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Re: TransformedDistribution, for a sum of M iid variables

  • To: mathgroup at smc.vnet.net
  • Subject: [mg120428] Re: TransformedDistribution, for a sum of M iid variables
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Fri, 22 Jul 2011 03:42:34 -0400 (EDT)

Clear[distSN]

distSN[1] = NormalDistribution[0, 1];

Assume that

distSN[n_] = NormalDistribution[0, Sqrt[n]];

Proof by induction

distSN[n + 1] == 
 TransformedDistribution[
  x + y, {Distributed[x, distSN[n]], 
   Distributed[y, NormalDistribution[0, 1]]}]

True

More generally,

Clear[distN]

distN[1] = NormalDistribution[m, s];

Assume

distN[n_] = NormalDistribution[n*m, Sqrt[n*s^2]];

Proof by induction

distN[n + 1] == 
  TransformedDistribution[
   x + y, {Distributed[x, distN[n]], 
    Distributed[y, NormalDistribution[m, s]]}] // ExpandAll

True

More general forms cannot be as proved as readily; however, the extensions are obvious.

Clear[dist]

dist[n_Integer?Positive] =
  NormalDistribution[Sum[m[k], {k, n}], Sqrt[Sum[s[k]^2, {k, n}]]];

dist[m_List, s_List] :=
  
  NormalDistribution[Total[m], Norm[s]] /; Length[m] == Length[s];

For example,

dist[4]

NormalDistribution[m[1] + m[2] + m[3] + m[4], 
 Sqrt[s[1]^2 + s[2]^2 + s[3]^2 + s[4]^2]]

With[{n = RandomInteger[{10, 100}]},
 Simplify[
  dist[n] == dist[Table[m[k], {k, n}], Table[s[k], {k, n}]],
  Thread[Table[s[k], {k, n}] > 0]]]

True

Testing consistency with the simpler cases:

With[{n = RandomInteger[{10, 100}]},
 (dist[n] /. {m[_] -> m, s[_] -> s}) == distN[n]]

True

With[{n = RandomInteger[{10, 100}]},
 (dist[n] /. {m[_] -> 0, s[_] -> 1}) == distSN[n]]

True


Bob Hanlon



---- paulvonhippel at yahoo <paulvonhippel at yahoo.com> wrote: 

=============
Using the TransformedDistribution function it is easy to show that the
sum of two normal variables is normal, e.g.,

Input:
TransformedDistribution[Z1+Z2,{Z1\
[Distributed]NormalDistribution[0,1],Z2\
[Distributed]NormalDistribution[0,1]}]

Output:
NormalDistribution[0, Sqrt[2]]


Likewise it wouldn't be hard to show that the sum of 3 normal
variables is normal, or 4.

But how do I show that the sum of M normal variables is normal, where
M is an arbitrary positive integer.

I'm thinking I should use the Sum function inside
TransformedDistribution, but I don't know how the notation would work.
I see that there's a UniformSumDistribution function, but that's
limited to uniform variables.

Later I will want to do similar calculations for nonnormal
distributions.

Thanks for any pointers.


--

Bob Hanlon



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