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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