Deriving the density of sum of iid Uniform distributions using Laplace Transforms.

Question

In Resnick's Adventures in Stochastic Processes, there's this example where the author derives the density of $\sum_i X_i$ where $X_i$ are iid uniform (0,1).

In the picture below, I don't understand 2 things:

1. What they mean (and why) by «$e^{-\lambda k}\lambda^{-n}$ is the transform of $\epsilon_k * g(x)$». Here, the $*$ is the symbol for convolution.
2. After deducing the form of the convolution between $\epsilon_k$ and $g(x)$, , the author states the form of the desired density. How does he get that?

Any help would be appreciated.

Answer 1

1) They mean that $$\int e^{-\lambda x} (\epsilon_k*g(x)) dx = e^{-\lambda k} \lambda^{-n}$$ This follows from the two equations (one right after "Now," and the other one right after "Furthermore,") ,and the fact that products in Laplace domain are convolutions in "normal" domain.

2) Plug the integral I just typed into the sum (right under (from Example 3.2.1)):

$$\sum_k \binom{n}{k} (-1)^k \int e^{-\lambda x} (\epsilon_k*g(x)) dx$$

And now swap sum and integral from Fubini's (Since the transform is defined, I am assuming it exists)

$$= \int e^{-\lambda x} \sum_k \binom{n}{k} (-1)^k (\epsilon_k*g(x)) dx$$

$$= E[e^{\lambda \sum X_i}]$$

$$= \int e^{-\lambda x} f(x) dx $$ where f is the density (if it exists). The result now follows from the uniqueness of transforms.