🚀 go-pugleaf

Georg W�tschner <georg.wuetschner@unibw-muenchen.de> wrote:

[...]
> Maybe something which could point in the same direction was the so
> called weather problem a friend told me of:
> If one weather station states that tomorrow it will rain with a
> probability of 60%, and a second, which operates which totally different
> methods (->independent), predicts rain wiht 70% prob., then one can
> somehow calulate that it will rain with a prob. of, say, 80% (higher
> than the higher of the two)... any idea how this works?

If I'm not mistaken, the combined probability (whatever scheme you
find for doing the combining) will be a convex linear combination
of the individual probabilities, i.e., a linear combination of the
form sum_i a_i p_i, with all a_i >= 0 and sum_i a_i = 1.
This implies that the combined probability must be somewhere between
the least and greatest individual probabilities.

I think the topic you want to look into is called "combining experts"
or "combining opinions". Here is a reference:

 Genest, C. and Zidek, J. V. (1986), "Combining Probability Distributions: 
 A Critique and an Annotated Bibliography," Statistical Science 1 114--135.

For what it's worth,
Robert Dodier
--
``Prediction is extremely difficult. Especially about the future.''
 -- Niels Bohr

Georg W�tschner <georg.wuetschner@unibw-muenchen.de> wrote in message news:<3EDF7781.4030100@unibw-muenchen.de>...
> Hi folk,
> I hope you allow me as an engineer a basic question:
> a random variable (RV) takes on a certain value, which I have to
> estimate. To this end I have a set of values, each of which was produced
> by a given function of the RV plus noise.
> Now an estimator E1 estimates the value of my RV with a certain variance V1.
> An _independant_ second estimator E2 does the same, but has a much worse
> set of values and therefore estimates with a bigger variance V2.
> 
> What would interest me is a statement on whether I can improve the
> overall performance of my estimation by somehow combining the 2
> estimators or if I allways make it worse. That is, is it possible to
> achieve a variance lower than V2 when combining the result of E1 and
> that of E2 (say weighted average or something more sophisticated)??
> Can this wanted relation between V1, V2 and Vcomb. be mathematically
> expressed? What are the conditions/constrictions?
> 
> [ snip ]
> 
> Thanks a lot,
> 
> Georg

To avoid confusion with the expected value operator E, 
call the estimates X1 and X2. Then we have 
E(X1|RV) = RV, Var(X1|RV) = V1, 
E(X2|RV) = RV, Var(X2|RV) = V2, 
with X1|RV independent of X2|RV. 
Let Xcomb = P1*X1 + P2*X2, where 
P1 = (1/V1)/(1/V1 + 1/V2), 
P2 = (1/V2)/(1/V1 + 1/V2). 
Then E(Xcomb|RV) = RV, 
and Var(Xcomb|RV) = V1*V2/(V1+V2), 
which is less than both V1 and V2. 
But note that V1 and V2 must be known 
(as opposed to estimated).

In article <242be6f5.0306081036.7e5fcaf1@posting.google.com>,
Ray Koopman <koopman@sfu.ca> wrote:

>To avoid confusion with the expected value operator E, 
>call the estimates X1 and X2. Then we have 
>E(X1|RV) = RV, Var(X1|RV) = V1, 
>E(X2|RV) = RV, Var(X2|RV) = V2, 
>with X1|RV independent of X2|RV. 
>Let Xcomb = P1*X1 + P2*X2, where 
>P1 = (1/V1)/(1/V1 + 1/V2), 
>P2 = (1/V2)/(1/V1 + 1/V2). 
>Then E(Xcomb|RV) = RV, 
>and Var(Xcomb|RV) = V1*V2/(V1+V2), 
>which is less than both V1 and V2. 
>But note that V1 and V2 must be known 
>(as opposed to estimated).


But using reasonably good estimates for V1 and V2 will probably still 
work.

Say Xcomb = t X1 + (1-t) X2

Var(Xcomb|RV) = t^2 V1 + (1-t)^2 V2

which is less than both V1 and V2 as long as

max(0,(V2-V1)/(V1+V2)) < t < min(1,(2 V2)/(V1+ V2))


Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2