Is this the most simplified version?

Mathematics

1 answer:

Nonamiya [84]3 years ago

6 0

Yes it is the most simplified version

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Let us analyze the following setting. You are given a circle of unit circumference. You pickkpoints on the circle independently

garik1379 [7]

Answer:

We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:

$\small \sum_{i = 1}^{k} L_i= 2\pi$

Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.

we get :

$\small \sum_{i = 1}^{k} C_i.l = 2\pi$

where C(i) is a constant coefficient obviously between 0 and 1.

$\small \sum_{i = 1}^{k} C_i= 2\pi/l$

All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]

So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.

We already know the sum so it is easy to compute the average :

$\small L_{Exp} = \frac{2\pi}{k}$