MHB Would Network Traffic Rate Estimates Follow a Pareto or Normal Distribution?

AI Thread Summary
Traffic generated from a Pareto distribution will maintain its Pareto characteristics for individual packet arrivals. However, when estimating the average traffic rate from a sample of these arrivals, the Central Limit Theorem indicates that the sample average will approximate a normal distribution under certain conditions. This means that while individual values remain Pareto distributed, their average converges towards normality. The discussion clarifies the relationship between the two distributions in the context of network traffic estimation. Understanding this distinction is crucial for accurate traffic analysis and modeling.
mloo01
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I have a question that is confusing me.

If traffic in a network is being generated from a Pareto distribution,
And I am estimating this traffic rate at each packet arrival,
If I was to take the previous x estimated rates as a sample of data,
Would the distribution of these estimated rates still be Pareto or would they be more of a Normal distribution?

Any comments would be great!
 
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if you take random samples from a pareto distribution, the individual values you draw will each be pareto distributed. However the central limit theorum says that the sample average will be approximately normal (if suitable conditions are met).ie, suppose you draw random values x_1, x_2, x_3, x_4...x_n.

Each value x_i is pareto distributed. However the sample average \bar{x} = \frac{\sum x_i}{n} is approximately normalDoes that make the distinction clearer?
 
Yes that's great thank you
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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