0

Suppose that $A, B, C$ are positive, independent random variables with distribution function $F$. I am trying to show that quadratic $Az^2 + Bz + C$ has real roots with probability $$\int_0^\infty \int_0^\infty F\left(\frac{x^2}{4y}\right) dF(x)dF(y).$$

I know that I need to calculate $P[B^2 \geq 4AC]$, which would give me the answer I want, but I'm not sure how to do this. I have tried looking for similiar problems, but the only examples I have been able to find, e.g. Probability that a quadratic polynomial with random coefficients has real roots, are where $A, B, C$ have a uniform distribution. How can I extend to the general case?

Jim
  • 31
  • Possible duplicate of Probability that a quadratic polynomial with random coefficients has real roots – lulu Oct 14 '18 at 14:23
  • @lulu That problem only deals with the case of a uniform distribution, whereas I am trying to work with a general distribution function $F$. – Jim Oct 14 '18 at 14:24
  • Ok, I'll retract the close vote. Though the method of solution ought to be the same....just weight the measure appropriately. – lulu Oct 14 '18 at 14:27
  • @lulu I was trying to extrapolate from those solutions, but was having trouble doing so. I couldn't get the proof to work out properly. They seem to take advantage of the concreteness of the situation, (for example by taking derivatives of functions) which I cannot do here. – Jim Oct 14 '18 at 14:29
  • I'll give you a hint to get you started. Clearly $x$ corresponds to $B$ because the appearance of $x^2$ in $F(x^2/(4y))$. Now there is a symmetry whether you choose $y$ to correspond to $A$ or $C$, say $A$. Then $F(x^2/(4y))$ corresponds to the probability of the other variable $C$ being .... – user10354138 Oct 14 '18 at 14:33
  • @user10354138 Is the second variable being taken to be $1$? How does that work? – Jim Oct 14 '18 at 14:35
  • No. In $F(x^2/(4y))=P(C\leq \dots)$ what is the "..."? Do you see how this all fits in $P(B^2\geq 4AC)$? – user10354138 Oct 14 '18 at 14:40
  • @user10354138 The ... is clearly $B^2 / 4A$. But how is the value of $C$ accounted for in the expression? Is it incorporated into the function $F$ - if so, how? This dependency doesn't seem clear, which I think is confusing me. – Jim Oct 14 '18 at 14:42
  • The cumulative distribution is the probability of $C$ at most that value. – user10354138 Oct 14 '18 at 15:11
  • I get it now, thank you so much for helping me! – Jim Oct 14 '18 at 15:12

1 Answers1

1

(Just for the record of discussion)

You want $\mathbf{P}(B^2\leq 4AC)$. Since $A,B,C$ are positive, i.i.d. random variables with cumulative distribution $F$, we have $$ \mathbf{P}(B^2\leq 4AC) = \int_0^\infty\int_0^\infty\int_0^\infty\mathbf{1}_{x^2\leq 4yc}(x,y,c)\,\mathrm{d}F_C(c)\,\mathrm{d}F_B(x)\,\mathrm{d}F_A(y) $$ where the subscripts $A,B,C$ are there to decorate which random variable we are referring to. Now $x^2\leq 4yc\iff c\leq x^2/(4y)$, so the innermost integral becomes $$ \int_0^\infty\mathbf{1}_{c\leq x^2/(4y)}(x,y,c)\,\mathrm{d}F_C(c) =\int_0^{x^2/(4y)}\,\mathrm{d}F_C(c) =F\left(\frac{x^2}{4y}\right) $$ which is the desired formula which you put the other integrals back.

user10354138
  • 33,239