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Topic: How to numerically invert Laplace transform to obtain mixture of probability distributions?
Replies: 7   Last Post: Jun 28, 2007 9:39 PM

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 Vista Posts: 259 Registered: 5/25/07
Re: How to numerically invert Laplace transform to obtain mixture of probability distributions?
Posted: Jun 28, 2007 9:31 PM

"Herman Rubin" <hrubin@odds.stat.purdue.edu> wrote in message
news:f5ueta\$3bqc@odds.stat.purdue.edu...
> In article <f5si5t\$p4u\$1@news.Stanford.EDU>, Vista <abc@gmai.com> wrote:
>>Dear all,
>
>>1. I have a question regarding a probability density function and its
>>Fourier transform(the characteristic function).

>
>>How to judge from the characteristic function that if the probability
>>density function is a mixture of continuous distribution and discrete
>>distributions? I am getting some peaks/spikes in using inverse FFT to
>>recover the
>>probability density function from the characteristic function;

>
> Are you considering the exact characteristic function
> or the sample characteristic function? From the sample
> function, the problem is hard, and I would suggest
> alternate methods. From the population function, a
> discrete part is equivalent to the existence of an
> almost periodic part of the characteristic function,
> and a density function requires that the characteristic
> function approaches 0; however, I cannot think of a
> simple way of deciding there is a density unless some
> power of the characteristic function is integrable.
> There are also the effects of a possible singular
> measure.
>
>

>>On the other hand, using the inverse Fourier integral(numerical) point by
>>point, I am getting not the peaks/spikes
>>but some oscilatory waves at the locations of peaks/spikes.

>
> This is to be expected. Roundoff errors will cause this.
>

>>I am not sure
>>how to judge which one gives me the correct result -- that's to say, I am
>>not sure if by theory there should be a peak/spike, or that's some
>>numerical error and there should be no peak/spikes . I hope to be able to
>>decide this from the behavior of the
>>characteristic function.

>
>>I actually think the FFT results are correct, i.e. there should be
>>spikes/peaks in the density function, which is a jump in the cumulative
>>distribution function.

>
>>2. What do I do with the spikes/peaks? Since my next step is try to
>>integrate numerically some function g(x) against this density function.
>>Suppose the results
>>of FFT was correct, and
>>by theory, there should be peaks/spikes there, which are a dirac delta
>>functions in theory. It's going to be
>>problematic for the next numerical integration step. What can I do to
>>handle this?

>
> If g is smooth, and its Fourier transform is computable,
> preferably in closed form, do so and interchange the order
> of integration. Otherwise, if g is continuous, with better
> numerical approximations, the error goes to zero.
>

>>3. Suppose the results of FFT was correct, ie. by theory, there should be
>>a discrete density component there and there should be the peaks/spikes
>>there, which are the dirac delta functions in theory. But a numerical
>>inverse Fourier transform gives oscilatory waves instead of peaks/spikes,
>>how do I correct for this? What's wrong with the numerical inverse Fourier
>>transform?

>
> Roundoff error. Unless the frequency of the inverse
> transform happens to conform precisely to the spike,
> this behavior will result.
>

>>4. If we think of the characteristic function to be more broadly the
>>Laplace
>>transform. How do I invert Laplace transform for a distribution which is a
>>mixture
>>of discrete distribution and continuous distribution and possibly singular
>>distributions? According to probability
>>theory, a distribution can be decomposed into three parts -- absolutely
>>continuous, discrete and singular parts. How to do the numerical inverse
>>Laplace transform reliably and stably for such complicated
>>characteristic/Laplace transforms?

>
> There are good procedures for the discrete part knowing the
> locations of the jumps. The density works well, as long as
> the discrete and singular parts are not too close. I have
> nothing good to say about computing the singular part.
>
> Computing the cdf can be done directly. Again, if there are
> jumps or much of a singular part near the point of computation,
> the precision goes down.
>

>>Thanks a lot!
>
>

Hi Herman,

Is there a way to decompose the characteristic function into discrete part
and the continous part of the distribution and do the inverse transforms
only on the continous part?

I got very strange behavior on my inversions.

Let's fix the following notations:

Probability Density Function: f(x)
Cumulative Distribution Function: P(x)
Complementary Distribution Function: Q(x)=1-P(x)
Characteristic Function: F(v), which is the Fourier Transform of f(x),
defined as F(v)=integrate(f(x)*exp(i * v* x), x from -infinity to
+infinity).

I use the numerical intergral in Matlab:

f_hat(x)=integrate(F(v) * exp(-i * v* x), v from -infinity to +infinity),

And I used truncation: for large L and Matlab function "quadl"

f_hat(x)=integrate(F(v) * exp(-i * v* x), v from -L to +L),

and also I tried transformation:

f_hat(x)=integrate(F(atanh(y)) * exp(-i * atanh(y)* x)/(1-y^2), y
from -1+epsilon to +1-epsilon),

---------------------------------

In all cases above, I got some of the recovered density function f_hat(x) to
be negative, which is wrong.

And when I sum the f_hat(x) up, and did sum(f_hat(x))*delta_x,

and I got this sum to be more than 1, which is impossible theoretically,
because I am actually approximating the cumulative probability distribution
function P(x), it should be between 0 and 1.

---------------------------------

I just don't understand how can a Fourier inverse transform of a
characteristic function, which is the probability density function, be
negative; and how can the cumulative function be more than 1?

It is worth noting that if I do the inverse Laplace transform to obtain the
cumulative distribution function directly from inverting F(v)/v, I got the
results agreeing with the above results. The cumulative function P(x) is
still non-monotonous and some places can go beyond 1.

It is really weird...

Anybody can help?

Date Subject Author
6/26/07 Vista
6/27/07 David Jones
6/27/07 Michael Press
6/27/07 Brendan Halpin
6/28/07 Michael Press
6/27/07 Herman Rubin
6/28/07 Vista
6/28/07 Vista