transform using cdf

**qwerty321** · January 1st 2010, 10:02 AM

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

**matheagle** · January 1st 2010, 02:47 PM

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...

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

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.

**Moo** · January 2nd 2010, 12:23 AM

Hello,

@ qwerty321 :
No they're not independent, so your working is unfortunately wrong.
First consider $P(Y_1\geq y_1,Y_2\leq y_2)$ (since it's continuous random variables, it doesn't matter whether is $\geq$ or >)

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

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

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

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

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

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

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.

**qwerty321** · January 2nd 2010, 12:43 AM

matheagle i cannot understand how you gor your formula
Moo I do not understand :

first why you took this case:
"
First consider $P(Y_1\geq y_1,Y_2\leq y_2)$ (since it's continuous random variables, it doesn't matter whether is $\geq$ or >)"

second:
then how did you pass from

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

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

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

thank you

**Moo** · January 2nd 2010, 05:52 AM

Originally Posted by qwerty321

Moo I do not understand :

first why you took this case:
"
First consider $P(Y_1\geq y_1,Y_2\leq y_2)$ (since it's continuous random variables, it doesn't matter whether is $\geq$ 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

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

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

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 ^^

**qwerty321** · January 2nd 2010, 01:30 PM

thank you it finally worked

Thread: transform using cdf

LinkBack

Thread Tools

Display

transform using cdf

Similar Math Help Forum Discussions

Laplace transform and Fourier transform what is the different?

MGF transform

How to transform double integral into a single integral using diamond transform

MAPLE: graph in hz, fourier transform, wavelet transform

how to transform??

Search Tags