# Probability that a quadratic polynomial with random coefficients has real roots

The following is a homework question for which I am asking guidance.

Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(0,1)$. What is the probability that the polynomial $Ax^2 + Bx + C$ has real roots?

That means I need $P(B^2 -4AC \geq 0$). I've tried calling $X=B^2 -4AC$ and finding $1-F_X(0)$, where $F$ is the cumulative distribution function.

I have two problems with this approach. First, I'm having trouble determining the product of two uniform random variables. We haven't been taught anything like this in class, and couldn't find anything like it on Sheldon Ross' Introduction to Probability Models.

Second, this strategy just seems wrong, because it involves so many steps and subjects we haven't seen in class. Even if I calculate the product of $A$ and $C$, I'll still have to square $B$, multiply $AC$ by four and then subtract those results. It's too much for a homework question. I'm hoping there might be an easier way.

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Hints: First consider $B^2 \geq 4AC$. Now, if $U$ is uniform$(0,1)$, then $-\log(U)$ is exponential$(1)$; further, the sum of two independent exponential$(1)$ random variables has pdf $x e^{-x}$, $x > 0$. Thus, using the law of total probability, the answer can be found by solving an elementary one dimensional integral. I've confirmed the result by simulations.

EDIT: Specifically, $${\rm P}(B^2 - 4AC \ge 0) = \int_{\log 4}^\infty {[1 - e^{ - (x - \log 4)/2} ]xe^{ - x} \,{\rm d}x} = \frac{{5 + 3\log 4}}{{36}} \approx 0.2544134.$$ The answer was confirmed using Monte Carlo simulations: $N=10^8$ runs gave average of $0.25444043$.

EDIT: Note that it is quite easy to determine, moreover, the distribution of the product of $n$ independent uniform$(0,1)$ random variables. Indeed, let $U_1,\ldots,U_n$ be independent uniform$(0,1)$ variables. Write $$U_1 \cdots U_n = \exp \Big[ - \sum\nolimits_{i = 1}^n { - \log U_i } \Big].$$ Since the $-\log U_i$ are independent exponential$(1)$ variables, $U_1 \cdots U_n$ is merely distributed as $e^{-X}$, where $X$ has gamma pdf $x^{n-1}e^{-x}/(n-1)!$, $x > 0$.

EDIT: Elaborating in response to the OP's request (see the first comment below).

Actually, the hint was supposed to send you in a slightly different direction, that is to consider the probability $${\rm P}\bigg( - \log B \le \frac{{( - \log A) + ( - \log C) - \log 4}}{2}\bigg),$$ or $${\rm P}\bigg(X \le \frac{{Y - \log 4}}{2}\bigg),$$ where $X$ is exponential$(1)$ and $Y$, independent of $X$, has gamma pdf $f_Y (x) = xe^{-x}$, $x > 0$. Then, by the law of total probability (and using that $X$ and $Y$ are independent), the above probability is given by $$\int_0^\infty {{\rm P}\bigg(X \le \frac{{Y - \log 4}}{2}\bigg|Y = x\bigg)f_Y (x)\,{\rm d}x} = \int_0^\infty {{\rm P}\bigg(X \le \frac{{x - \log 4}}{2}\bigg)xe^{ - x} \,{\rm d}x},$$ and so substituting the exponential$(1)$ cdf yields the desired integral.

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Taking log on both sides, I want $P(-2(-\log B)\geq\log 4-\left[-\log A+\left(-\log C\right)\right])$, right? With $-\log B$ exponential(1) and $\left[-\log A+\left(-\log C\right)\right]$ a gamma distribution. I'm still having trouble getting to the integral you wrote. I'm just starting to study about pdfs, so a lot is new to me. Where does $\log 4$ goes? How do I compare two random variables to see the prob. that one is greater? I was hoping you could walk this newbie a bit further along the path -- or provide me with some reference on manipulating random variables. Thanks! –  Pedro d'Aquino Mar 27 '11 at 17:05
@Pedro: I'll give more details later on. –  Shai Covo Mar 27 '11 at 17:55
Thanks a lot. I'll still need to think about it, but now I'm sure you've given me all the basis I needed. –  Pedro d'Aquino Mar 27 '11 at 21:48

Hint: You are looking for the volume of the $(a,b,c) \in [0,1]^3$ such that $b^2 \geq 4ac$.

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Just wondering...why isn't he interested in the volume of the solid (i.e. the interior of an elliptic cone) in all of $\mathbf{R}^3$? –  Isaac Solomon Mar 27 '11 at 5:27
because $A,B,C$ are constrained to be between zero and one. –  Zarrax Mar 27 '11 at 5:33
Aha. I glanced over that. Thanks. –  Isaac Solomon Mar 27 '11 at 5:43
Thanks for answering. I upvoted your answer, but didn't select it as answer because I ended up doing it Shai Covo's way. Yours would probably be easier if my calculus wasn't so rusty, though. –  Pedro d'Aquino Mar 27 '11 at 21:50
That's ok. I did do the integral and it agreed with Shai Covo's answer. –  Zarrax Mar 28 '11 at 0:03

Hints:

It is not impossible to find the cumulative distribution of $D=AC$, and so easily of $E=4D=4AC$ and $F=\sqrt{E}=\sqrt{4AC}$ [you know $4AC \ge 0$].

Then all you have to do is find the probability that $F<|B|$ which is not difficult if you know or can work out that $\int x^2 \log(x) \, dx = x^3(3 \log(x)-1)/9$.

As a check, you should be getting a result slightly more than 0.25.

Incidentally as a comment on the question, the requirement that $A$ and $C$ have the same sign makes a big difference to the result, as $B^2-4AC$ is always positive if $A$ and $C$ have opposite signs.

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Thanks for answering. I'm having a bit of trouble calculating all those distributions. Do you have any references that could help me out? Preferably online, but a good book will do as well. I'm just starting to learn about probability distributions, so I'm still not sure how to manipulate random variables algebraically. Again, thank you. –  Pedro d'Aquino Mar 27 '11 at 16:49