Probabilities involving Mins and Maxes

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The discussion focuses on calculating the probability of the minimum of two independent random variables, specifically P(Min(X, Y) < 1). It is clarified that P(Min(X, Y) < 1) cannot simply be expressed as P(X < 1) + P(Y < 1) due to the nature of minimum values. Instead, the correct approach involves using the formula for the union of events, leading to P(Min(X, Y) < 1) = 1 - P(X ≥ 1)P(Y ≥ 1). Participants emphasize the relevance of order statistics for understanding the distribution of minimum and maximum values in random samples. The conversation highlights the importance of foundational probability concepts in solving such problems.
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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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