrm dx=\lambda\int_0^\infty e^{-x \lambda (1-a) } \;\mathrm dx$$ Changing the variable to $t=\lambda (1-a)x$ one can get $$\mathrm{E}\{e^{a\lambda X}\}=\frac{1}{1-a}\int_0^\infty e^{-t}\;\mathrm dt=\frac{1}{1-a}$$ if $\Re{(a)}<1$.
If $a\in\mathbb{C}$ then $$ \begin{eqnarray} \mathrm{E}\{e^{a\lambda X}\}=\lambda\int_0^\infty e^{\mathrm i \Im{(a)}\lambda x}e^{\Re{(a)}\lambda x}e^{-\lambda x} \;\mathrm dx= \lambda\int_0^\infty \cos(\underbrace{\Im{(a)}\lambda}_{\omega} x)e^{-\overbrace{(1-\Re{(a)})\lambda}^p x} \;\mathrm dx&+&\\ +\lambda\int_0^\infty \sin(\underbrace{\Im{(a)}\lambda}_{\omega} x)e^{-\overbrace{(1-\Re{(a)})\lambda}^p x} \;\mathrm dx \end{eqnarray} $$ which can be treated as the unilateral Laplace transforms of $\sin(\omega x)$ and $\cos(\omega x)$. $$\mathcal{L}\left\{ \sin(\omega x) \right\}={ \omega \over s^2 + \omega^2 }, \qquad \mathcal{L}\left\{ \cos(\omega x) \right\}={ s\over s^2 + \omega^2 } $$ After simplifications one can get $\mathrm{E}\{e^{a\lambda X}\}=\frac{1}{1-\left(\Re{(a)}+\mathrm i\Im{(a)}\right)}$ if $\Re{(a)}<1$.