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.

  • $\begingroup$ This seems doubtful. Which source? $\endgroup$ – Did Dec 28 '13 at 18:54
  • $\begingroup$ @Did I replied. Hope it worked. $\endgroup$ – triomphe Dec 28 '13 at 19:19
  • $\begingroup$ It did. Will have a look in a while. $\endgroup$ – Did Dec 28 '13 at 19:21
  • $\begingroup$ @Did I found another paper which does something similar. It is available on arxiv. arxiv.org/pdf/1103.2177.pdf Page 9 equation (15). But this new paper has a Poisson point process mixed up in the equations too, but nevertheless they also use a similar result to the previous paper. $\endgroup$ – triomphe Dec 28 '13 at 19:25
  • $\begingroup$ @Did (re-comment) It was Appendix A equation (22) in the previous paper to which I sent the link. $\endgroup$ – triomphe Dec 28 '13 at 19:39

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...


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...

  • $\begingroup$ Yes the sum of independent RVs has the distributions convoluted. But I don't see how the probability comes. Thanks. $\endgroup$ – triomphe Dec 28 '13 at 19:32

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