Thread: transform using cdf

1. transform 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

2. 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.

3. 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.

4. 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

5. Originally Posted by qwerty321
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 >)"
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).

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 $\displaystyle P(Y_2\leq y_2,Y_1\leq y_1)=P(Y_2\leq y_2)-P(Y_2\leq y_2,Y_1\geq y_1)$ "
erm... a=b+c ===> b=a-c ?

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

thank you
Of course ^^

6. thank you it finally worked