• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter mattclgn
  • Start date
  • #1
19
0

Homework Statement


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?

Homework 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).

The Attempt at a Solution


[/B]
.....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.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,503
5,433
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.
 

Related Threads on Ross ch.6 problem 26 Joint Distribution of Random Variables.

Replies
10
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
1
Views
599
Replies
2
Views
1K
Replies
3
Views
1K
Replies
7
Views
734
Replies
2
Views
451
Top