1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ross ch.6 problem 26 Joint Distribution of Random Variables.

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose that A, B, C are independent random variables, each being uniformly distributed over (0, 1). )

    What is the probability that all the roots of the equation Ax2 + Bx + C = 0 are real?

    2. Relevant equations
    (b) What is the probability that all the roots of the equation Ax2 + Bx + C = 0 are real? Recall that the roots of Ax2 + Bx + C = 0 are real if B2 − 4AC ≥ 0, which occurs with probability Z 1 0 Z 1 0 Z 1 min{1, √ 4ac} 1dbdcda = Z 1 0 Z min{1,1/(4a)} 0 Z 1 √ 4ac 1dbdcda = Z 1/4 0 Z 1 0 Z 1 √ 4ac 1dbdcda + Z 1 1/4 Z 1/(4a) 0 Z 1 √ 4ac 1dbdcda = Z 1/4 0 Z 1 0 (1 − 2a 1/2 c 1/2 )dcda + Z 1 1/4 Z 1/(4a) 0 (1 − 2a 1/2 c 1/2 )dcda = Z 1/4 0 c − 4 3 a 1/2 c 3/2 c=1 c=0 da + Z 1 1/4 c − 4 3 a 1/2 c 3/2 c=1/(4a) c=0 da = Z 1/4 0 1 − 4 3 a 1/2 da + Z 1 1/4 1 12 a −1 da = a − 8 9 a 3/2 a=1/4 a=0 + 1 12 log(a) a=1 a=1/4 = 5 36 + 1 12 log(4).

    3. The attempt at a solution

    .....okay, so the solution is actually listed above from this http://www.math.duke.edu/~hadams/teaching/math151win2012/Hw6Sol.pdf. I guess my problem is more the understanding of it....I get the part where it asks me the roots and I have to use the quadrilateral to get them. I don't get fully why the first integral is Z min{1,b2/(4c)} 0 Also, despite having looked it up on wolfram's, I am still not quite sure what I am supposed to do with min{}? Why does it simplify to something where the parameters of integration change? Why does it break into two separate integration's? It doesn't seem to me as though we are doing Int by Parts. I'm pretty much lost on the answer.
     
  2. jcsd
  3. Feb 22, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If you were to take the range for b as ##\sqrt{4ac}## to 1 in the first integral, what would go wrong? The first integral fixes the problem by changing the range for b. The second fixes it by changing the range for c instead. The third resolves the min function by breaking it into two ranges, one in which 1 is the smaller value and the other in which 1/(4a) is the smaller.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted