I'm having trouble solving the below equation question, the idea is to solve it by using convolution,

please provide me some guidance , what am I missing?

Find the probability of the sum of two random independent equally distributed logistic variables:

unsolved equation

First clue: It will be useful to use y=exp(-x/β)when developing the term

Second clue: second clue

Any help will be greatly appreciated.

  • $\begingroup$ Do you know the convolution? Convolution $f(x)*g(x)=\int f(\tau)g(x-\tau)d\tau$. So you have to calculate the integral. I believe above equations are given to help you in the solution of the integral. $\endgroup$ – tempx Mar 27 '17 at 6:28
  • $\begingroup$ I used the following convolution equation 'P( X_g + X_f <= a) = integral g(s) F(a - s) dx when Xg is G distributed and Xf is F distributed and not the one you've noted above when trying to solve the question. It got messy when I developed the integral and I didn't find out how to simplify the equation using clue 1 and clue 2 $\endgroup$ – DeJaVo Mar 27 '17 at 6:56
  • $\begingroup$ My formula is Convolution of probability distributions $\endgroup$ – DeJaVo Mar 27 '17 at 7:39
  • $\begingroup$ They are the same formulations, except you have to write $ds$ instead of $dx$ in the integral. I believe you have to use change of variables as stated in the first clue but I am not sure. (It may be easier if you think $\alpha=0$) $\endgroup$ – tempx Mar 27 '17 at 8:05
  • 1
    $\begingroup$ @dejava I am not sure but I am not sure whether the integrand is correct. The integral at the bottom of the page assumes that the integral is over x. Use the integral command itself to make sure that the integral is over the convolution parameter. I have used the following command $$\text{integrate} (1/(1+e^-((x-α)/β)))(1/(1+e^-((x-y-α)/β))) dy, y=-\text{infinity to infinity} $$ $y$ being the convolution parameter. But the solution time exceeded, thus could not reach a solution. $\endgroup$ – tempx Mar 30 '17 at 8:12

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