Probability $X_1+X_2+X_3\leq a$ by Laplace transform If $X_1,X_2,X_3$ are independent random variables then can we write the probability $$P(X_1+X_2+X_3\leq a),$$ as the product of the Laplace transforms of the random variables $$P(X_1+X_2+X_3\leq a)=\mathcal L_{X_1}(a)\mathcal L_{X_2}(a)\mathcal L_{X_3}(a)$$where $a$ is a constant and $\mathcal L_{X_i}(a)$ is the Laplace transform of $X_i$ evaluated at $a$. I saw  it in a paper but they do not provide the derivation. The random variables in that paper are exponential but not sure if they use that fact in this probability.

Mixing somewhat the notations in the paper with the notations in your question, one considers a random variable 
$$\gamma=\frac{h_0}{X_1+X_2+X_3},
$$ where each $X_k$ is almost surely positive and $h_0$ is independent of $(X_k)$ and standard exponential. Thus, for every $x\gt0$, $P[h_0\gt x]=\mathrm e^{-x}$ and, for every $a\gt0$,
$$
P[\gamma\gt a]=P[h_0\gt a(X_1+X_2+X_3)]=E[P[h_0\gt a(X_1+X_2+X_3)\mid X_1+X_2+X_3]].
$$
First by definition of the distribution of $h_0$ and second by independence of $(X_k)$, this yields
$$
P[\gamma\gt a]=E[\mathrm e^{-a(X_1+X_2+X_3)}]=E[\mathrm e^{-aX_1}]\,E[\mathrm e^{-aX_2}]\,E[\mathrm e^{-aX_3}],
$$
which is the formula in the post. Later on, the specific distributions of the $X_k$ are used, but this is another story...
