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## Friday, July 04, 2008

### Continuation of TOD commentary

TheOilDrum.com doesn't allow comments on posts after a certain amount of time, so I will use this post to continue the discussion. For this subthread:
Vitalis on July 4, 2008 - 3:36am

Yes, if L is a sum of independent exponentials with means Lo and L1, then L has density

(*) 1/(L1-L0)*( exp(-x/L1)-exp(-x/L0) ), for x >= 0.

(Of course, when L0->L1 this will collapse to a Gamma distribution with 2 degrees of freedom).

In the formulas above, I assumed that S is also Gamma(2,s) distributed, which perhaps is harder to give a physical interpretation than for L. If we assume S ~ Exp(lambda) and L density (*), then

E[D(S,L)] = lambda*(1 - 1/(L1-Lo)*(1/(1/lambda + 1/L1) - 1/(1/lambda + 1/Lo)) ).

But perhaps all this fiddling around with different distibutions is of moderate interest: it is probably most important to recognice that the most important thing is to have some kind of uncertainty ("dispersion") in lambda and L. Whether this is the case, I would say, comes down to how sensitive the "output" is to the various assumptions on the distributions, the "output" being whatever the model help us to say about the real world. (In this case, perhaps the expected volume of discoveries in the next few years.)

Another thing we could try to adress is the sample variability: we have focused on the mean of D(S,L), but it should be straight forward to compute condfidence intervals for future discoveries, given assumptions on the distributions of S and L. (Maybe you already did this).

--

Another (minor) comment: i don't think the single dispersive model corresponds to a uniformly distributed L --- in that model, L = Lo deterministically (i.e. point mass 1 at Lo in probability lingo).

In fact, L uniformly distributed on [0,2*L0] (to make E[L] = L0) gives

E[D(S,L)] = labmda*( 1 - 1/(2*Lo) + 1/(2*Lo)*exp(-2*Lo/lambda) ).

I agree with Vitalis about the statement "it is probably most important to recognice that the most important thing is to have some kind of uncertainty ("dispersion") in lambda and L."

The following figure shows how incremental dicoveries basically trend to the same asymptote, independent of the volume distribution.

I would also note that, contrary to Vitalis' suggestion, the Delta "thin seam" volume situated at L0 would have a cumulative discovery profile that looks like this:

Professor  Anonymous said...

Nice graphs!
Regarding the "DeltaDiscovery" we are probably misunderstanding each other somewhere.
The way I understand it, the single dispersive model is the result of the assumptions

S ~ Exp(lambda), i.e. density 1/lambda*exp(-x/lambda), and L = Lo deterministically.
(Or, equivalently, Prob(L = Lo) = 1.)
So, calulating E[D(S,L)] = E[D(S,Lo)] =
\int_0^Lo x*1/lambda*exp(-x/lambda) dx + \int_Lo^\infty Lo*1/lambda*exp(-x/lambda) dx

= lambda*(1 - exp(-Lo/lambda).

Is this right?

Vitalis

4:25 PM
Professor  WHT said...

I believe the first integral equals zero because it never quite gets to Lo. Another way to look at it -- the Delta function is an identity function that produces the original profile with the delta position replacing the variable. So it can't have an additional expression clause.

Could it be that you are "double counting" integration limits?

Or is it this one as well?
Indefinite Integral(1/lambda*exp(-x/lambda) dx) = -exp(-x/lambda)

due to the chain rule.

Now take an absurd situation as a sanity check, where we sit on an ocean of oil, at delta depth=0. Evaluating from 0 to infinity, this gives 1, which checks that probabilities sum to 1 immediately. All oil would be discovered instantaneously in this case.

The other sanity check is if delta depth was at infinity. Then the discovery depth is infinity as well since we can never reach it.

Thanks for trying to desk-check these equations, as I like to approach it from every conceivable angle. And I could still be wrong, but that is life.

8:09 PM
Professor  Anonymous said...

