# Math Help - Joint Uniform Distrubtion of 3 variables problem

1. ## 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 $U_1, U_2, U_3$ be uniformly distributed on $(0,1)$ . Find the prob that $U_1x^2 + U_2x +U_3$ 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:

$P(U_2^2 -4U_2U_3 >0)$...... 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:

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

2. Hello,

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

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

Your integral is that of $E[U_1^2-4U_2U_3]$ 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 $u_1^2-4u_2u_3>0$

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

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

3. thanks Moo! will try that tomorrow... bit late where i am now!

4. I'm 1 hour ahead of you

5. I guess its easy for a prob queen like you to solve these things late at night, us mere moretals though ....

Originally Posted by Moo
Hello,

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

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

Your integral is that of $E[U_1^2-4U_2U_3]$ 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 $u_1^2-4u_2u_3>0$

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

So this gives you a new boundary for the integral with respect to $u_3$... continue and you will find the answer
I don't follow Moo.... If I integrate over this region, with $du_1,du_2$ first, dont I just end up with $\int_0^1 1 du_3$ and is it here that I make the upper limit of integration to $\int_0^{u_1^2 -4u_2u_3} du_3$ ?? I am not sure... maybe I need to look back at my old multivariable integration notes first!

6. I'm not a queen, merely a cow

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

So since the integral is 0 if $u_1$ doesn't satisfy this inequality, we'll have $\displaystyle \int_0^1 \int_0^1 \int_{\sqrt{u_2u_3}}^1 1~du_1du_2du_3$, which gives you $\displaystyle \int_0^1\int_0^1 1-\sqrt{u_2u_3} ~ du_2du_3$, 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)

7. cheers for stepping me through that moo. I don't get 1/9.... this is my working:

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

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

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

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

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

$= \frac{5}{9}$

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!