Naked before God: HELP ME SKLANSKY!

[RubbleRobble]

I'm not sure how quickly Mr. Sklansky and Mr. Malmuth find/respond to post directed mostly at them, but I have a question in the realm of probability theory that I'm afraid is a little too tricky for us mere mortals...

This question was proposed to me by my instructor in probability theory at the University of Maryland, Professor Michael Brin.

You die and arrive at the gates of heaven. St. Peter informs you that, due to budget cuts, people no longer are admitted to heaven based on the merit of good deeds. The new system goes like this:

God has a distribution of random numbers in mind. (No psychology here folks... we cannot deduce the type of distribution, i.e. uniform, gamma, normal, etc; nor can we deduce the nature of the parameters.) From this distribution of numbers, God takes a number, X, and gives it to you to look at, touch, hug, etc. You KNOW the value of X. God then takes another number from the same distribution, called Y, and puts it in his pocket (I forgot to mention... God has a pocket).

In order to enter heaven, you must 'guess' whether the number He handed to you is greater than or less than Y, God's number, which resides in his pocket.

Stressing again that the only things you have to tackle this problem are your wits/knowledge of probability and statistics plus the knowledge of the value of X. GIVE A STRATEGY THAT WILL GET YOU INTO HEAVEN WITH PROBABILITY STRICTLY GREATER THAN 50%.

I would greatly appreciate the help of Mr. Sklansky or Mr. Malmuth (or anyone that can competently and succinctly explain the solution to this 'Naked before God' problem.) Class is at 1 pm EST on Weds, so a solution by then would be awesome, but I'll take what I can get... Thanks in advance.

[chopchoi]

Since we cannot deduce the nature of the parameters, we must assume that they are infinite. Thus, if God gives you a positive number, then you should guess that his number is less than yours, since there are (in theory) more numbers less than yours than there are numbers greater than yours. Likewise, if God gives you a negative number, then you should guess that God's number is greater than yours.

Or, if all of the numbers are positive, then you should always guesss that god's number is greater than yours, since the distance between your number and zero will always be shorter than the other way.

[Mr Mojo Risin]

[ QUOTE ]
Since we cannot deduce the nature of the parameters, we must assume that they are infinite. Thus, if God gives you a positive number, then you should guess that his number is less than yours, since there are (in theory) more numbers less than yours than there are numbers greater than yours. Likewise, if God gives you a negative number, then you should guess that God's number is greater than yours.

Or, if all of the numbers are positive, then you should always guesss that god's number is greater than yours, since the distance between your number and zero will always be shorter than the other way.

[/ QUOTE ]

This theory is very flawed in my opinion because there are an infinate amount of negitive numbers and an infinate amunt of positive numbers. Due to the fact that the number line goes on forever in either direction, the number being negitive or positive should have no effect on the outcome.

[Moonsugar]

some infinities are bigger than others

[jason1990]

[ QUOTE ]
From this distribution of numbers, God takes a number, X...

You KNOW the value of X. God then takes another number from the same distribution, called Y...

In order to enter heaven, you must 'guess' whether the number He handed to you is greater than or less than Y

[/ QUOTE ]
You must be assuming that God's distribution is continuous. If it were allowed to be discrete, then God could choose the distribution satisfying P(X=0)=P(X=1)=1/2. If He did that, then 50% of the time, X and Y would be equal. Since you are only apparently allowed to guess higher or lower, you would automatically lose in those cases. Thus, you could never get into heaven more than 50% of the time, no matter what you did.

If we can assume that X and Y are never equal, then you can find your solution in this post, which summarizes someone else's solution that you can read about in that long, long thread.

[reubenf]

[ QUOTE ]
If we can assume that X and Y are never equal

[/ QUOTE ]

Is it sufficient to assume that the probability that x=y is zero?

[jason1990]

[ QUOTE ]
[ QUOTE ]
If we can assume that X and Y are never equal

[/ QUOTE ]

Is it sufficient to assume that the probability that x=y is zero?

[/ QUOTE ]
I believe it is necessary and sufficient to have the CDF of X be continuous, which happens if and only if f(y):=P(X=y)=0 for all y. Do a little logical song and dance to show that f(y)=0 for all y if and only if f(Y)=0 almost surely. Then note that since f(Y) is nonnegative, f(Y)=0 almost surely if and only if E[f(Y)]=0. Finally, observe that

P(X=Y) = E[P(X=Y|Y)] = E[f(Y)].

So the CDF of X is continuous if and only if P(X=Y)=0.

[RubbleRobble]

Thanks to chop and jason for responses. One edit that I should make is, yes the distribution is continuous, I neglected to mention that fact... sowwy.

However, I looked at the post and replies on the thread indicated by jason, and my first reaction (though I could always be wrong here) is to say that this in insufficient to answer the 'naked before god' problem.

First, the question posed by RMJ in his thread is trying deciding to stay on/switch from a number "A" taken from the list of all real numbers. That you are taking two numbers from the distribution of all real numbers implies knowledge of the distribution. Tomcollins shows that a solution can be found with P(success)>0.5 (and I was able to understand his proof there), but I can't understand what his actual strategy is. What if the distribution is f(x)=x when 0<=x<=1, f(x)=(-x)+2 when 1<=x<=2 and 0 otherwise (a ligitimate pdf); and the number you are given by God is 0.5? 0.5 is a positive number, so tomcollin's solution would have you not switching (in the cotext of this problem, this means telling God that your number is bigger). Again, I could be wrong, but if you follow this strategy, chances are you won't be winning.

Abstract problem I got here, maybe an explanation tailored to the setting of this problem would help me understand things better... then again maybe I've already been given the correct answer, but im too dumb to know it...

[jason1990]

Okay, let me try to get my head on straight before I reply. The idea of TomCollins is this: take an appropriate f (like the one I linked you to). Look at X and compute f(X). Then, with probability f(X), tell God that you think X is bigger than Y. One way to facilitate this is to choose a random number U uniformly on [0,1]. Then, if U < f(X), you tell God you think X is bigger than Y. Otherwise, of course, you guess that Y is bigger than X. The probability you will be right is

P(U < f(X), Y < X) + P(U > f(X), Y > X).

Conditioned on X, this becomes

P(U < f(X))P(Y < X) + P(U > f(X))P(Y > X)
= f(X)F(X) + (1 - f(X))(1 - F(X)),

where F is the cumulative distribution function of God's chosen distribution. Since we conditioned on X, we need to take expectations, so the probability you will be right is

E[f(X)F(X) + (1 - f(X))(1 - F(X))].

What you have to show is that this will always be strictly greater than 1/2, regardless of F.

[RubbleRobble]

Jason,

Thank you VERY much for the help with this problem. I was able to follow the logic in your more recent posts and the story checks out.

Just a point of clarification though... When tomcollins has that we "must find an increasing function that maps from R-> [0,1]," what does it mean to map R-> [0,1]? He says later that his chosen function "has a range within [0,1]." What does that mean?

I want to have a commanding grasp on all points of the proof before I speak up on this tomorrow. Thanks.