Steven G
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- Dec 30, 2014
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Let X1, X2 and X3 be three independent, identically distributed random variables each density function f(x) = 3x^2, 0< x < 1, 0 elsewhere.
Let Y = Max {X1, X2, X3}. Find P(Y> 1/2).
I am at a loss here.
I will (hopefully) show that I know what I.I.D. means by this example.
Suppose you have an urn with two balls in it--1 red and 1 blue. You pick two balls, one at a time w/o replacement.
Let X = the color of the 1st ball and Y = the color of the 2nd ball.
The two outcomes are {red, blue} and {blue, red}.
P(X=red) = 1/2 and (coincidentally) P(X= blue) = 1/2
P(Y=red) = 1/2 and (coincidentally) P(Y= blue) = 1/2
So X and Y are identically distributed.
Now if we know the outcome of the 1st ball we then know the outcome of the 2nd ball. That is, for example, P(X=red|Y=B) = 1\(\displaystyle \neq\) P(X=red)=1/2. So X and Y are NOT independent and hence X and Y are not i.i.d.
Since X1, X2 and X3 are i.i.d. I would think there is no max for them--after all they are identical. Am I missing something here????
So I assume that Y= Max {X1, X2, X3} = X1
Then p(X1 > 1/2) 1- p(X1 < 1/2) = 1- (1/2)3 = 7/8. My solution manual disagrees with this result.
However, if I compute p( X11/3> 1/2) = p( X1 > 1/8) I get the solution manual's answer (511/512)
Let Y = Max {X1, X2, X3}. Find P(Y> 1/2).
I am at a loss here.
I will (hopefully) show that I know what I.I.D. means by this example.
Suppose you have an urn with two balls in it--1 red and 1 blue. You pick two balls, one at a time w/o replacement.
Let X = the color of the 1st ball and Y = the color of the 2nd ball.
The two outcomes are {red, blue} and {blue, red}.
P(X=red) = 1/2 and (coincidentally) P(X= blue) = 1/2
P(Y=red) = 1/2 and (coincidentally) P(Y= blue) = 1/2
So X and Y are identically distributed.
Now if we know the outcome of the 1st ball we then know the outcome of the 2nd ball. That is, for example, P(X=red|Y=B) = 1\(\displaystyle \neq\) P(X=red)=1/2. So X and Y are NOT independent and hence X and Y are not i.i.d.
Since X1, X2 and X3 are i.i.d. I would think there is no max for them--after all they are identical. Am I missing something here????
So I assume that Y= Max {X1, X2, X3} = X1
Then p(X1 > 1/2) 1- p(X1 < 1/2) = 1- (1/2)3 = 7/8. My solution manual disagrees with this result.
However, if I compute p( X11/3> 1/2) = p( X1 > 1/8) I get the solution manual's answer (511/512)
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