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From: Dan Stark <interlock.public@gmail.com>
To: OCaml Mailing List <caml-list@inria.fr>
Subject: [Caml-list] How does OCaml std test Random
Date: Sun, 29 Jun 2014 12:11:48 +0100	[thread overview]
Message-ID: <CANQp=sT3a+1LynnsGfOhP0kF7K6B2-PN9A4YMqTci13j6G=fpg@mail.gmail.com> (raw)

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Hi all

I am looking at the source of Random module

https://github.com/ocaml/ocaml/blob/master/stdlib/random.ml

I found that it is tested via chi-square test, here is the test code inside:


(* Return the sum of the squares of v[i0,i1[ *)
let rec sumsq v i0 i1 =
  if i0 >= i1 then 0.0
  else if i1 = i0 + 1 then Pervasives.float v.(i0) *. Pervasives.float v.(i0)
  else sumsq v i0 ((i0+i1)/2) +. sumsq v ((i0+i1)/2) i1
;;

let chisquare g n r =
  if n <= 10 * r then invalid_arg "chisquare";
  let f = Array.make r 0 in
  for i = 1 to n do
    let t = g r in
    f.(t) <- f.(t) + 1
  done;
  let t = sumsq f 0 r
  and r = Pervasives.float r
  and n = Pervasives.float n in
  let sr = 2.0 *. sqrt r in
  (r -. sr,   (r *. t /. n) -. n,   r +. sr)
;;


I understand how the chi-square is calculated there.

What I don't understand is this comment:

(* Test functions.  Not included in the library.
   The [chisquare] function should be called with n > 10r.
   It returns a triple (low, actual, high).
   If low <= actual <= high, the [g] function passed the test,
   otherwise it failed.
*)


From my knowledge, if I get a chi-square value, I should check it against a
table with the degree of freedom and then decide whether the null
hypothesis fails or not.

Why (r -. sr, (r *. t /. n) -. n, r +. sr) can be used to check? What's the
theory behind?

thanks

Dan

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             reply	other threads:[~2014-06-29 11:11 UTC|newest]

Thread overview: 2+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2014-06-29 11:11 Dan Stark [this message]
2014-07-02 19:42 ` Damien Doligez

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