# Convolution of Time Distributions

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1. Feb 17, 2016

### SSGD

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. Feb 17, 2016

### SSGD

Assuming the above is correct could I also combine time distributions for different process that each had N1 or N2 convolutions.

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

$P(z) = P_1(\tau_1)*P_2(\tau_2)=\frac{1}{N_1}P_1(\frac{z}{N_1})*\frac{1}{N_2}P_2(\frac{z}{N_2})$

3. Feb 21, 2016

### 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.

4. Feb 21, 2016

### SSGD

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.