Fourier Transfer and a game?!?!
- To: mathgroup at smc.vnet.net
- Subject: [mg54106] Fourier Transfer and a game?!?!
- From: "elparedblanco" <cire1611 at gmail.com>
- Date: Thu, 10 Feb 2005 02:47:56 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
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?
Below is the CODE and the ANSWER...I'm close, but close don't cut it.
I think it has something to do with setting the Fourier Parameters.
CODE:
In[149]:=
severity1={.35,.35,.15,.15,0,0,0,0,0,0,0,0,0,0,0,0}
In[150]:=
Length[severity1]
In[151]:=
lossList={2500,5000,7500,10000,0,0,0,0,0,0,0,0,0,0,0,0}
In[161]:=
freq={.4,.35,.15,.1,0,0,0,0,0,0,0,0,0,0,0,0,0}
In[153]:=
fftSeverity = Fourier[severity1]
In[154]:=
one = fftSeverity*Sqrt[1]*.35
In[155]:=
two = fftSeverity^2*Sqrt[1]*.15
In[156]:=
three = fftSeverity^3*Sqrt[1]*.1
In[157]:=
totalFourier = one + two + three
In[158]:=
finalDistribution = InverseFourier[totalFourier]
In[159]:=
Total[finalDistribution]
=======================================
Here's the answer..
ANSWER = {.4000, .1225, .1409, .0935, .0995, .0499, .040, .0257, .016,
.0074 .0034, .0010, .0003}
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