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06-13-2012 , 05:39 PM
Quote:
Originally Posted by Clue
yes I think what masque de Z said

edit:

so this is part of the lecture notes:



This part of the notes lead me to wonder about the Prob(K = k/sqrt(n)) type distribution

And also I forgot an (n Choose k) up there obv
Even if you let

Xn = ( B(n,p) - np) / Sqrt[np(1-p)]

then the comments in A and B still hold. Notice in A and B he's talking about densities. Xn is discrete so does not have a conventional density function, thus A. Also, for any real x, P(Xn=x) => 0 so B holds.

What is true is that Xn => Z in distribution (or "Law") where Z ~ N(0,1). What this means is that the cumulative distribution functions of Xn converge to the cumulative distribution function of Z.

ie. P(Xn <= x) converges to P(Z <=x) for every real x where the CDF of Z is continuous (which for Z is all x).


See Wiki - Convergence of random variables.

They give an example of random variables which do have density functions and which converge in distribution to a nice uniform distribution but where their density functions do not converge at all.




PairTheBoard
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06-13-2012 , 10:45 PM
Huh, feels weird a cdf can converge in distribution, but not density. What you say makes sense to me though, ty
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06-14-2012 , 02:11 PM
Quote:
Originally Posted by Clue
Huh, feels weird a cdf can converge in distribution, but not density. What you say makes sense to me though, ty

The example they give in Wiki involves continuous density functions which oscillate with increasing frequency between the values 0 and 2 on the interval (0,1). They clearly don't converge pointwise but their integrals over any subinterval of (0,1) converge to the length of the subinterval - the oscillations average out in the integral. So their cdf's converge to the uniform distribution on (0,1).

fn = (1 - cos(2pi*nx)* 1_{x in (0,1)}

where 1_{x in (0,1)} is the indicator function on the interval (0,1)
ie.
1_{x in (0,1)} = 1 if 0<x<1 and 0 otherwise.


PairTheBoard
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06-14-2012 , 10:49 PM
Hands down greatest SMP thread of all time and in the running for all time 2+2 best IMO. The knowledge ITT is excellent but what makes Jason1990 and this thread so bad ass is the occasional poster that decides to step up by saying he doesn't understand the question or thinks they have some brilliant pithy post. He bitch slaps them back in their place more politely than anyone I have ever seen. Thanks for the thread Jason1990 :-)
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06-15-2012 , 07:23 AM
Let X be a partially ordered set and P and Q be probability measures on X. Suppose that for each principle upset U of X, we have that P(U)<=Q(U). Can you give an example where P is not stochastically smaller than Q?
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06-15-2012 , 04:40 PM
Quote:
Originally Posted by mSed84
Let X be a partially ordered set and P and Q be probability measures on X. Suppose that for each principle upset U of X, we have that P(U)<=Q(U). Can you give an example where P is not stochastically smaller than Q?
I don't recall having seen these concepts of "upsets" and "stochastically smaller" before. Looking at Wiki I see "stochastically smaller" defined for random varibles. ie. If A and B are random variables then A is smaller than B means that for every real x, P(A > x) <= P(B > x). Conceptually, A has more probability weight on smaller real numbers relative to B.

However, in your question you don't have any random variables. You have probablity measures on X but no function from X to R defining random variables. So I assume you mean to have a function, say A, from X to R defining two random variables via P and Q, say Ap and Aq. And you then want Ap not smaller than Aq.

It seems to me that you would only expect Ap to be smaller than Aq if the partial ordering on X somehow agreed with the natural ording on R under the map A. So look for a counter example where the map A does not respect the ordering on X. How about letting X = R with the natural ordering, assume you have your P striclty smaller than Q, and let the map A:X --> R be A(x) = -x.

Or assume you have a Q and P on R with Q strictly smaller than P, then let X = R with reverse ordering, and A the identity map.


Or am I missing something?


PairTheBoard
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06-18-2012 , 10:11 PM
Quote:
Originally Posted by PairTheBoard
I don't recall having seen these concepts of "upsets" and "stochastically smaller" before. Looking at Wiki I see "stochastically smaller" defined for random varibles. ie. If A and B are random variables then A is smaller than B means that for every real x, P(A > x) <= P(B > x). Conceptually, A has more probability weight on smaller real numbers relative to B.

However, in your question you don't have any random variables...

Or am I missing something?


PairTheBoard
Let (X,F,P) and (X,F,Q) be two probability spaces. P is said to be stochastically smaller than Q if P(U) <= Q(U) for every upset U of X. Other terminology for upset is 'ideal', in case you've heard of that. A subset U of a partially ordered set X is said to be an upset if a in U and a<=b implies b is in U. There is no need to define a random variable for this. Additionally, I don't think the problem can be solved using a totally ordered set for X.
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06-18-2012 , 10:13 PM
I should also mention that a principle upset is an upset generated by a single element in a partially ordered set.
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06-19-2012 , 03:54 AM
At least it is easy to find orders with lots of P,Q s.t. P(U)=Q(U)=0 for any principal subset.

e.g. on R with the trivial ordering (x<=y iff x=y), any P,Q that do not charge single points will do.

If you add the requirement that P(U)>0 for all principal subset then it is more difficult.
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06-19-2012 , 09:54 AM
Quote:
Originally Posted by checktheriver
At least it is easy to find orders with lots of P,Q s.t. P(U)=Q(U)=0 for any principal subset.

e.g. on R with the trivial ordering (x<=y iff x=y), any P,Q that do not charge single points will do.

If you add the requirement that P(U)>0 for all principal subset then it is more difficult.
Thanks--that works. I kept trying to solve the problem using finite partially ordered sets (I'm sure it's possible but I'm not being creative enough I guess).
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06-24-2012 , 12:38 PM
I play gin rummy. The idea is to collect melds: ranking (888 or KKKK of three or four cards) or sequential (A23 upto QKA no round the corner le 2AK). Sequential melds could comprise ten cards, which is the number of cards in a gin rummy hand. Any combination of melds that uses up all ten cards is said to be "gin" hence the name.

Would you be able to give me a probabilistic idea of what the average dealt ten-card gin rummy hand would like?
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07-03-2012 , 07:21 PM
http://forumserver.twoplustwo.com/18...lling-1204267/

Jason...there is alively discussion in this thread regarding the bolder clause

I contend that this clause lowers the avg ROI of the package while others think its unaffected

What do you say?
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