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Two independent random variables X and Y are uniformly distributed in the
interval [−1,1]. The probability that max[X,Y] is less than 1/2 is
interval [−1,1]. The probability that max[X,Y] is less than 1/2 is
- a)3/4
- b)9/16
- c)1/4
- d)2/3
Correct answer is option 'B'. Can you explain this answer?
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Two independent random variables X and Y are uniformly distributed in ...
Probability density function of uniformly distributed variables X and Y is shown
as
as
Most Upvoted Answer
Two independent random variables X and Y are uniformly distributed in ...
0,1]. To find the expected value of the maximum of X and Y, we can use the fact that the probability density function of the maximum is given by:
f(max(X,Y)) = 2x
where x is the value of the maximum. This can be derived by considering the area of the region where the maximum is x, which is a square with side length x and therefore area x^2, and dividing by the total area of the unit square, which is 1.
Using this probability density function, we can then find the expected value of the maximum as follows:
E[max(X,Y)] = ∫(from 0 to 1) x*f(max(X,Y)) dx
= ∫(from 0 to 1) x*2x dx
= 2∫(from 0 to 1) x^2 dx
= 2*(1/3)
= 2/3
Therefore, the expected value of the maximum of X and Y is 2/3.
f(max(X,Y)) = 2x
where x is the value of the maximum. This can be derived by considering the area of the region where the maximum is x, which is a square with side length x and therefore area x^2, and dividing by the total area of the unit square, which is 1.
Using this probability density function, we can then find the expected value of the maximum as follows:
E[max(X,Y)] = ∫(from 0 to 1) x*f(max(X,Y)) dx
= ∫(from 0 to 1) x*2x dx
= 2∫(from 0 to 1) x^2 dx
= 2*(1/3)
= 2/3
Therefore, the expected value of the maximum of X and Y is 2/3.
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Two independent random variables X and Y are uniformly distributed in theinterval [−1,1]. The probability that max[X,Y] is less than 1/2 isa)3/4b)9/16c)1/4d)2/3Correct answer is option 'B'. Can you explain this answer?
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