Even in college, changing variables (as was often done in advanced calculus) confused the heck out of me. Now I'm taking this Probability Theory class for my master's degree, and I'm having a hard enough time understanding how if X is distributed somehow and Y is some function of X, then the density function of Y is such-and-such; I have no idea what I'm doing when U is distributed somehow and X is some function of U! I could try to just memorize formulas but I'd rather understand what I'm doing, and it's clear I don't understand.
I think I'm done for tonight, my brain is fried. I'm gonna go watch football.
This may be way off base, but your issue may be related to the Monty Hall problem. Setup: 3 doors, 1 with a prize. You pick 1. Host opens another and asks if you want to switch. 2/3 of the time, switching pays off.
ReplyDeleteIt's easier to understand with more doors. Suppose there are 10. Pick a door; 10% chance it's right, meaning the other doors, collectively, are 90% likely to have the prize. Host opens 8 doors showing no prize. Switch?
Your door still has its original 10% probability. The one remaining door is now 90% likely to win, because all the other dud doors are gone.
Back to your problem: X, the probability of win, is distributed randomly among Y elements (the door). U is the % chance of all the unpicked doors. As doors open, U shifts among the remaining doors, eventually reaching maximum when 1 door is still closed (in addition to your pick, of course).
Did that explain it? Or did I get the problem set totally, completely wrong?
I get the Monty Hall problem. Here's what I'm dealing with:
ReplyDeleteZ is normally distributed
Y=e^Z
What is the density function for Y?
U is uniformly distributed on [0,1]
What is the density function for U^2?
I really think the Monty Hall problem solution is wrong. Why wouldn't it still be 10%?
ReplyDeleteThe solution is well established, everyone agrees Marilyn vos Savant was correct, it's not 10% for the same reason 3+2 is not equal to 7--because it's not.
ReplyDeleteYeah, but why?
ReplyDeleteI'm sure it's well explained on Wikipedia or something so I don't have to go into it here.
ReplyDeleteThis may be a bit too late, but have you tried just playing around with them a bit? A pair of dice give an approximate normal distribution. Roll dice several times and look at a table of the exponentials of the values. Similarly for the uniform distribution example (mark one die so you can get 36 distinct values). Depending on how comfortable you are with programming, writing a simple program to do this on a larger scale than is feasible with physical dice might be an alternative.
ReplyDeleteBuild a couple concrete examples, and then think of it in terms of how a range of values in the initial distribution gets turned into a different range of values in the transformed distribution.
Re, the Monty Hall Problem, by the rules, the "Host [is forced to open] 8 doors showing no prize." Without a rule saying the host will open 8 dud doors and offer a choice, the probability doesn't work. If the host has a choice, then matters of psychology, not probability dominate.
ReplyDelete