Laplace functional of sum of independent uniformly distributed random variables I'm doing some of the exercises in Cinlar's "Probability and Stochastics" to better understand the material. This exercise (VI.1.17) is taken from page 247:


  Fix an integer $n \geq 1$. Let $X_1,\ldots,X_n$ be independent and uniformly distributed over $E = [0,1]$. Define M by $$Mf(\omega) = \sum_i f \circ X_i(\omega), \qquad \omega \in \Omega, $$ where $f$ is a Borel function on E. Note that this makes M a random measure on $(E,\mathcal{E})$. Compute the Laplace functional of M.


The Laplace functional is defined as a mapping $$f \mapsto \mathbb{E}[e^{-Mf}].$$
I'm having a little trouble calculating it, here's my current approach:

$$\begin{align}\mathbb{E}[e^{-Mf}] &amp;= \mathbb{E}[e^{\sum_{i=1}^n -f(X_i)}]\\
&amp;= \mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]\end{align}$$
I think the next step would be to use independence to put the product outside, and then "insert" the uniform distribution - but in the book it's not really mentioned what the uniform distribution "means", so I don't know what exactly to insert.

Some good hints or the next steps of the calculation would be appreciated.
I'm assuming:


$f$ is not a Borel set but a Borel-measurable function $\mathbb R \to \mathbb R$.
$\mathcal{E} = \mathscr{M}(E) \ or \ \mathscr{B}(E)$




$$\mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]$$

$$= \prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

Now

$$\mathbb{E}[e^{-f(X_i)}] = \int_{\Omega} e^{-f(X_i)} d\mathbb P$$

$$= \int_{E} e^{-f(x_i)} d \mathcal{L}_{X_i}(t)$$

$$= \int_{E} e^{-f(x_i)} f_{X_i}(x_i) dx_i$$

Now insert the uniform distribution $f_{X_i}(x_i) = 1_{E}(x_i)$

$$= \int_{E} e^{-f(x_i)} 1 dx_i$$

$$= \int_{E} e^{-f(x_i)} dx_i$$

Hence,

$$\prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

$$= \prod_{i=1}^n [\int_{E} e^{-f(x_i)} dx_i]$$
I'm assuming:


$f$ is not a Borel set but a Borel-measurable function $\mathbb R \to \mathbb R$.
$\mathcal{E} = \mathscr{M}(E) \ or \ \mathscr{B}(E)$




$$\mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]$$

$$= \prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

Now

$$\mathbb{E}[e^{-f(X_i)}] = \int_{\Omega} e^{-f(X_i)} d\mathbb P$$

$$= \int_{E} e^{-f(x_i)} d \mathcal{L}_{X_i}(t)$$

$$= \int_{E} e^{-f(x_i)} f_{X_i}(x_i) dx_i$$

Now insert the uniform distribution $f_{X_i}(x_i) = 1_{E}(x_i)$

$$= \int_{E} e^{-f(x_i)} 1 dx_i$$

$$= \int_{E} e^{-f(x_i)} dx_i$$

Hence,

$$\prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$

$$= \prod_{i=1}^n [\int_{E} e^{-f(x_i)} dx_i]$$