Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the probability that $x^2 - 2ax + b$ has complex roots if the coefficients $a$ and $b$ are random variables with the common density

  1. uniform, that is $1/h$, and
  2. exponential, that is $ae^{-ax}$

This comes down to finding $P(a^2 \lt b)$. But since $a$ and $b$ are both random variables, would it be $P(a^2\lt b) = P(x\lt k)P(y \lt k^2)$? That doesn't seem particularly correct.

share|improve this question
2  
I've tried to fix your TeX. Please check if I have introduced any mistakes. Moreover, you probably mean non-real roots, otherwise it would be trivially $1$ :) –  t.b. Apr 18 '11 at 22:22
1  
For 1, is the range of $a,b \ [0,h] \text{ or } [\frac{-h}{2},\frac{h}{2}]$ or what? –  Ross Millikan Apr 18 '11 at 22:25
1  
I guess $a$ and $b$ are independent? –  Mariano Suárez-Alvarez Apr 18 '11 at 22:33
    
the range of 1) is 0<x<h and for 2) it is x>0. And yes, a and b are uncorrelated –  user9752 Apr 18 '11 at 22:37
    
The $a$ in the quadratic is probably not the same $a$ in $ae^{-ax}$ so you should change the latter $a$ to something else. As it is, sampling $a$ from the density $ae^{-ax}$ doesn't make much sense in this context. –  JasonMond Apr 19 '11 at 1:22
show 1 more comment

3 Answers

I have seen several similar questions here. The idea is to use the joint density function of $a$ and $b$, which is (assuming independence), (1) $f(x,y)=\frac1{h^2}$ in the square $[0,h]^2$ and 0 otherwise; (2) $f(x,y)=\lambda^2e^{-\lambda(x+y)}$ in the first quadrant and 0 otherwise (I replace the parameter $a$ by $\lambda$ because it's confusing with a r.v. $a$.)

In both cases: $$P(a^2<b)=\iint_{\{(x,y):x^2<y\}} f(x,y)dxdy.$$ I only solve (1) when $h\le 1$. The case when $h>1$ should be similar. $$P(a^2<b)=\int_0^h\int_{x^2}^h dydx=h^2-\frac{h^3}3,$$ which is the area of the region within the square $[0,h]^2$ and above the parabola $y=x^2$.

Case (2) can be solved similarly with a different integral, and I'll leave it to you.

share|improve this answer
add comment

I'm guessing that you need to add "uncorrelated" to the description of the random variables a and b. In that case, you will have to integrate over the possible values. For the uniform case this is easy, the probability is proportional to the area. For the other, I'm going to let you think about it and see what you come up with.

share|improve this answer
    
My idea for the exponential is to do a double integral, with paramters -b^1/2 to b^1/2 and parameters for b from 0 to infinity. Would I then just plug in the probability density functions for a and b as the integrand? Thanks so much! –  user9752 Apr 18 '11 at 22:57
    
@zamenek; And I guess you realize that you only want to integrate over the area where the result is whatever it is you want (i.e. complex). –  Carl Brannen Apr 18 '11 at 23:06
    
And I guess that by uncorrelated what you really mean is independent. –  Did Apr 19 '11 at 5:55
    
@zamenek Did you notice that the beginning of @GWu's post applies to any distribution of $(a,b)$? This answers your question to @Carl. –  Did Apr 19 '11 at 5:56
    
@Didier; Yes! That's a problem with having taken the class back in the 1970s. –  Carl Brannen Apr 19 '11 at 23:00
add comment

The polynomial has complex roots if and only if (2a)^2 - 4b <0. That means $a^2 -b <0.$ Now all you have to calculate $P(A^2 < B)$ with both of those distributions.

That I will leave to you (or another poster) to do.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.