I don't know if this distribution has a name or not so the best I can do is describe how it is obtained. You start with two arrays of n uniform random variables U(0,1) where n>2. I will provide a numeric example for the case of n=5 to make it easier to explain; however, n can take any value above 2.

So, for the case of n=5, the random variables for the first array may take the values

[0.8147 0.9058 0.1270 0.9134 0.6324]

and for the second array,

[ 0.0975 0.2785 0.5469 0.9575 0.9649].

You then subtract the minimum value of each array from it's respective array and obtain for the first array

[0.6877 0.7788 0 0.7864 0.5054]

and for the second,

[0 0.1810 0.4493 0.8600 0.8673].

You then divide by the new maximum, yielding for the first array

[0.8746 0.9904 0 1.0000 0.6426]

and for the second,

[0 0.2086 0.5181 0.9915 1.0000].

From henceforth I will refer to the arrays as A and B respectively. You now perform the operation 1-abs(A-B) yielding

[0.1254 0.2182 0.4819 0.9915 0.6426].

You then calculate the mean of the above array, yielding the final variable for which the probability density function of interest is obtained. I will refer to this variable as Z. For this case Z=0.4919, however, Z can of course take on any value between 0 and 1 for any n>2.

I would like to know the mean and variance of Z as a function of n.

On a separate but related note, I think it's worth mentioning that when I simulate 1,000,000 trials of the process for obtaining Z for n=3 in MATLAB and plot a histogram of the values obtained for Z to visualize the distribution, I obtain a distribution that is the same form as the first figure shown in this Wikipedia article about the Central Limit Theorem:


The histogram I obtain is below enter image description here

and the figure from the Wikipedia article is shown below as well enter image description here

As you can see they are very similar in shape, and I doubt this is just a coincidence. Does anyone know if this particular pdf has a name that I could look up and do more research on? Sorry I know I deviated from my original question about the mean and variance as a function of n (which is still the primary thing I would like answered), but I couldn't resist pointing out the similarity between the two distributions.

  • $\begingroup$ From that wiki article you can see the density is just one they have choosen arbitrarily as a starting point! So the similarity is indeed, just a coincidence. As for your quation, I cannot see a (fast) way to find the expectation and variance analytically. Most probably you must bwe content with approximations. $\endgroup$ – kjetil b halvorsen Nov 4 '14 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.