hello I need help

I spent 2 hours without getting the right solution

=

[1-[1-P(X1<=y1)[1-P(X2<=y1)][P(X1<=y2)P(X2<=y2)]

but the book gives a different aswer from what I got

I think I cannot assume that Y1 and Y2 are independant right?

Thank you

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- Jan 1st 2010, 10:02 AMqwerty321transform using cdf
hello I need help

I spent 2 hours without getting the right solution

=

[1-[1-P(X1<=y1)[1-P(X2<=y1)][P(X1<=y2)P(X2<=y2)]

but the book gives a different aswer from what I got

I think I cannot assume that Y1 and Y2 are independant right?

Thank you - Jan 1st 2010, 02:47 PMmatheagle
This is the joint distribution of two order statistics from a sample of two.

You can find that in many books.

I think it's in Wackerly, I used to use Hogg and Craig.

since you only have two observations...

$\displaystyle f( x_{(1)},x_{(2)})={2!\over 1!1!}f(x_{(1)})f(x_{(2)})I(x_{(1)}<x_{(2)})$

There are 2! ways this can happen.

Either Y1=X1 or Y1=X2.

The reason it's still a valid density is that we now have a restriction on the space.

The order statistics are dependent. - Jan 2nd 2010, 12:23 AMMoo
Hello,

@ qwerty321 :

No they're not independent, so your working is unfortunately wrong.

First consider $\displaystyle P(Y_1\geq y_1,Y_2\leq y_2)$ (since it's continuous random variables, it doesn't matter whether is $\displaystyle \geq$ or >)

If the min is superior to a y1 and the max is inferior to y2, it simply means that $\displaystyle X_1,X_2 \in [y_1,y_2]$

So $\displaystyle P(Y_1\geq y_1,Y_2\leq y_2)=P(X_1\in[y_1,y_2],X_2\in[y_1,y_2])$

And then you can use the independence between X1 and X2 to calculate this probability.

Then $\displaystyle P(Y_2\leq y_2)=P(Y_2\leq y_2,Y_1\leq y_1)+P(Y_2\leq y_2,Y_1\geq y_1)$

Hence $\displaystyle P(Y_2\leq y_2,Y_1\leq y_1)=P(Y_2\leq y_2)-P(Y_2\leq y_2,Y_1\geq y_1)$

the missing part is $\displaystyle P(Y_2\leq y_2)=P(\max(X_1,X_2)\leq y_2)=P(X_1\leq y_2,X_2\leq y_2)$

can you finish it ?

You have to visualize that if the min of two values is > to something, then the two values are > to something. If the max of two values is < to something, then the two values are < to something. - Jan 2nd 2010, 12:43 AMqwerty321
matheagle i cannot understand how you gor your formula

Moo I do not understand :

first why you took this case:

"

First consider $\displaystyle P(Y_1\geq y_1,Y_2\leq y_2)$ (since it's continuous random variables, it doesn't matter whether is $\displaystyle \geq$ or >)"

second:

then how did you pass from

"

$\displaystyle P(Y_2\leq y_2)=P(Y_2\leq y_2,Y_1\leq y_1)+P(Y_2\leq y_2,Y_1\geq y_1)$

to this $\displaystyle P(Y_2\leq y_2,Y_1\leq y_1)=P(Y_2\leq y_2)-P(Y_2\leq y_2,Y_1\geq y_1)$ "

and for the missing part i guess you mean i could use the independance thing for X1 and X2

thank you - Jan 2nd 2010, 05:52 AMMoo
Because it's easy to deal with min if it's > to something, and to deal with max if it's < to something (see my last sentence in the first post).

Quote:

second:

then how did you pass from

"

$\displaystyle P(Y_2\leq y_2)=P(Y_2\leq y_2,Y_1\leq y_1)+P(Y_2\leq y_2,Y_1\geq y_1)$

to this $\displaystyle P(Y_2\leq y_2,Y_1\leq y_1)=P(Y_2\leq y_2)-P(Y_2\leq y_2,Y_1\geq y_1)$ "

Quote:

and for the missing part i guess you mean i could use the independance thing for X1 and X2

thank you

- Jan 2nd 2010, 01:30 PMqwerty321
thank you it finally worked:D