# Confirmation of MGF of Shifted Exponential Distribution

Consider the probability distribution $$D \sim X - a$$ where $$X \sim \text{Exp}(\lambda)$$. My task is to find the MGF of probability distribution $$D$$. I think I have a solution but it contradicts what I thought would be the intuitive solution, so I would like to confirm my work:

For $$D \sim X - 5$$ where $$X \sim \text{Exp}(\lambda)$$, the PDF of $$D$$ is given by $$f_D(x) = \lambda e^{-\lambda(x+a)} \quad \text{ for } x \geq -a$$ and so \begin{align*} M_D(t) &= \text{E}_D\left[ e^{tx} \right] \\ &= \int_{-a}^{\infty} e^{tx} \lambda e^{-\lambda(x+a)} \text{d}x \\ &= \lambda e^{-\lambda a} \int_{-a}^{\infty} e^{-x(\lambda - t)} \text{d}x \\ &= \lambda e^{-\lambda a} \left( \frac{1}{t - \lambda} \right) \left( 0 - e^{a(\lambda - t)} \right) \\ &= e^{-at} \left( \frac{\lambda}{\lambda-t} \right) \end{align*} This result seems believable, but (before doing any actual math) I had originally suspected $$M_D(t) = \frac{\lambda}{\lambda - (t+a)}$$. Are there flaws with my solution method, or does $$M_D(t) = e^{-at} \left( \frac{\lambda}{\lambda-t} \right)$$ seem reasonable? I am having a hard time convincing myself.

What is the effect of change of origin and scale on moment generating function? Specifically if $$X\mapsto \frac{X-a}{h}=: U,$$ then what is $$M_U(t)$$ and is it related to $$M_X(t)$$ in any way?
$$M_U(t) =\mathbb E\exp(tU)=\mathbb E\exp\left[\frac{-at}{h}+\frac{tX}{h}\right]=\exp\left[\frac{-at}{h}\right]\mathbb E\exp\left[\frac{tX}{h}\right]=\exp\left[\frac{-at}{h}\right] M_X\left(\frac{t}{h}\right).\tag 1$$
Using this, you can check if $$X\sim\textrm{Exp}(\lambda),$$ then $$M_D(t) =\exp\left[{-at}\right] M_X\left({t}\right)= \exp\left[{-at}\right] \left(\frac{\lambda}{\lambda-t}\right).$$