Suppose there are two boxes with money in them. The amount of money is a positive real number of dollars, but is chosen completely randomly. All you know is that one of the boxes has twice as much money as the other.
Which box is the optimal choice? (Think before checking your answer!)
[spoiler]Answer: Box X. There is a 50% chance that x = 2y and a 50% chance that x = y/2, so E(x) = 0.5*(2y)+0.5(y/2) = 5y/4 > y.[/spoiler]
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No such distribution exists that could set that up as there will always be a median amount for a box's value
Answer: Box Y. There is a 50% chance that y = 2x and a 50% chance that y = x/2, so E(y) = 0.5*(2x)+0.5(x/2) = 5x/4 > x.
proved you wrong, OP.
No, the expected value would be undefined since y is undefined. You can't just say a random real number without defining the distribution. Define the distribution properly and the problem goes away.
It doesn't matter what the distribution is, though. Pick any one you like and you'll still have E(x) = 0.5*(2y)+0.5*(y/2).
Distribution: x+y = 3
If x =1 then y=2
If x=2 then y=2
Yes 50% of the time x=2y and 50% of the time x= y/2 but in the first case y must be 1 and in the second y must be 2. So E(x) = 1.5. Problem solved! y must be defined to calculate E(x) with it.
*if x=2 then y=1 of course
Is OP coming back or was he BTFO?
That's not a probability distribution. Those are added constraints I did not specify. Examples of probability distributions include "normal", "exponential", and so on.
yeah thats what i thought too. the choice is trivial because both have the same odds - i.e. there is nothing different between the two other than the random chance that one is double the other. pick either, the chances are the same.
Of course it's a distribution. x+y=3 with probability 100%. And as you said, it doesn't matter what the distribution is. The flaw remains the same, which is that by leaving y undefined you pretend to calculate an expected value where y is a singular variable when in fact it isn't.
If it's a distribution, then what is its probability density function?
And yet E(x) = 1.25*E(y).
To make it more clear, regardless of the distribution, half the time x=2y AND y is the smaller number, half the time x=y/2 AND y is the larger number. You cannot combine into one term two different values of y.
E(x) = 0.5*2*(x+y)/3+0.5*0.5*2(x+y)/3
E(x) =(x+y)/2
QED
Is off is the sane a the distribution,just like any other discrete pdf.
What on earth are you doing? Why are you dividing things by 3?
>Is off is the sane a the distribution
Are you okay? Are you having a stroke?
Its pdf is the same as the distribution, just like any other discrete distribution.
Surely you understand that the larger number is 2/3*(x+y) while the smaller is (x+y)/3. Surely you aren't retarded or pretending to be?
>Think before checking your answer!
Grate bayt meit
Your distribution is not fully defined. What is the probability that x = 1?
But your extra assumptions are unnecessary. Clearly if x = 2y and both are positive, y is the smaller number.
Your assumptions also seem to lead to incorrect conclusions, since we know that E(x) = 5/4*y in contradiction to your assertion that E(max(x,y)) = 2*(x+y)/3. Clearly this cannot be true since x is on average larger than y, and so any formula for E(max(x,y)) cannot be symmetric.
Thank you I made it myself.
Discrete distributions do not have a pdf. They have a pmf.
The dollar values are allowed to range over all of x and y, so the distribution is continuous.
Over all positive reals, I mean.
The probability that x=1 is 0.5.
>But your extra assumptions are unnecessary. Clearly if x = 2y and both are positive, y is the smaller number.
What extra assumptions? Everything I wrote follows from your statement of the problem, specifically that one amoumy is double the other.
>Your assumptions also seem to lead to incorrect conclusions, since we know that E(x) = 5/4*y
No, I already explained this is false since you combined two unequal variables. The fact that they are unequal follows directly from the premise you gave.
This is like arguing that 1=-1 since -1=sqrt(x)=1
Not in the distribution I gave. Read the thread.
Why does it matter. If the amount of money varies from zero to infinity, the probability that you'll get less than a trillion dollars in either box is arbitrarily small, so just take your money and be happy being the richest man on earth.
