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rubymathprobabilityfinance
Marcos
Marcos

Reputation: 4930

using probability for rounding decimals

What might be a simple Ruby way to round numbers using probability, i.e., based on how close the value is to one boundary or the other (floor or ceiling)?

For example, given a current price value of 28.33, I need to add 0.014.

Equivalent to starting with 28.34 and needing to add 0.004, but the final value must be rounded to two decimal places(which can be provided as parameter, or fixed for now).

The final value should therefore be:

The reason it occured to me this could serve best is that the application is stateless and independent across runs, but still needs to approximate the net effect of accumulating the less significant digits normally rounded into oblivion (eg. micropenny values that do have an impact over time). For example, reducing a stop-loss by some variable increment every day (subtraction like -0.014 above instead).

It would be useful to extend this method to the Float class directly.

Upvotes: 0

Views: 468

Answers (2)

Marcos
Marcos

Reputation: 4930

Wound up using this method:

class Float
  def roundProb(delta, prec=2)
   ivalue=self
   chance = rand  # range 0..1, nominally averaged at 0.5
#   puts lower=((ivalue + delta)*10**prec -0.5).round/10.0**prec  # aka floor
#   puts upper=((ivalue + delta)*10**prec +0.5).round/10.0**prec  # ceiling
   ovalue=((ivalue + delta)*10**prec +chance-0.5).round/10.0**prec  # proportional probability
   return ovalue
  rescue  
    puts $@, $!
  end
end

 28.33.roundProb(0.0533)
=> 28.39

Maybe not the most elegant approach but seems to work for the general case of any precision, default 2. Even works on Ruby 1.8.7 I'm stuck with in one case, which lacks a precision parameter to round().

Upvotes: 0

Joe Pym
Joe Pym

Reputation: 1836

How about:

rand(lower..upper) < current ? lower.round(2) : upper.round(2)

EDIT:

The above will only work if you use Ruby 1.9.3 (due to earlier versions not supporting rand in a range).

Else

random_number = rand * (upper-lower) + lower
random_number < current ? lower.round(2) : upper.round(2)

Upvotes: 2

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