How Can We Minimize Prediction Errors in Binary Variable Analysis?

AI Thread Summary
The discussion revolves around analyzing a binary variable Y with equal probabilities for outcomes and a random variable X that takes different uniform distributions depending on Y's value. The means of X are calculated as 2.5 when Y = 0 and 7.5 when Y = 1. The conditional probability P(Y = 1|X = x) is determined based on the value of X, with clear distinctions for ranges of X. A critical challenge identified is the inability to predict Y without error when X falls between 4 and 5, leading to uncertainty in classification. The minimum probability of rejection for accurate prediction remains unresolved.
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Homework Statement



Let Y be binary variable P(Y = 1) = P(Y = 0) = 0.5 and X a random variable uniform on [0,5] when Y = 0 and uniform [4, 9] when Y = 1. Draw mean of X and P(Y = 1|X = x) as functions of x. What is the minimum probability of rejection to predict Y from X without mistake.


Homework Equations





The Attempt at a Solution



Since Y acts like some "switch", I considered two independant distributions of X given Y: p = 1/5 on [0,5] (for Y = 0) and the other one p = 1/5 on [4,9] (for Y = 1). So two means are: 2.5 (for Y = 0) and 7.5 for (for Y = 1).

For P(Y = 1|X = x):
if X in [0, 4]: P(Y = 1|X = x) = 0
if X in [5, 9]: P(Y = 1|X = x) = 1
if X in [4, 5]: P(Y = 1|X = x) = 0.5

for this "What is the minimum probability of rejection to predict Y from X without mistake." I have no idea.
 
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Luksdoc said:

Homework Statement



Let Y be binary variable P(Y = 1) = P(Y = 0) = 0.5 and X a random variable uniform on [0,5] when Y = 0 and uniform [4, 9] when Y = 1. Draw mean of X and P(Y = 1|X = x) as functions of x. What is the minimum probability of rejection to predict Y from X without mistake.


Homework Equations





The Attempt at a Solution



Since Y acts like some "switch", I considered two independe[/color]nt distributions of X given Y: p = 1/5 on [0,5] (for Y = 0) and the other one p = 1/5 on [4,9] (for Y = 1). So two means are: 2.5 (for Y = 0) and 7.5 for (for Y = 1).
I think the problem is asking you for E[X] as a function of x, similar to what you did for P(Y=1|X=x). If x is in [0,4], what is E[X]? And so on.
 
Luksdoc said:

Homework Statement



Let Y be binary variable P(Y = 1) = P(Y = 0) = 0.5 and X a random variable uniform on [0,5] when Y = 0 and uniform [4, 9] when Y = 1. Draw mean of X and P(Y = 1|X = x) as functions of x. What is the minimum probability of rejection to predict Y from X without mistake.


Homework Equations





The Attempt at a Solution



Since Y acts like some "switch", I considered two independant distributions of X given Y: p = 1/5 on [0,5] (for Y = 0) and the other one p = 1/5 on [4,9] (for Y = 1). So two means are: 2.5 (for Y = 0) and 7.5 for (for Y = 1).

For P(Y = 1|X = x):
if X in [0, 4]: P(Y = 1|X = x) = 0
if X in [5, 9]: P(Y = 1|X = x) = 1
if X in [4, 5]: P(Y = 1|X = x) = 0.5

for this "What is the minimum probability of rejection to predict Y from X without mistake." I have no idea.

I don't see any way of predicting Y exactly from X in all cases. If we happen to observe a value of X between 4 and 5, Y is allowed to be 0 or 1, and there is no way to be sure which is correct.

RGV
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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