Prob of Real Roots

maccaman

New Member
Hi TalkStats,
This is my first post. If someone could help me solve this problem I would be very grateful.

Question:
A,B,C are independent RV's. Consider the quadratic equation:

A(x^2) + Bx + C = 0.

Find the probability that the given equation has real roots when A,B,C have a U(0,1) distribution?

My work so far:

The probability that the roots are real is the probability that B^2 - 4AC >= 0. i.e. the discriminant is positive.

I have worked out the following from my text.....please see image.

but I am now confused with how to get the solution. Can anyone please show me what to do from here?

vinux

Dark Knight
Nice problem.. And you should apply the condition in the range... B2-4AC >=0.
So the ranges could be

A ranges 0 to 1
C ranges 0 to 1
B ranges from -2sqrt(AC) to 2sqrt(AC) => 0 to min(2sqrt(AC) ,1)

maccaman

New Member

I am lost with the how to apply the condition. Given the pdf of a uniform distribution, what would the actual integration be??

This is where im having trouble

vinux

Dark Knight
Yes. it seems a difficult problem. Applying the condition is tricky...

Let us split this into multiple sum..
because you have constraint( B2-4AC >=0 ) here and lower and upper bounds ( 0 and 1). So this would make multiple cases.

case1 2sqrt(AC) <=1 and 4C <=1
the above condition we can include in the limit as follows

C ranges 0 to 1/4
A ranges 0 to min(1/(4C),1) => 0 to 1
and B ranges 0 to 2sqrt(AC)

So the area is = int(C) int(A) int(B,0 to 2sqrt(AC)) fC fA fC
= int(C) int(A , 0 to 1 ) 2sqrt(AC) fA fC
= int(C, 0 to 1) {2sqrt(AC) A *2/3}at A = 1 fC
= int(C, 0 to 1) {sqrt(C) *2/3}fC
= {2/3 sqrt(C)C *2/3 }at C=1/4
= 1/18

case2 2sqrt(AC) <=1 and 4C >=1
the above condition we can include in the limit as follows

C ranges 1/4 to 1
A ranges 0 to 1/(4C)
and B ranges 0 to 2sqrt(AC)

So the area is = int(C) int(A) int(B,0 to 2sqrt(AC)) fC fA fC
= int(C) int(A , 0 to 1 ) 2sqrt(AC) fA fC
= int(C, 0 to 1) {2sqrt(AC) A *2/3}at A = 1/4C fC
= int(C, 0 to 1) {1/4C *2/3}fC
= {1/6 log(C) }at C=1 - {1/6 log(C) }at C=1/4

case3 2sqrt(AC) >=1 and C<=1/4
the above condition we can include in the limit as follows

C ranges 0 to 1/4
A ranges 1/4C to 1 => 1 to 1
and B ranges 0 to 2sqrt(AC)

area =0

case4 2sqrt(AC) >=1 and C >=1/4
the above condition we can include in the limit as follows

C ranges 1/4 to 1
A ranges 1/(4C) to 1
and B ranges 0 to 1
calculate area..

You can sum the four cases and get the proabability.

Please cross check the integral, because i haven't done integration for long time
please let me know if you want to have clarification on any steps apart from integration.