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

  1. uniform, that is $1/h$, and
  2. exponential, that is $\alpha e^{-\alpha x}$

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.

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    Nitpick: The probability is 1. The polynomial always has complex roots, as the reals are$a$subset of the complex numbers. Perhaps the question should be "non-real" roots...2011-04-19

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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 I only solve (1) when $h\le 1$. The case when $h>1$ should be similar. $P(a^2 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.

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    I recently was going through the same problem. I proceeded in a similar way. However the answer given was h/3 for 02017-03-19
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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.

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    @Didier; Yes! That's a problem with having taken the class back in the 1970s.2011-04-19
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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.