On a niblet of wisdom

November 17th, 2012 Leave a comment Go to comments

So I was doing something rather dull at work, basically I had a handful of expense distributions, like, 25% of the expense goes to this category and 75% to that category, and I had to aggregate them in some way, so I averaged each category out only to find out that the resultant average of buckets summed to one!  In other words, the finite average of discrete distributions (over finite, comparable categories) is a distribution itself.  I thought this was a little bit surprising, so I'll just elucidate the proof here.

Claim:  Take n \in \mathbb{Z_+} discrete distributions over m \in \mathbb{Z_+} comparable categories.  They are distributions because \forall n, \sum_m p_n(x_{m}) = 1.  Then \sum_m \frac{\sum_n p_n(x_{m})}{n} = 1.  In other words, the (finite) average of each of the categories yields a new (probability) distribution, with \sum_m Av_n(p_n(x_{m})) = 1.

Proof:  We know \forall n, \sum_m p_n(x_{m}) = 1.  Thus, n such distributions must sum to n:  \sum_m p_1(x_{m}) + \ldots + \sum_m p_n(x_{m}) = n.  In other words, \sum_n \sum_m p_n(x_{m}) = n.  Absolute convergence of the sums (they are, after all, finite sums) allows us to switch the order of the sums, and \sum_m \sum_n p_n(x_{m}) = n.  Finally, a division by n yields the desired result: \sum_m \frac{\sum_np_n(x_{m})}{n} = 1.

This can be extrapolated to a finite number of continuous distributions (averaging finitely point-wise across the distributions).

Claim:  Take n \in \mathbb{Z_+} continuous probability distributions f_n(x), so that \int_\mathbb{R} f_n(x) dx = 1,  \forall n.  Then \int_{\mathbb{R}} \frac{\sum_nf_n(x)}{n} dx = 1.

Proof:  Again n distributions sum to n, and we have \int_\mathbb{R} f_1(x) dx + \ldots + \int_\mathbb{R} f_n(x) dx = n.  The linearity of the integral operator allows us to exchange the sum within the argument, thus \int_\mathbb{R} \left( f_1(x) + \ldots + f_n(x) \right) dx = n.  Finally, dividing by n yields the desired result:  \int_{\mathbb{R}} \frac{\sum_nf_n(x)}{n} dx = 1.

One has to wonder if one cannot average infinitely, as by if there were point-wise sequences of the distributions such that they converge.  This is an interesting thought in my mind at present.

  1. Luza
    November 18th, 2012 at 21:57 | #1

    You are gonna drive me crazyyyyyy
    Ahora no voy a poder dormir por pensar en esto :p. Te voy a deber mis ojeras... GRACIAS :p

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