Good evening! At least in Sweden ;)

The first integral is not zero
(sanity: it is an integral of a strictly positive function over
the positive-length interval [0,Lo]). In fact,

I1 := \int_0^Lo x*1/lambda*exp(-x/lambda) dx =
[-x*exp(-x/lamba)]_0^Lo + \int_0^Lo exp(-x/lambda)dx
= (-Lo*exp(-Lo/lambda) - (-0) ) +
+ [-lambda*exp(-x/lambda)]_0^Lo =
-Lo*exp(-Lo/lambda) +
+ (-lambda*exp(-Lo/lambda) - (- lambda) ) =
-Lo*exp(-Lo/lambda) + lambda*(1 - exp(-Lo/lambda) ).

The second integral
I2 := \int_Lo^\infty Lo*1/lambda*exp(-x/lambda) dx =
Lo*[-exp(-x/Lo)]_Lo^\infty =
Lo*(0 - (-exp(-Lo/lambda) ).

The sum is, precisely as you got in the singel dispersive model,
I1 + I2 = lambda*(1 - exp(-Lo/lambda) ).

I'm not sure what you mean by IndefiniteIntegral(1/lambda*exp(-x/lambda) dx).
But if you mean \int_0^\infty 1/lambda*exp(-x/lambda) dx, then surely this equals 1.
(Integrating the density of an exponential variable over the whole real line gives 1.)

As I see it, all oil is contained in the depth [0,L] --- "evenly" spread out (I do not want to say "uniformly" in order not to confuse with the probabilistic meaning of that word). If L = Lo deterministically, then so be it: all oil is contained in [0,Lo]. It does not mean that all oil sits exactly at depth Lo.

With my interpretation of the model, the sanity checks look like this.

If Lo = 0, then we find all oil immediately, but the total oil volume is 0.

If Lo = \infty, then yes, we never reach Lo, but the total volume of oil is \infty.

All the above assumes that the oil density alpha := (total oil volume)/Lo = 1,
i.e. everywhere we search there is nothing but oil. This is analogous to a haystack made of needles (a needlestack ;) Taking 0< alpha<1 does not change the sanity checks, and all equations are multiplied by alpha.

I think our misuderstanding is somewhere around here: what Lo actually means. Will be interesting to hear what you say! I can for sure be wrong too, it happened before and will happen again:)

2:15 PM
Professor  Vitalis said...

/Vitalis

4:09 PM
Professor  WHT said...

It turns out that all these cumulative discovery distributions turn into the set of Laplace Transforms (or characteristic functions in probability-speak) after applying dispersion. So the uniform or "even distribution" is Laplace(u(t)-u(t-a))= (1-exp(-sa))/s. The shifted delta is Laplace(delta(t-a))=exp(-as).

So mathematically these all look sound because I can cross-check against a Laplace transform table. So the problem may lie in the differences in interpretations we have with the premise or problem statement.

Now why this looks exactly like a Laplace transform I think has a somewhat intuitive basis. A Laplace or Fourier transform is essentially an integration across all the time/spatial frequencies (or time/spatial constants) in the spectrum. And that is what we are doing with the dispersion, generating a spread in values that use a damped exponential generating function (ala the characteristic function). It just so happens that this turns it into the set of Laplace or characteristic transforms.
http://www.vibrationdata.com/Laplace.htm

The higher-order gamma distributions become trivial in this case, as well -- Laplace(t^n*exp(-at)) = (n-1)!/(s+a)^(n-1)

Sounds kind of bizarre, but the work of dispersive cumulative discovery generates the equivalent of an FFT of the underlying distribution.

I think my problem also may stem from labeling my first analysis "single dispersion" and the second "double dispersion". Both of these actual fall under a "general dispersion" analysis and the first and second cases are a specific instance of general dispersion. It is both a dispersion in amounts of progress in discovery, and a dispersion in depths at which discoveries are made. That is where the "double" qualifier prefix comes in. The case of the "evenly distributed" depths actually does show a degree of dispersion, whereas the delta discovery at depth Lo is the case of no dispersion. I think if it wasn't for you questioning this premise, that whole notion of a generalized approach would have slipped past me. So thanks for the insight.

8:55 AM

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