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I am studying frame delay variation (FDV) in packet networks. I will explain what that is (or what I interpret it to be) in more detail in a second. I'm assuming the time difference between two packets follows a normal distribution for various reasons. My empirical tests are not matching what my intuition was expecting. Allow me to explain further.

I am taking 10,000 samples from a Gaussian distribution with mean 6 and standard deviation 1. This is my latency. So for example:

5.5, 6.5, 7.0, 5.0

For each sample I use the absolute value of the difference from the previous sample to get my FDV. So for the above it would be:

1.0, 0.5, 2.0

I then average the result. For this simple case I get:

3.5/3 ~ 1.17

When I run this against 10,000 samples I was expecting the FDV to be equal to the standard deviation or the variance (both of which are 1) but instead it appears I am getting 1.25. Does this make sense to anyone? Is there some way to calculate whatever this is for a given normal distribution?

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  • $\begingroup$ what's a "FDV"? $\endgroup$ – Glen_b -Reinstate Monica Feb 6 '15 at 7:14
  • $\begingroup$ @Glen_b: Probably "frame delay variation", as mentioned above. $\endgroup$ – Aleksandr Blekh Feb 6 '15 at 7:25
  • $\begingroup$ Yep, I am trying to calculate the average frame delay variation (FDV) where I am defining FDV as the absolute value of the difference between two samples. $\endgroup$ – Pace Feb 6 '15 at 7:26
  • $\begingroup$ @AleksandrBlekh, Pace I should have been able to figure that out. Sorry to be thick. $\endgroup$ – Glen_b -Reinstate Monica Feb 6 '15 at 7:36
  • $\begingroup$ @Glen_b: No big deal, as far as I'm concerned. We're all human... $\endgroup$ – Aleksandr Blekh Feb 6 '15 at 7:41
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You are calculating $\overline{(|x_t-x_{t-1}|)}$.

But $E(|X_t-X_{t-1}|)$ is not the same thing as $\sigma_X$.

If $X_j\stackrel{iid}{\sim} N(6,1)$ then $X_t-X_{t-1}\sim N(0,2)$.

So $\frac{1}{\sqrt{2}}(X_t-X_{t-1})\sim N(0,1)$, and the absolute value of that has a $\chi_1$ distribution, with expectation $\sqrt\frac{2}{\pi}$.

Consequently $E(|X_t-X_{t-1}|)=\frac{2}{\sqrt{\pi}}\approx 1.128379$

More generally, if $X_t\sim N(\mu,\sigma^2)$ then $E(|X_t-X_{t-1}|)=\frac{2\sigma}{\sqrt{\pi}}\approx 1.128379\sigma$

Check (this is in R):

> mean(abs(diff(rnorm(10000000,6,1))))
[1] 1.128571

Pretty close. (I took a bigger sample than 10000 but that won't hurt anything)

If you're really getting around $1.25$ I think there must be some mistake in your code. [I just ran a couple of dozen sets of size 10000 and none of the averages were bigger than $1.15$, so I'm not sure how you're getting up around $1.25$. Were you maybe getting $1.\color{red}{1}\color{}25$?]

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  • $\begingroup$ You are correct, it was 1.125. Thank you. Also, thanks for the great explanation. $\endgroup$ – Pace Feb 6 '15 at 9:27

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