|
|
The Workshop
Diablo II Mechanics and Statistics
Knowledge is power. In this forum, we discuss detailed game mechanics and statistics, from how Diablo II functions, to the probabilities of events happening.
Questions? Read the Forum FAQ and please obey the Forum Rules.
Spirea
Charter Member
|
29-May-01, 08:38 PM (GMT) |
   |
1. "RE: General Stats Question (PDFs) "
In response to message #0
| |
Er .. you mean P(xy=a), not P(xy), because that will be meaningless or at least contradict your notation. (; P(xy=a) = IntegralOF( f(k)g(a/k) dk )
P(x/y=a) = IntegralOF( f(k)g(k/a) dk ) Not sure what you mean by P(x op y).
|
|
|
|
Printer-friendly page | Top |
|
|
 |
|
|
|
chaos_preacher
Charter Member
|
01-Jun-01, 11:19 PM (GMT) |
|
|
5. "RE: General Stats Question (PDFs) "
In response to message #0
| |
Yes, there is a very general technique
that allows to obtain the density of
functions of random variables. It comes
from a general theorem of change of variables
under integrals. First of all some basic facts:if the random
variables(r.v.) X and Y have
densities, then the probabilities of the events
{X=a},{Y=b} are ZERO,whatever a and b are.This implies
that all the probabilities P(X+Y=a),P(X-Y=a),P(XY=a)
P(X/Y=a) are zero: that is, the prob of taking any
SPECIFIC value is zero. Now, this sounds contradictory
as they must take SOME value??!! For these continuous
(as opposed to discrete r.v. which do not have densities)
the only events that have non zero probs. are ones involving
intervals. This ain't no prob. crash course so I will not
continue this lecture. If you are interested in the topic
I suggest the following book: Hogg & Craig, Introduction
to Mathematical Statistics, Prentice Hall,1984
or any elementary prob. text Now, if f(x) and g(y) are the respective densities of X and Y,
then you are right(even if you make strong abuse of the notation)
the density of Z=X+Y is h(z)=integral OF< f(z-y)g(y) dy>
and similarly for Z=X-Y. The formulas for the densities of
Z=XY and Z=X/Y are respectively
h(z)=integral OF, and
h(z)=integral OF
where | | stands for absolute value.
With some theorical justification we can even handle the possibility that the abobe domain of integration
could contain 0.chaos_preacher |
|
|
|
Printer-friendly page | Top |
|
|
|
whyBish
Charter Member
|
02-Jun-01, 00:31 AM (GMT) |
 |
6. "Wow, thanks for that"
In response to message #5
| |
your formulas for Z=XY and Z=X/Y have disappeared. I've been thinking about this a lot lately and see that I need to ensure that the areas of the transformed g(y) must be equated to one and the area of a region preserved. That is P(x=a)=f(a)
P(y=b)=g(b) and for P(h(x,y)=a) transform into y=j(x) then P(h(x,y)=a)=IntegralOf(f(x)g(j(x))) Where j(x) has been normalised such that IntegralOF(g(x)) from a to b = IntegralOf(g(j(x))) from j(a) to j(b) Am I getting warm?  |
|
|
|
Printer-friendly page | Top |
|
|
|
|
chaos_preacher
Charter Member
|
03-Jun-01, 06:10 PM (GMT) |
|
|
7. "RE: Wow, thanks for that"
In response to message #6
| |
Oups, here they are again for Z=XY, the density is
h(z)=integral OF(f(z/y)g(y)|1/y| dy) and for Z=X/Y
h(z)=integral OF(f(zy)g(y)|y| dy) where f,g are the (marginal)densities of X,Y
respectively. The above formulas are valid only
if X and Y are independent(that is their joint
PDF is the product of the marginals.) When only averages matter, it is not always necessary
to obtain the entire PDF provided we have independency.
Indeed, in that case the expectation(average) of the
product(resp. quotient) is the product of expectations:
E(XY)=E(X)E(Y)
(resp.E(X/Y)=E(X)E(1/Y); beware as E(1/Y) is not in
general the same as 1/E(Y)!)
The same holds for the variance. hope this helps chaos_preacher |
|
|
|
Printer-friendly page | Top |
|
|
|
|
|
|
Printer-friendly copy
Email this topic to a friend
