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JeeJee
Canada5652 Posts
On December 02 2009 03:27 Mastermind wrote: I guess you have never heard of science or economics. edit: calculus is also used in probability lol. yeah yeah and you can trace it all back to addition oops, namechange | ||
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Meta
United States6225 Posts
On December 02 2009 02:56 JeeJee wrote: =( =( i loooove stats its by far the most fun and the most USEFUL area of math that i've learned in like 3+ years of uni. unlike, say, calculus where applications of that are limited to pretty much.. more calculus :S hell, once i get to pick some courses, i'm taking more stats, no questions asked Try studying engineering. It's pure calculus applications... pretty much every machine or structure made for general public use in the past hundred years has been a product of the application of calculus. | ||
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Freaky[x]
Canada995 Posts
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BlackJack
United States10568 Posts
There are two envelopes of money with one envelope having exactly twice as much money as the other. You choose one envelope and then decide if you want to switch envelopes and keep what is in the other envelope. The "trick" is that most people will always want to switch because they think they stand to gain more than they stand to lose. For example, you open the first envelope and it has $100 in it. You know the second envelope either has $200 or $50, since one envelope is double the other. So if you switch you risk to gain $100 or lose $50 which seems like a nice risk. | ||
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Slithe
United States985 Posts
Here's what I've done with the problem. I really don't know if I'm correct or not though. + Show Spoiler + Given that grandma put X in one envelope and 2X in another envelope: v = quantity that we observed in envelope 1 p1 = probability that v=X given envelope 1 has v, i.e. P(v=X | e1=v) EV[keep] = x EV[trade] = p1*0.5*x + (1 - p1)*2*x trade if : x < p1*2*x + (1 - p1)*0.5*x x < 1.5*p1*x + 0.5*x 0.5*x < 1.5*p1*x p1 > 1/3 Now to determine p1: q(x) = the probability distribution from which Grandma decides X Using Bayes' Theorem: P(v=X | e1=v) = P(v=X, e1=v) / P(e1=v) Solving for P(e1=v): P(e1=v) = 1/2 * q(v) + 1/2 * q(v/2) Solving for P(v=X, e1=v) P(v=X, e1=v) = 1/2 * P(v=X) P(v=X) = q(v) P(v=X, e1=v) = 1/2 * q(v) Putting it all together: P(v=X | e1=v) = P(v=X, e1=v) / P(e1=v) P(v=X | e1=v) = 1/2 * q(v) / (1/2 * q(v) + 1/2 * q(v/2)) p1 = 1/2 * q(v) / (1/2 * q(v) + 1/2 * q(v/2)) Going back to the original inequality: p1 > 1/3 1/2 * q(v) / (1/2 * q(v) + 1/2 * q(v/2)) > 1/3 3/2 * q(v) > 1/2 * q(v) + 1/2 * q(v/2) 2 * q(v) > q(v/2) This means that when we observe the amount v in envelope 1, we should switch if 2*q(v) > q(v/2), where q(x) is the probability distribution that grandma uses. I don't know what's being asked with the second bullet point. | ||
Empyrean
16992 Posts
On December 02 2009 06:53 BlackJack wrote: I think the problem in layman terms is this: There are two envelopes of money with one envelope having exactly twice as much money as the other. You choose one envelope and then decide if you want to switch envelopes and keep what is in the other envelope. The "trick" is that most people will always want to switch because they think they stand to gain more than they stand to lose. For example, you open the first envelope and it has $100 in it. You know the second envelope either has $200 or $50, since one envelope is double the other. So if you switch you risk to gain $100 or lose $50 which seems like a nice risk. Actually that's just the two envelopes problem. The major way in which this problem differs is that the amount X is generated by a probability distribution. It changes the problem completely. And as for applications, there's so much calculus in statistics you wouldn't even believe it. Calculus is such a useful field of mathematics and finds itself in every day life. For example, you could use calculus to argue that if it rains at a constant rate, you'll get the same amount of wet no matter what speed you move through it (though you'll probably end up colder if you walk leisurely). The argument can basically be summed up as the fact that you're displacing the same amount of water, but it takes some calculus to show exactly why. | ||
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538
Hungary3932 Posts
It should be easy from this on. | ||
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Black Gun
Germany4482 Posts
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Empyrean
16992 Posts
On December 02 2009 08:01 Black Gun wrote: i could definitely solve it, but i agree that its a bit challenging for an introducing question of a first course on probability. (im a statistics major currently writing my m.sc. thesis^^) What's your thesis about? If you don't mind my asking :D What was your experience as an undergraduate statistics major like? I find it really interesting. | ||
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Black Gun
Germany4482 Posts
On December 02 2009 08:02 Empyrean wrote: What's your thesis about? If you don't mind my asking :D What was your experience as an undergraduate statistics major like? I find it really interesting. well, its a really cool subject, but the part of statistics which is taught to non-majors is rather boring.  here in germany, we start with these "ordinary statistics topics" in the first 2 semesters, which include only the basic results of probability, and only in the 3rd and 4th we are taught the rigorous foundations of probability, measure theory, and so on. the really problematic courses are analysis I and II and linear algebra which are hold during the first 2 semesters. there, about 50% of all freshmen fail already, and about half of them (25% of the freshmens) fail so hard that they give it up already. (the math basics are taught on a really high lvl here...) basically, they intentionally want to purge the weak by these math courses. therefore the introduction into statistics and probability is rather easy and practically-oriented during the first 2 semesters. edit: whats ur major btw?^^ i havent started yet with my master´s thesis, but im currently getting into the literature and finding a topic. i plan to write about the estimation of response densities in regression settings via boosting techniques. if that tells u anything | ||
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Empyrean
16992 Posts
