"Chia C Chong" <[EMAIL PROTECTED]> wrote in message
a145qk$qfq$[EMAIL PROTECTED]">news:a145qk$qfq$[EMAIL PROTECTED]...
> Hi!
>
> I have a series of observations of 2 random variables (say X and Y)
from my
> measurement data. These 2 RVs are not independent and hence f(X,Y) ~=
> f(X)f(Y). Hence, I can't investigate f(X) and f(Y) separately. I tried
to
> plot the 2-D kernel density estimates of these 2 RVs and from the it
looks
> like Laplacian/Gaussian/Generalised Gaussian shape in one side and the
other
> side looks like Gamma/Weibull/Exponential shape.
>
> My intention is to find the joint 2-D distribution of these 2 RVs so
that I
> can reprenseted this by an equation (so that I could regenerate this
plot
> using simulation later on). I wonder whether anyone has come across
this
> kind of problem and what method that I should use??
>
> Thanks...
>
> Regards,
> CCC
>

In plotting the distributions of these two RVs, were you looking at the
MARGINAL distributions?  If so, it might be more useful to look at a
range of CONDITIONAL distributions for each variable, since it is the
conditional distributions that you ultimately need to define in order to
arrive at a joint distribution. One variable's conditional distribution
could conceivably change substantially over the range of the other
variable's values.

By looking at how each variable's conditional pdf shape changes at
different values of the other variable, you may be able to select a
distributional form (Weibull, Gamma, etc.) that is able to represent the
varying shape of one variable's pdf by a change of parameter values.
Whichever variable has a conditional pdf form that seems best suited to
representation by a known distributional form (with varying parameters),
is the one you can choose as the "dependent" variable.

For example, let's say that, in looking at the conditional distributions
for each variable, you decide that the pdf for one of the variables can
be represented pretty well by a Gamma distribution, with parameters b
and c.  Let Y be the variable whose pdf can be represented by the Gamma
distribution, and call the other variable X.  Then f(Y) = Gamma[Y,b,c],
where Gamma[Y,b,c] denotes the Gamma probability density as a function
of Y, with parameters b and c.  By changing b and c, you are able to
obtain the different shapes that f(Y) assumes over the range of values
of X.

Thus, you can fit a different Gamma distribution for Y, AT EVERY VALUE
OF X.  This will give you a set of b and c parameter values for each X.
If you plot the different b and c values as functions of X, you can get
some idea of what the functional form of the dependence might be.  For
the sake of simplicity, let's say that it turns out to be linear for
both b and c.  Then...

  Gamma parameter b = P0 + P1*X
  Gamma parameter c = Q0 + Q1*X

You can now do regressions to determine the coefficients.  Of course,
the functional form will probably NOT be linear.  And the functional
form may also not be the same for both parameters.

With the parameters expressed as a function of X, you can write...

  f(X,Y) = Gamma[Y,b(X),c(X)].

And this is, in fact, the joint distribution you are looking for!

WARNING! You will need a LOT of data.  You first need to determine a
conditional distribution for Y, at every value of X, which is one set of
regressions (but, hopefully, you have software that will do the
distribution fits automatically for you).  Then you have to do another
regression for each distribution parameter.  And you will probably need
fairly good fits to do a reasonable job of reproducing the overall joint
pdf.

The difficult part of this will probably be trying to find a single
distributional form (Weibull, or Gamma, or whatever) that can represent
all of the conditional pdf shapes for one of the variables.  Of course,
if you can't, then you could define several intervals for one of the
variables, and apply a different distributional form for each interval.
But things can get very messy very quickly!

This is probably not the only way to approach the problem, but I hope
this helps.

--
T. Arthur Wheeler
MathCraft Consulting
Columbus, OH 43017







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