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.
