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Convolution of Time Distributions

  1. Feb 17, 2016 #1
    I need some help to make sure my reasoning is correct. Bear with me please.

    I have a time distribution for a process and I want to construct a distribution for the time it takes to perform two processes. So I would define

    ##\tau = t + t##

    This would create a new distribution with is a convolution of the process performed twice.

    ##P(\tau) = P(t)*P(t)##

    Now could I do the same for performing the process N times

    ##\tau = t + t + ... + t = Nt##

    ##P(\tau) = P(t)*P(t)*...*P(t)##

    Could the N convolutions be performed with a change of variables instead

    ##P(\tau) = P(t)\frac{dt}{d\tau}##

    ##P(\tau) = P(\frac{\tau}{N})\frac{1}{N}##
     
  2. jcsd
  3. Feb 17, 2016 #2
    Assuming the above is correct could I also combine time distributions for different process that each had N1 or N2 convolutions.

    ##z = \tau_1+\tau_2=N_1t_1+N_2t_2##

    ##P(z) = P_1(\tau_1)*P_2(\tau_2)=\frac{1}{N_1}P_1(\frac{z}{N_1})*\frac{1}{N_2}P_2(\frac{z}{N_2})##
     
  4. Feb 21, 2016 #3

    Dale

    Staff: Mentor

    Is that correct. I know that you could form a joint distribution and then project the joint distribution down onto lines of constant sum, but I didn't know that would give the same result as a convolution. If it does, then that is convenient.

    Or you could take the Fourier transform and multiply. That would be my approach.
     
  5. Feb 21, 2016 #4
    Dale you that is a great idea!!! I didn't even think about the convolution being a product in the transformed domain. When I get a chance I'm going to do the transforms on a few different distributions and see if the above ideas work out for convolution and change of variables. Thanks.
     
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