# Math Help - An integration problem

1. ## An integration problem

Hi! I'm new here and sure that you guys are able to help me out. I've got stuck with this assignment that I don't know how to solve. My problem is that I don't know what limits I should use for the integrals.

You can see the problem here:
http://img103.imagevenue.com/img.php..._122_916lo.jpg

a) Find P(Y1 < 2, Y2>1)
I found the right answer to this one, but there are also a B- and a C-assignment, which I don't know how to solve.

b) Find P(Y1> or equal to 2Y2)

c) P(Y1-Y2> or equal to 1)

The correct answer in this is e^-y1

Can you please show how to limit the integrals as well?

Thanks a lot!

2. Originally Posted by mirrormirror
Hi! I'm new here and sure that you guys are able to help me out. I've got stuck with this assignment that I don't know how to solve. My problem is that I don't know what limits I should use for the integrals.

You can see the problem here:
http://img103.imagevenue.com/img.php..._122_916lo.jpg

[snip]
b) Find P(Y1> or equal to 2Y2)

[snip]
$\Pr(Y_1 \geq 2 Y_2) = \int_{y_2 = 0}^{+\infty} \int_{y_1 = 2 y_2}^{+\infty} e^{-y_1} \, dy_1 \, dy_2$.

3. Originally Posted by mirrormirror
Hi! I'm new here and sure that you guys are able to help me out. I've got stuck with this assignment that I don't know how to solve. My problem is that I don't know what limits I should use for the integrals.

You can see the problem here:
http://img103.imagevenue.com/img.php..._122_916lo.jpg

[snip]

c) P(Y1-Y2> or equal to 1)

The correct answer in this is e^-y1 Mr F says: The correct answer is actually e^-1.

Can you please show how to limit the integrals as well?

Thanks a lot!
$\Pr(Y_1 - Y_2 \geq 1) = \Pr(Y_1 \geq Y_2 + 1) = \int_{y_2 = 0}^{+\infty} \int_{y_1 = y_2 + 1}^{+\infty} e^{-y_1} \, dy_1 \, dy_2 = e^{-1} = \frac{1}{e}$.

4. Thanks a lot! But why can y2 go from 0 - infinity when there is a restriction that y2 must be smaller than or equal to y1? And can you please show how to calculate the B-assignment as well? I don't get it right anyway. Once again, thanks a lot for your help. And you were right about the answer in C, i read it wrong from the book.

5. Originally Posted by mirrormirror
Thanks a lot! But why can y2 go from 0 - infinity when there is a restriction that y2 must be smaller than or equal to y1? Mr F says: The restriction means that the pdf is non-zero only when you're to the right of the Y1 axis and above the line Y1 = Y2 .....

Draw the region of integration for the question ..... The region is the area to the right of the Y1 axis and above the line ${\color{red}y_1 = 2 y_2}$. Do you have experience calculating double integrals over a region (it's possible that you don't, but you need to know how to do it for this statistics subject .....)

And can you please show how to calculate the B-assignment as well? I don't get it right anyway. Once again, thanks a lot for your help. And you were right about the answer in C, i read it wrong from the book.
$\int_{y_2 = 0}^{+\infty} \int_{y_1 = 2 y_2}^{+\infty} e^{-y_1} \, dy_1 \, dy_2 = \int_{y_2 = 0}^{+\infty} \left[ -e^{-y_1} \right]_{2 y_2}^{+\infty} \, dy_2 = \int_{y_2 = 0}^{+\infty} e^{-2 y_2} \, dy_2 = \, ....$

6. Thanks again. I've got two other problems that I've been stuck with for a couple of days. I'd be so glad if someone could help me out with these. Since I'm not very good at writing down the problems, I took a photo of the two assignments that you can see here: http://img121.imagevenue.com/img.php..._122_582lo.jpg

The problem is the same here, I'm not sure how to limit the integrals, and when I start calculate I always go wrong. So I'd be very grateful for a quite detailed explanation. It'd make my day!

7. Originally Posted by mirrormirror
Thanks again. I've got two other problems that I've been stuck with for a couple of days. I'd be so glad if someone could help me out with these. Since I'm not very good at writing down the problems, I took a photo of the two assignments that you can see here: http://img121.imagevenue.com/img.php..._122_582lo.jpg

The problem is the same here, I'm not sure how to limit the integrals, and when I start calculate I always go wrong. So I'd be very grateful for a quite detailed explanation. It'd make my day!
New questions should be posted in a new thread.

For (a) and (b) does the notation mean $\Pr\left(-\infty \leq y_1 \leq \frac{1}{2}, ~ -\infty \leq y_2 \leq \frac{1}{2}\right)$ and $\Pr\left(-\infty \leq y_1 \leq \frac{1}{2}, ~ -\infty \leq y_2 \leq 2 \right)$.

For all three questions you first need to draw on a set of Y1-Y2 axes the region over which you're doing the double integral. Can you show the region represented by $y_1 - 1 \leq y_2 \leq -y_1 + 1$ and $0 \leq y_1 \leq 1$.

(a) and (b) Can you show the appropriate part of the above region?

(c) Can you then show the region corresponding to $y_1 > y_2$?

What experience do you have in setting up and solving double integrals? Doing this looks like your real problem - my advice is that you first learn/consolidate/revise this material before trying these sorts of questions.

8. Thanks for your help. I've managed to solve a and b. And I'm quite good at solving the integrals, my problem is that I don't always know how to limit them. Now I don't know how to limit the integral to solve P(Y1>Y2) I've drawn a picture of the integration area, which you can see here:

http://img203.imagevenue.com/img.php..._122_494lo.jpg

The red area is where Y1>Y2, but I need some help to limit the integrals. Thanks in advance.

9. Originally Posted by mirrormirror
Thanks for your help. I've managed to solve a and b. And I'm quite good at solving the integrals, my problem is that I don't always know how to limit them. Now I don't know how to limit the integral to solve P(Y1>Y2) I've drawn a picture of the integration area, which you can see here:

http://img203.imagevenue.com/img.php..._122_494lo.jpg

The red area is where Y1>Y2, but I need some help to limit the integrals. Thanks in advance.
The first thing you need to do is get the coordinates of the intersection point of the lines $y_2 = y_1$ and $y_2 = -y_1 + 1$. Let them be (a, b). Then one possible double integral is:

$\Pr(Y_1 > Y_2) = \int_{y_2 = y_1 - 1}^{y_2 = y_1} \int_{y_1 = 0}^{y_1 = a} f(y_1, \, y_2) \, dy_2 \, dy_1 + \int_{y_2 = y_1 - 1}^{y_2 = -y_1 + 1} \int_{y_1 = a}^{y_1 = 1} f(y_1, \, y_2) \, dy_2 \, dy_1$.

It's no good being able to integrate if you can't set up the appropriate integrals with the required integral terminals - you obviously have to work on this.

10. Yeah, I know that I need to practice. Do you know any good internet pages or books that teaches you how to put up double integrals?