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sam
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Joined: December 5th, 2001, 12:04 pm

Fourier Transform of Probability distribution

October 19th, 2005, 8:30 pm

Hi,Can anyone please give me an outline for the derivation of the probability function by inverting its fourier transform, i.e.where f is the characteristic function.Basically, I do not understand where the 1/2 comes from. My approach was to calculate the fourier transform of the probabity function:and this reduces to being a function of the characteristic function as shown above (f/i theta). I then inverted the fourier transform and got the integral above. But I don't see where the 1/2 would come from. Thanks in advance
 
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Rez
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Joined: May 28th, 2003, 9:27 pm

Fourier Transform of Probability distribution

October 20th, 2005, 1:38 am

sam, my sketch which is probably wrong:* g(u) is the characteristic function* f(x) is the pdf* F(x,y) is the Pr(x<X<y) your F(x) = P(x>X) = F(x,inf)The density function (pdf) is given as the Fourier inverse of g(u)f(x) = 1/pi int{ exp(-i u x) g(u), u in R }The cumulative F(x) = int{ f(s) ds, s in (x,y) }Which makes F(x) equal to the double integral F(x) = 1/pi int2{ exp(-i u s) g(u), u in R, s in (x,y) }Changing the integration order will give F(x) = 1/pi int2{ exp(-i u s) g(u), s in (x,y), u in R }and applying the integral with respect to s over (x, y) producesint{ exp(-i u s) g(u), s in (x,y) } = int{ exp(-i u s), s in (x,y) } g(u)Nowint{ exp(-i u s), s in (x,y) } = - exp(-i u y)/(i u) + exp(-i u x)/(i u)thusF(x,y) = I(y) + 1/pi int{ exp(-i u x)/(i u) g(u), u in R } where I(y) = 1/pi int{ exp(-i u y)/(i u) g(u), u in R }Your F(x) would be retrieved at the limit y->inf, where automagically I(y)->1/2 because g(u) is the Fourier transform of a pdf. Actually, it can't be anything else I guess... if we try F(-inf)=0, F(+inf)=1 etc.
 
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sam
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Joined: December 5th, 2001, 12:04 pm

Fourier Transform of Probability distribution

October 20th, 2005, 9:21 am

Hi Rez,Thanks for the reply.I'm with you right until the end. Where does the 1/2 come from again? How does setting y = infinity change the integral into a 1/2??Regards,
 
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Rez
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Joined: May 28th, 2003, 9:27 pm

Fourier Transform of Probability distribution

October 20th, 2005, 8:02 pm

Kia Ora sam:1. First of all, I think that all 1/pi in my sketch should be 1/(2 pi)... Always get confused with that stuff! I am in Samoa at the moment and I never thought bringing my Kendall+Stuart along :-~ Also there is a minus missing: should be -I(y) instead of I(y).2. If we call the integral: I(x) = 1/(2 pi) int{ exp(-i u x)/(i u) g(u), u in R } as in my post below, and the limit: I = lim{ I(x), x->inf }Then we can write Prob{x<X<y} = F(x,y) = I(x) - I(y)end Prob{X>x} = F(x) = I(x) - INow Prob{X>-inf} = 1, thus F(-inf) = 1 => I(-inf) - I = 1 and I(-inf) = -I(inf) = -I, therefore 1 = -2 I => I = -1/2which I think makes the formula work. I have not considered whether the limits or the integrals are actually finite, but I guess they should since we deal with probabilities.I am sure that there is a more elegant way, perhaps in Kendal+Stuart or Cramer.K.
 
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spursfan
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Joined: October 7th, 2001, 3:43 pm

Fourier Transform of Probability distribution

October 20th, 2005, 9:17 pm

read the bakshi and madan paperthink of these as similar to Nd1 and Nd2 and also N(d=0)=0.5
 
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Rez
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Joined: May 28th, 2003, 9:27 pm

Fourier Transform of Probability distribution

October 20th, 2005, 9:37 pm

QuoteOriginally posted by: spursfanread the bakshi and madan paperthink of these as similar to Nd1 and Nd2 and also N(d=0)=0.5True. Of course you would need the mean to be at d=0. But even in this case, wouldn't N(d=0)=0.5 hold only if the distribution is symmetric? The Fourier inversion formula will hold for any distribution.K.
 
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spursfan
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Fourier Transform of Probability distribution

October 21st, 2005, 5:43 am

all i was trying to do was give some intuition as to why, roughly speaking, it's 0.5 plus the integral in the very simple BS case
 
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sam
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Joined: December 5th, 2001, 12:04 pm

Fourier Transform of Probability distribution

October 21st, 2005, 9:44 am

Hi,Thank you all for yoru comments and insights.I have moved a step further and started to implement the integral numerically. I am setting the model to the B-S case for testing purposes (i.e. vol of vol =0) and I am comparing the probabilities from my numerical integration with the closed form values and there is some noise in there due to discretisation etc. In practice, how do people solve this integral?1. FFT?2. Clever substitution (change of integration variable) so that we dont have to integrate to infinity?3. Any thing else?Many Thanks,
 
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sunya
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Joined: February 1st, 2003, 9:09 pm

Fourier Transform of Probability distribution

November 1st, 2005, 11:32 pm

Sorry not to help, but why do you need that ? I ask because if you play with it ans then invert it to get back distribution, you are very likely o realise that it is not positive. I know no practical way of imposing a condition on the fourier transform so that the direct function is positive.Is relevent to you, and if yes does anyone know such a condition ?
 
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Rez
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Joined: May 28th, 2003, 9:27 pm

Fourier Transform of Probability distribution

November 2nd, 2005, 1:36 am

In theory the resulting pdf is of course positive. In practice when you numerically integrate over a discrete subset you will experience negative values of the pdf (around the tails areas). You can increase the sampling interval with a trapezoidal rule to inrease the accuracy until you make these negative values insignificant (and then set them to zero).The numerical recipes books have a section dedicated to the inversion of characteristic functions.FFT is useful since you will recover the whole pdf in one go, rather than inegrating each point individually. Of course in FFT the smaller the discretization interval of the c.f., the larger the output interval of the pdf (since the product of the two has to be equal to a constant -- 2*pi). This means that you end up with pdf poimts that are too far aart, and of no interest. The fractional FFT remedies this problem.Using an adaptive fractional FFT can take care of the infinite boundary problem.K.