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Archive for November, 2012

## On a niblet of wisdom

November 17th, 2012 1 comment

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! Â In other words, the finite average of discrete distributions (over finite, comparable categories) is a distribution itself. Â I thought this was a little bit surprising, so I'll just elucidate the proof here.

Claim: Â Take $n \in \mathbb{Z_+}$ discrete distributions over $m \in \mathbb{Z_+}$ comparable categories. Â They are distributions because $\forall n, \sum_m p_n(x_{m}) = 1$. Â Then $\sum_m \frac{\sum_n p_n(x_{m})}{n} = 1$. Â 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: Â We knowÂ $\forall n, \sum_m p_n(x_{m}) = 1$. Â Thus, $n$ such distributions must sum to $n$: Â $\sum_m p_1(x_{m}) + \ldots + \sum_m p_n(x_{m}) = n$. Â In other words, $\sum_n \sum_m p_n(x_{m}) = n$. Â Absolute convergence of the sums (they are, after all, finite sums) allows us to switch the order of the sums, and $\sum_m \sum_n p_n(x_{m}) = n$. Â Finally, a division by $n$ yields the desired result:Â $\sum_m \frac{\sum_n p_n(x_{m})}{n} = 1$.

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

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

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

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

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## my soul is Yours to sow

my soul is Yours to sow

in fertile land, not fallow

in richest earth to grow

and root quite deep, not shallow

i'll germinate to know

my foliage green, not sallow

so sow me head to toe,

but sow me in Your Heart to make me hallowed

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