For my school, you aren't even allowed to become a statistics major until you pass multivariate calculus and linear algebra. My path through the major looks like 1st semester: Probability, Regression Analysis 2nd semester: Statistics, Bayesian/Modern Statistics, Survey Methodology 3rd semester: Statistical Decision Analysis, Design and Analysis of Causal Studies 4th semester: Independent Study, Computational Data Analysis 5th semester: Independent Study (continued). After your independent study (a year long project), if you still have time, you can take some higher level electives. Some examples are Probability/Measure Theory, Applied Stochastic Processes, Generalized Linear Models, Data Mining, Spatial Statistics, etc., it really depends on what they feel like teaching. | ||
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nttea
Sweden4353 Posts
* For the naive estimate of 1/2 to be correct for that probability, what would Granny's distribution p(x) have to be? Any problems with that? * If Granny used an exponential distribution with mean $100, what strategy optimizes the expected value of the amount of money you keep? sorry, i don't understand most of this since i've had no formal edumication. However to me it seems like it always makes sense to always swap the envelope. Since you have 50% chance of gaining double the amount of your current envelope, and 50% chance of gaining only half. I'm not sure if this is the only thing you ask for though (: | ||
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JeeJee
Canada5652 Posts
On December 02 2009 06:56 Empyrean wrote: And as for applications, there's so much calculus in statistics you wouldn't even believe it. Calculus is such a useful field of mathematics and finds itself in every day life. For example, you could use calculus to argue that if it rains at a constant rate, you'll get the same amount of wet no matter what speed you move through it (though you'll probably end up colder if you walk leisurely). The argument can basically be summed up as the fact that you're displacing the same amount of water, but it takes some calculus to show exactly why. is that actually true? doesn't make intuitive sense to me.. lets say its raining out, and you have to walk from your car door to apartment door (say 5 meters). you can sprint that, and you'll get some raindrops but you won't be soaked otoh, walk very slowly (say 3 hours for those 5m..) and you'll def. be soaked (and probably suffering from extreme hypothermia) or am i misunderstanding what you mean? (the "same amount of wet" seems .. ambiguous) | ||
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Fontong
United States6454 Posts
On December 02 2009 11:38 JeeJee wrote: is that actually true? doesn't make intuitive sense to me.. lets say its raining out, and you have to walk from your car door to apartment door (say 5 meters). you can sprint that, and you'll get some raindrops but you won't be soaked Yeah this doesn't really make any sense to me either. The way empyrean explained it only makes sense if the raindrops were stationary, and there were a constant amount of raindrops per volume of air. In the real world situation, taking 5 hours to walk 5 meters would end up with a lot more rain landing on your head than taking 5 seconds to walk 5 meters. The amount of rain landing on your head would be constant no matter what speed you are moving at though, it just matters how much time you are in the rain. Not really important though, most likely just a mistake from empyrean | ||
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Black Gun
Germany4482 Posts
imagine 2 extreme cases: u stand still, ie zero velocity. only ur head and ur shoulders get wet. (and ur belly, depending on how fat u are  ).the other extreme would be moving so fast that the velocity of the raindrops is negligible compared to it, in this case ur whole front would get wet while the back half of ur head and shoulders wouldnt. empyrean is right though that the amount of raindrops per exposed area is unaffected by ur velocity. edit: omfg, whats going on? my english is so horrible and full of mistakes today. wtf! | ||
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Empyrean
16992 Posts
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JeeJee
Canada5652 Posts
On December 02 2009 14:05 Black Gun wrote: ur own velocity changes the effective angle in which the raindrops hit u, the faster u are the more skewed the angle is. therefore a greater percentage of ur total body surface gets directly hit by raindrops when running through the rain instead of walking. imagine 2 extreme cases: u stand still, ie zero velocity. only ur head and ur shoulders get wet. that's not true, consider the rain at a slight angle. it still falls under empyrean's assumptions of constant rate yet renders this argument moot and even if it's purely vertical, it'll still trickle down and you get completely soaked if you just stand there ^_^ edit: empyrean is right though that the amount of raindrops per exposed area is unaffected by ur velocity. ok, this i could see as being true not the original "same wetness" claim though and yes your english is very good :O i don't know what you're complaining about :p | ||
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50bani
Romania480 Posts
On December 02 2009 11:10 nttea wrote: * Under what conditions should you trade, after knowing how much money your envelope contains? * For the naive estimate of 1/2 to be correct for that probability, what would Granny's distribution p(x) have to be? Any problems with that? * If Granny used an exponential distribution with mean $100, what strategy optimizes the expected value of the amount of money you keep? sorry, i don't understand most of this since i've had no formal edumication. However to me it seems like it always makes sense to always swap the envelope. Since you have 50% chance of gaining double the amount of your current envelope, and 50% chance of gaining only half. I'm not sure if this is the only thing you ask for though (: No. Granny has either 3Y or 1.5Y between the two envelopes. The thing is one value is more likely than the other. If you work out how more likely one is than the other you can make a decision. Has been solved in a previous post I think. 1.5Y has to be less than 2 times as likely as 3Y for the switch to be profitable. | ||
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gyth
657 Posts
For example, you could use calculus to argue that if it rains at a constant rate, you'll get the same amount of wet no matter what speed you move through it (though you'll probably end up colder if you walk leisurely). The argument can basically be summed up as the fact that you're displacing the same amount of water, but it takes some calculus to show exactly why. The amount of water that hits your front (assuming vertical rain and stomach) is always the same. But the amount that hits your top depends on how long you're in the rain. | ||
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