# Joint Uniform Distrubtion of 3 variables problem

• Jan 6th 2011, 01:43 PM
Joint Uniform Distrubtion of 3 variables problem
Hi all,

I am working through a problem but I cant seem to see the method... the answer is in the back of the text book but no method!

Q: Let $\displaystyle U_1, U_2, U_3$ be uniformly distributed on $\displaystyle (0,1)$ . Find the prob that $\displaystyle U_1x^2 + U_2x +U_3$ have real roots. The answer in the back of the book is 1/9.

So using the quadratic eqn, I deduce taht the prob distribution we need to find is given by:

$\displaystyle P(U_2^2 -4U_2U_3 >0)$...... I am not sure how to evaluate this at all.

Looking through the usual methods in the text book usually points to evaluating the area via direct integration but when I integrate:

$\displaystyle \int_0^1 \int_0^1 \int_0^1 U_2^2 -4U_2U_3 du_1 du_2 du_3$ I get a negative answer which is obviously wrong... I suspect the limits need to be changed, but not sure where to start.....

• Jan 6th 2011, 02:04 PM
Moo
Hello,

(by the way, you wrote $\displaystyle U_2^2$ instead of $\displaystyle U_1^2$)

And in order to do this problem, you have to assume that the three rv's are independent.

Your integral is that of $\displaystyle E[U_1^2-4U_2U_3]$ so this is not what you want.

In order to find the probability you're looking for, you need to integrate the joint pdf of the three rv's over the region where $\displaystyle u_1^2-4u_2u_3>0$

So this gives $\displaystyle \displaystyle \int_0^1 \int_0^1 \int_0^1 \bold{1}_{\{u_1^2-4u_2u_3>0\}} du_3 du_2 du_1$

So this gives you a new boundary for the integral with respect to $\displaystyle u_3$... continue and you will find the answer
• Jan 6th 2011, 02:16 PM
thanks Moo! will try that tomorrow... bit late where i am now!
• Jan 6th 2011, 02:30 PM
Moo
I'm 1 hour ahead of you :D
• Jan 8th 2011, 05:50 AM
I guess its easy for a prob queen like you to solve these things late at night, us mere moretals though ....(Nod)

Quote:

Originally Posted by Moo
Hello,

(by the way, you wrote $\displaystyle U_2^2$ instead of $\displaystyle U_1^2$)

And in order to do this problem, you have to assume that the three rv's are independent.

Your integral is that of $\displaystyle E[U_1^2-4U_2U_3]$ so this is not what you want.

In order to find the probability you're looking for, you need to integrate the joint pdf of the three rv's over the region where $\displaystyle u_1^2-4u_2u_3>0$

So this gives $\displaystyle \displaystyle \int_0^1 \int_0^1 \int_0^1 \bold{1}_{\{u_1^2-4u_2u_3>0\}} du_3 du_2 du_1$

So this gives you a new boundary for the integral with respect to $\displaystyle u_3$... continue and you will find the answer

I don't follow Moo.... If I integrate over this region, with $\displaystyle du_1,du_2$ first, dont I just end up with $\displaystyle \int_0^1 1 du_3$ and is it here that I make the upper limit of integration to$\displaystyle \int_0^{u_1^2 -4u_2u_3} du_3$ ?? I am not sure... maybe I need to look back at my old multivariable integration notes first!
• Jan 8th 2011, 10:45 AM
Moo
I'm not a queen, merely a cow :D

If you integrate wrt $\displaystyle u_1$ first, you have $\displaystyle u_1^2-u_2u_3>0 \Leftrightarrow u_1>\sqrt{u_2u_3}$ ($\displaystyle U_1$ is uniform(0,1) so $\displaystyle u_1>0$ almost surely).

So since the integral is 0 if $\displaystyle u_1$ doesn't satisfy this inequality, we'll have $\displaystyle \displaystyle \int_0^1 \int_0^1 \int_{\sqrt{u_2u_3}}^1 1~du_1du_2du_3$, which gives you $\displaystyle \displaystyle \int_0^1\int_0^1 1-\sqrt{u_2u_3} ~ du_2du_3$, which, I guess, you can finish off !
(I just hope it really makes 1/9 lol)

You can do it in any order you wish to do (thanks to Fubini-Tonelli's theorem)
• Jan 10th 2011, 12:44 PM
cheers for stepping me through that moo. I don't get 1/9.... this is my working:

$\displaystyle \displaystyle \int_0^1\int_0^1 1-\sqrt{u_2u_3} ~ du_2du_3$

$\displaystyle \displaystyle = \int_0^1\ [u_2 - \frac{2}{3}u_2^{3/2}u_3^{1/2}]_0^1 du_3$

$\displaystyle \displaystyle = \int_0^1\ 1 - \frac{2}{3}u_3^{1/2} du_3$

$\displaystyle \displaystyle = [ u_3 - \frac{2}{3}.\frac{2}{3}u_3^{3/2} ]_0^1$

$\displaystyle = 1- \frac{4}{9}$

$\displaystyle = \frac{5}{9}$

I follow your working and theory so I have either made a mistake in the working here or theres a misprint in the book!

In any case, thank you very much for the help in this!