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.

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...
