What is the joint probability of X and Y on a unit interval of (0,1)?

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In summary, the conversation discusses the determination of probabilities, joint CDF, and joint density function for two uniform distributions (X and Y) over the unit interval (0,1). The problem considers various cases to determine the probability that X is greater than a given value and Y is less than a given value. The solution is found to be (1-x)y. The CDF of the joint probability is determined to be xy, and it is mentioned that if U1,...,Un are independent, then X and Y are also independent.
  • #1
cse63146
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Homework Statement



Let U1...Un be independant and uniformly distributed over the unit interval (0,1). Let X be the minimum of U1...Un and Y be the maximum

a) Determine P(X > x, Y < y). Consider the following cases: 1) 0< x < y < 1 2) 0 < y 1, x < 0
3) 0 < x < 1, y > 1 4) x < 0, y > 1. 5) all remaining possibilites
b) Determine the joint CDF of X and Y
c) using b), determine a joint density funtion of X and Y

Homework Equations





The Attempt at a Solution



for a), is the only possible case that can occur is (1)? since it's on the interval (0,1) so X/Y cannot be smaller than 0, and cannot be bigger than 1? And Y also has to be greater than X, since X is the minimum and Y the maximum.

[tex]\int^1_0 \int^Y_0 dx dy[/tex]

it doesn't seem right. Any hints?
 
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  • #2
Anyone?
 
  • #3
cse63146 said:

Homework Statement



Let U1...Un be independant and uniformly distributed over the unit interval (0,1). Let X be the minimum of U1...Un and Y be the maximum

a) Determine P(X > x, Y < y). Consider the following cases: 1) 0< x < y < 1 2) 0 < y 1, x < 0
3) 0 < x < 1, y > 1 4) x < 0, y > 1. 5) all remaining possibilites
b) Determine the joint CDF of X and Y
c) using b), determine a joint density funtion of X and Y

Homework Equations





The Attempt at a Solution



for a), is the only possible case that can occur is (1)? since it's on the interval (0,1) so X/Y cannot be smaller than 0, and cannot be bigger than 1? And Y also has to be greater than X, since X is the minimum and Y the maximum.

[tex]\int^1_0 \int^Y_0 dx dy[/tex]

it doesn't seem right. Any hints?
There is something wrong with your problem. You say X and Y are independent. If so, then there is no reason to "Consider the following cases". The probability that X> x is 1- x. The probability that Y< y is y. The probability that X> x and Y<y is (1- x)y.

And, the CDF of the joint probability if just P(X>x and Y>y)= xy.
 
  • #4
if U1,U2,...,Un are independant, then so are X & Y?
 
  • #5
Oh, blast! I misread the problem. I thought you were saying X and Y were independent.
 
  • #6
Any ideas?
 

What is the meaning of P(X > x, Y < y)?

P(X > x, Y < y) is a mathematical notation for the probability that the random variable X is greater than a specific value x, while the random variable Y is less than a specific value y.

How is P(X > x, Y < y) calculated?

P(X > x, Y < y) can be calculated by finding the joint probability of X being greater than x and Y being less than y, and then dividing it by the total probability space.

What is the significance of P(X > x, Y < y) in statistical analysis?

P(X > x, Y < y) is commonly used in statistical analysis as it allows for the calculation of the probability of two events occurring simultaneously. This helps in making predictions and understanding the relationship between two variables.

Can P(X > x, Y < y) have a value greater than 1?

No, the probability of an event cannot be greater than 1. P(X > x, Y < y) will always have a value between 0 and 1, where 0 means there is no chance of the event occurring and 1 means the event is certain to occur.

How can P(X > x, Y < y) be used in real-life situations?

P(X > x, Y < y) can be used in various real-life situations, such as predicting the likelihood of a stock price being higher than a certain value while the market is down, or the chance of a student scoring above a certain grade on a test while studying for a specific amount of time.

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