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.
 A: 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...
A: Isn't this just using the fact that the Laplace transform of a convolution is the product of Laplace transform (the distribution function of a sum is the convolution of the distribution functions of the summands...)? Now, granted there is a Laplace transform missing on the LHS somewhere...
