Probabilities involving Mins and Maxes

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Discussion Overview

The discussion revolves around the probabilities associated with the minimum and maximum of two random variables, specifically exploring the expressions for P(Min(X, Y) < 1) and P(Max(X, Y) < 1). Participants are examining theoretical aspects of probability, particularly in the context of independent random variables.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant states that P(Max(X, Y) < 1) = P(X < 1)P(Y < 1) for independent random variables.
  • Another participant questions whether P(Min(X, Y) < 1) could equal P(X < 1) + P(Y < 1), suggesting a potential misunderstanding of the union of events.
  • A participant mentions the formula for P(A or B) = P(A U B) and indicates that the minimum case does not follow the same intuitive logic as the maximum case.
  • One participant proposes using the expression P(Min(X, Y) < t) = 1 - P(Min(X, Y) >= t) and assumes it leads to 1 - P(X >= t)P(Y >= t).
  • Another participant agrees with the previous point and elaborates on the relationship between the probabilities of the events A and B, suggesting a formula involving their intersection.
  • Several participants recommend studying order statistics to better understand the distribution of minimum and maximum values in samples of random variables.

Areas of Agreement / Disagreement

Participants express differing views on the correct expression for P(Min(X, Y) < 1), with some proposing additive relationships and others suggesting a more complex formulation. The discussion remains unresolved with multiple competing views on the topic.

Contextual Notes

Participants reference the need for understanding order statistics to tackle more complex scenarios involving minimum and maximum values, indicating a potential gap in knowledge regarding this topic.

muzihc
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I know for sure that P(Max(X, Y) < 1) = P(X < 1)P(Y < 1) if the two RVs are independent, but what does P(Min(X, Y) < 1) equal?

Would it be P(Min(X, Y) < 1) = P(X < 1) + P(Y < 1)?

Thanks
 
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muzihc said:
what does P(Min(X, Y) < 1) equal?
Would it be P(Min(X, Y) < 1) = P(X < 1) + P(Y < 1)?
The probability of "X < 1 or Y < 1" involves the union of two sets. Have you sudied the formula for P(A \cup B) ?
 
muzihc said:
I know for sure that P(Max(X, Y) < 1) = P(X < 1)P(Y < 1) if the two RVs are independent, but what does P(Min(X, Y) < 1) equal?

Would it be P(Min(X, Y) < 1) = P(X < 1) + P(Y < 1)?

Thanks

Hey muzihc and welcome to the forums.

Have you ever come across order statistics?

http://en.wikipedia.org/wiki/Order_statistic
 
Hi,

I've studied P(A or B) = P(A U B) - I had a class in probability/statistics. I've never formally studied order statistics, though maybe I've overlapped with it at some point.

The maximum case is kind of intuitive - if the max is less than t, everything else is. We can treat it as P(A and B), in the independent case. On the other hand, if the minimum is less than t, the other random variable isn't necessarily.

I guess we could use P(Min(X, Y) < t) = 1 - P(Min(X, Y) >= t), in which case I assume it's 1 - P(X >= t)P(Y >= t).
 
muzihc said:
Hi,

I've studied P(A or B) = P(A U B) - I had a class in probability/statistics. I've never formally studied order statistics, though maybe I've overlapped with it at some point.

The maximum case is kind of intuitive - if the max is less than t, everything else is. We can treat it as P(A and B), in the independent case. On the other hand, if the minimum is less than t, the other random variable isn't necessarily.

I guess we could use P(Min(X, Y) < t) = 1 - P(Min(X, Y) >= t), in which case I assume it's 1 - P(X >= t)P(Y >= t).

Hey muzihc.

Take a look at order statistics. Order statistics helps you calculate the distribution of the minimum and maximum of a value given the number of samples that you have. It will help you in the first part of your question.
 
muzihc said:
I've studied P(A or B) = P(A U B) - I had a class in probability/statistics.

Let A = the events where X < t [/itex]
Let B = the events where Y < t [/itex]

P(A \cup B) = P(A) + P(B) - P(A \cap B)

So for independent events A and B

P(A \cup B) = P(A) + P(B) - P(A) P(B)


I guess we could use P(Min(X, Y) < t) = 1 - P(Min(X, Y) >= t), in which case I assume it's 1 - P(X >= t)P(Y >= t).

I agree.

1 - (1 - P(A)) (1-P(B)) = 1 - (1 - P(A) - P(B) + P(A)P(B)) = P(A) + P(B) - P(A)P(B)

As chiro says, "order statistics" are the topic you should study if you want to tackle more complicated versions of the situation. For example: "In a sample of 4 values drawn from 4 independent random variables A,B,C,D, what is the probability that the second largest value is less than 1?"
 

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