# Transform of Variable Using CDF Tech

• Feb 25th 2009, 03:37 PM
B_Miner
Transform of Variable Using CDF Tech
Problem:

Let x1, x2 be independent random variables representing lifetimes (in hours) of two key components of a device, which fails when and only when both components fail. Say Xi has the exponential distribution with mean 1000. Let Y1 = min(X1,X2) and Y2 = max(X1,X2); so the space of Y1,Y2 is 0<y1<y2< inf+

Find G(y1,y2) = P(Y1<=y1, Y2<=y2)

Steps Taken:

1) First, I think I solved this correctly as:

P(Y1<=y1, Y2<=y2)
=
P(Y1<=y1)* P(Y2<=y2)
=
[1-[1-P(X1<=y1)[1-P(X2<=y1)][P(X1<=y2)P(X2<=y2)]

but the book gives a much different form of an answer.

2) I multiply two exponential distributions together to get
1/1000000 * exp( (-x1-x2)/1000). This is the joint pdf.

3) I wanted to draw a picture of the first quadrant (Y1 on the horizontal and Y2 on the vertical) and shade everything in Q1 above the line y2=y1.

Then I'd integrate the joint pdf wrt Y2 first and then Y1, with the limits y1 to y2 and 0 to y1 respectively to get the asked for cdf.

The book gives an answer with a 2 multiplied by the joint pdf I dont have and I cant figure out why (i.e I am missing the important logic). Anyone help?

Thanks!
• Feb 25th 2009, 09:20 PM
matheagle
The orginal observations, X's, may be independent, but your order stats certainly aren't.
Hence
P(Y1<=y1, Y2<=y2)=P(Y1<=y1)* P(Y2<=y2) is FALSE.
You need to play the cdf of the Y's and switch them to the X's
THEN the X's joint distribution function can factor.
• Feb 26th 2009, 05:31 AM
B_Miner
Are you saying I just need change

P(Y1<=y1, Y2<=y2)
=
P(Y1<=y1)* P(Y2<=y2)
=
[1-[1-P(X1<=y1)[1-P(X2<=y1)][P(X1<=y2)P(X2<=y2)]

TO

P(Y1<=y1, Y2<=y2)
=

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

If so, what do you think about the ultimate answer from the book? I am trying to make sure I get the required logic..

• Feb 26th 2009, 09:28 AM
meymathis
You have probabilities in your probabilities which is probably not good. (Wink)

$\mathbf{P}(Y_1\leq y_1, Y_2 \leq y_1) = \mathbf{P}(\min(X_1,X_2)\leq y_1, \max(X_1,X_2) \leq y_2)$

Now we try to whittle this down to CDF's of X1 and X2.

$= \mathbf{P}\left( (X_1\leq y_1 \cup X_2\leq y_1) \cap (X_1\leq y_2 \cap X_2\leq y_2) \right)$ where $\cup$ is "or" and $\cap$ is "and".

Distribute:
$= \mathbf{P}\left( (X_1\leq y_1 \cap X_1\leq y_2 \cap X_2\leq y_2) \cup (X_2\leq y_1 \cap X_1\leq y_2 \cap X_2\leq y_2) \right)$

Simplify:
$= \mathbf{P}\left( (X_1\leq \min(y_1,y2) \cap X_2\leq y_2) \cup (X_2\leq \min(y_1,y_2) \cap X_1\leq y_2) \right)$

Break the probability on the "or":
$= \mathbf{P}\left( X_1\leq \min(y_1,y2) \cap X_2\leq y_2)\right) + \mathbf{P}\left( X_2\leq \min(y_1,y_2) \cap X_1\leq y_2 \right) -$
$\mathbf{P}\left( (X_1\leq \min(y_1,y2) \cap X_2\leq y_2) \cap (X_2\leq \min(y_1,y_2) \cap X_1\leq y_2) \right)$

Simplify:
$= \mathbf{P}\left( X_1\leq \min(y_1,y2) \cap X_2\leq y_2\right) + \mathbf{P}\left( X_2\leq \min(y_1,y_2) \cap X_1\leq y_2 \right) -$
$\mathbf{P}\left( X_1\leq \min(y_1,y2) \cap X_2\leq \min(y_1,y_2) \right)$

Now you have "and"s with X1 and X2, which are independent. So you can break those up and now you have a whole bunch of CDF's running around! Can you finish it up?
• Feb 26th 2009, 06:09 PM
B_Miner
Thanks very very much for this! I will sit down and wade through to make sure I understand. Do you think this is the process the book used to arrive at their answer ?

G(y1,y2) =int [ int [ 2*(1/100^2)* exp (-(u+v)/1000) dv] du
• Feb 27th 2009, 12:19 PM
meymathis
It's hard to say since I don't have the book nor any real context. I honestly can't think of another way of doing it. Were you able to get any further?
• Feb 27th 2009, 04:52 PM
matheagle
Quote:

Originally Posted by B_Miner
Thanks very very much for this! I will sit down and wade through to make sure I understand. Do you think this is the process the book used to arrive at their answer ?

G(y1,y2) =int [ int [ 2*(1/100^2)* exp (-(u+v)/1000) dv] du

Why don't you scan or type some of the work from your book.
• Feb 27th 2009, 08:12 PM
mr fantastic
Quote:

Originally Posted by matheagle
Why don't you scan or type some of the work from your book.

I'd suggest type and give it a prominent reference. Scanning might create copyright problems.

Quote:

Originally Posted by B_Miner
Thanks very very much for this! I will sit down and wade through to make sure I understand. Do you think this is the process the book used to arrive at their answer ?

G(y1,y2) =int [ int [ 2*(1/100^2)* exp (-(u+v)/1000) dv] du

Nevertheless, I would sincerely hope that you could comprehend what's been posted, compare it with what your book does and then answer your own question.