# Thread: Transform of Variable Using CDF Tech

1. ## 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!

2. 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.

3. 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..

4. You have probabilities in your probabilities which is probably not good.

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

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

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

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

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

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

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

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?

5. 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

6. 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?

7. 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.

8. 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.

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.

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