Re: Re: Fourier Transfer and a game?!?!

• To: mathgroup at smc.vnet.net
• Subject: [mg54238] Re: [mg54207] Re: Fourier Transfer and a game?!?!
• From: DrBob <drbob at bigfoot.com>
• Date: Mon, 14 Feb 2005 00:57:53 -0500 (EST)
• References: <cuf5kg\$god\$1@smc.vnet.net> <cukb23\$lmi\$1@smc.vnet.net> <200502130521.AAA03736@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

>> With the default setting of FourierParameters
>> F[Convolution[f, g]] == Sqrt[Length[f]]*F[f]*F[g]

I think you mean something like this:

F = Fourier;
IF = InverseFourier;
SetOptions[#, FourierParameters -> {0, 1}] & /@ {F, IF};
Module[{n, L, s, powers, payoffs},
payoffs = Union[Plus @@@ Distribute[Table[{0, 2500,
5000, 7500, 10000}, {3}], List]];
n = Length@payoffs; s = Sqrt@n;
L = s PadRight[{0., .35, .35, .15, .15}, n];
powers = Table[F[L]^i, {i, 0, 3}];
Transpose@{payoffs, Chop at IF[{.4, .35, .15, .1}.powers/s]}]

{{0, 0.4}, {2500, 0.1225}, {5000, 0.140875}, {7500, 0.0935375}, {
10000, 0.0994875}, {12500, 0.049875}, {15000, 0.03995}, {17500,
0.02565}, {20000, 0.015975}, {22500, 0.007425}, {25000,
0.003375}, {27500, 0.0010125}, {30000, 0.0003375}}

I'm not sure that calculation for "payoffs" would always work, however.

Bobby

On Sun, 13 Feb 2005 00:21:36 -0500 (EST), Maxim <ab_def at prontomail.com> wrote:

> On Sat, 12 Feb 2005 07:25:55 +0000 (UTC), Maxim <ab_def at prontomail.com>
> wrote:
>
>> On Thu, 10 Feb 2005 08:22:40 +0000 (UTC), elparedblanco
>> <cire1611 at gmail.com> wrote:
>>
>>> let's say we have a probabability of picking an amount of money from a
>>> vat.  I can go into the vat either 0,1,2,3 times.  The probability with
>>> which I pick from the vat is described by the vector {.4, .35,.15,.1}.
>>> I call this frequency.
>>>
>>> I can pull only four amounts of money from the vat.  The amounts are
>>> {\$2500,\$5000,\$7500,\$10000}.  The probability of picking each amount is
>>> described by the vector {.35,.35,.15,.15}. I call this severity1.
>>>
>>> Process is this.  First I choose how many times I can go into the vat.
>>> Then I go in that many times.  I always replace what I pick out.  so it
>>> is possible to win \$30000.
>>>
>>> The Question is what's the probabililty of winning certain amounts of
>>> money, such as 15,000 or 7,500 or any number, given the fact that I can
>>> pick multiple times?
>>>
>>
>>>
>>> ANSWER = {.4000, .1225, .1409, .0935, .0995, .0499, .040, .0257, .016,
>>> .0074 .0034, .0010, .0003}
>>>
>>
>> Suppose we have a model with three vats which open independently with
>> probabilities p1, p2, p3 respectively. Then the generating function (GF)
>> for the amount of money taken from the first vat is ((1 - p1)
>> + p1*(.35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000)), and similarly
>> for the other vats. Of course then the solution would be simple, because
>> to find the GF for the sum taken from the three vats we just have to
>> multiply the three GFs. So if we could find p1, p2, p3 such that the
>> probability of all three vats opening, which is p1*p2*p3, equals .1 and
>> the equations for other combinations hold as well, we would be done.
>>
>> In[1]:=
>> sol = Solve[
>>    {(1 - p1)*(1 - p2)*(1 - p3) == .4,
>>     p1*(1 - p2)*(1 - p3) + (1 - p1)*p2*(1 - p3) +
>>       (1 - p1)*(1 - p2)*p3 == .35,
>>     p1*p2*(1 - p3) + p1*(1 - p2)*p3 + (1 - p1)*p2*p3 == .15},
>>    {p1, p2, p3}][[1]]
>>
>> Out[1]=
>> {p1 -> 0.253124 - 0.401596 I, p2 -> 0.253124 + 0.401596 I,
>>   p3 -> 0.443751}
>>
>> Other solutions are just the permutations of the first one. The solution
>> is complex-valued, but what if we substitute it in the GF?
>>
>> In[2]:=
>> Module[
>>    {gf = .35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000},
>>    ((1 - p1) + p1*gf)*((1 - p2) + p2*gf)*((1 - p3) + p3*gf) /.
>>      sol // Expand // Chop //
>>        Replace[List @@ #,
>>          {k_.*z^p_. -> {p, k}, k_ -> {0, k}},
>>          {1}]&
>> ]
>>
>> Out[2]=
>> {{0, 0.4}, {2500, 0.1225}, {5000, 0.140875}, {7500, 0.0935375}, {10000,
>> 0.0994875}, {12500, 0.049875}, {15000, 0.03995}, {17500, 0.02565},
>> {20000,
>> 0.015975}, {22500, 0.007425}, {25000, 0.003375}, {27500, 0.0010125},
>> {30000, 0.0003375}}
>>
>> It would be interesting to give a rigorous justification for this trick.
>>
>> Maxim Rytin
>> m.r at inbox.ru
>>
>
> Alternatively, we can simply use the formula for the total probability
>
> P(amount = k) =
>    Sum(P(amount = k | openvats = i)*P(openvats = i), i = 0..3).
>
>  From this we can infer that the GF for amount is .4 + .35*gf + .15*gf^2
> +.1*gf^3, arriving at the same expression for GF as above without the
> intermediate step of finding p1, p2, p3.
>
> In[1]:=
> Module[
>    {gf = .35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000},
>    .4 + .35*gf + .15*gf^2 +.1*gf^3 // Expand //
>      Replace[List @@ #,
>        {k_.*z^p_. -> {p, k}, k_ -> {0, k}},
>        {1}]&
> ]
>
> Out[1]=
> {{0, 0.4}, {2500, 0.1225}, {5000, 0.140875}, {7500, 0.0935375}, {10000,
> 0.0994875}, {12500, 0.049875}, {15000, 0.03995}, {17500, 0.02565}, {20000,
> 0.015975}, {22500, 0.007425}, {25000, 0.003375}, {27500, 0.0010125},
> {30000, 0.0003375}}
>
> Here's how we can apply Fourier to this problem:
>
> In[2]:=
> F = Fourier;
> IF = InverseFourier;
> SetOptions[#, FourierParameters -> {1, 1}]& /@ {F, IF};
>
> Module[{L},
>    L = PadRight[{0., .35, .35, .15, .15}, 13];
>    IF[.4*F[L]^0 + .35*F[L]^1 + .15*F[L]^2 + .1*F[L]^3] // Chop
> ]
>
> Out[5]=
> {0.4, 0.1225, 0.140875, 0.0935375, 0.0994875, 0.049875, 0.03995, 0.02565,
> 0.015975, 0.007425, 0.003375, 0.0010125, 0.0003375}
>
> With the default setting of FourierParameters
>
> F[Convolution[f, g]] == Sqrt[Length[f]]*F[f]*F[g].
>
> Maxim Rytin
> m.r at inbox.ru
>
>
>
>

--
DrBob at bigfoot.com
www.eclecticdreams.net

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