What if I told you one box had a million dollars and the other two million? Would you keep switching? Or would you realize that the symmetry makes switching pointless and that OP calculated E(x) incorrectly?
>No, I already explained this is false since you combined two unequal variables. The fact that they are unequal follows directly from the premise you gave.
Are you saying y is not equal to y?
The problem specifies that x and y can range over all positive real numbers. Your distribution breaks this assumption.
>Are you saying y is not equal to y?
y in the first case cannot be equal to y in the second. In the first case x = 2y and in the second x = y/2. Now since x = x we get 2y = y/2, which means y = 0. But you defined y as a positive number. This a contradiction.one of your assumptions must be wrong. That assumption is that when you split the problem into two cases, the x in one case is still equal to the x in the other. This is clearly false since the x in one is smaller than the x in the other. The same goes for y.
>The problem specifies that x and y can range over all positive real numbers. Your distribution breaks this assumption.
You then said that the problem works for any distribution. The problem fails for the same reason for every distribution. You then asserted that my simple discrete distribution is not a distribution because it doesn't have a pdf, which is nonsense. Now you are pretending like none of this happened. You lose. Make another bait thread.
You are all brainlets, in this case the value of a expected value is the arithmetic mean. And that value is a invalid one.
E(x) = x1*(1/x)+x2*(1/x)...
Since [math]x\in\mathbb{N}[/math], and you are not taking the box randomly, but the money, the experiment is going there.
So, this is stupid.
That doesn't actually explain why his answer is wrong though. The primary fault is not in the distribution he chose. The primary fault is that he splits the distribution into two different cases of x and y but pretends that the two ys are still equal.
>y in the first case cannot be equal to y in the second. In the first case x = 2y and in the second x = y/2.
No, they are absolutely the same thing. It's x that's different between cases, because we're trying to calculate E(x). For example, suppose you gain the additional knowledge that y = 4. Then either x = 2 or x = 8, and the expected value is E(x) = 0.5*2+0.5*8 = 5. But y = 4 is a constant throughout this calculation. Nowhere am I assuming multiple different values of y and combining them.
Ruining the fun...
With no survivors
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the problem with all of this is that there is no uniform distribution over the positive reals. it's a fallacious problem to begin with
>uniform distribution
I never specified such a thing. It's just that without any prior knowledge, it is irrational to assign different probabilities to x = 2y as opposed to y = 2x. You have to assume a 50% chance of each.
Veeky Forumsmon bomb!
(0)!>0
then when you say "chosen randomly" you have to specify how
It can be any distribution, as long as that distribution ranges over the positive real numbers. You just don't know what the distribution is when choosing between the boxes.
It's not box x. It is box A with x amount of dollars.
E(x) is defined based on E(y) using the equations you wrote, not y, but E(y) is infinity, hence you get screwy results.
>The amount of money is a positive real number of dollars, but is chosen completely randomly.
The impossible nature of the game I am being asked about triggers my scam sense. I tell the strange man I left my wife on in the kitchen and that I have to go check on her. Phew, bullet dodged.
>And yet E(x) = 1.25*E(y).
No, it's 1.25*y, not 1.25 times expectation of y.
And E(y) = 1.25*x as well, so the choice is trivial.
This also proves that expectation fails with this premise.
Well, exactly. The expectation only works if y ISN'T variable. So it doesn't work.
>the strange man was using a gamma distribution to determine the amount of money
>it wasn't impossible
Guess you just missed out on $10000000 user.
For any possible value of y, you can easily compute that E(x) = 5*y/4. So why does it matter that you don't know exactly what y is?
>For any possible value of y, you can easily compute that E(x) = 5*y/4. So why does it matter that you don't know exactly what y is?
Because if you don't know what y is, the problem becomes symmetric and E(x) is not 5*y/4. It's (x+y)/2. Of course knowing what one of the envelopes contains changes the dependency of E(x), since you break symmetry.
depends on money.
imagine instead of 2 the koefficient is 1000
you open box and there is $10, it is obvious that you should check another box for $10000 and if other box is $0.01 - no big deal
but if you open box and there is $1000000000, I would keep it despite chance if 1000 billion, because $1 billion is enough for me
Logic vs math
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