Probability that a quadratic equation has real roots

Problem

The premise is almost the same as in this question. I’ll restate for convenience.

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

Note: The distribution is now $-1$ to $+1$ instead of $0$ to $1$.

My Attempt

Preparation

When the coefficients are sampled from $\mathcal{U}(0,1)$, the probability for the discriminant to be non-negative that is, $P(B^2-4AC\geq0) \approx 25.4\% $. This value can be obtained theoretically as well as experimentally. The link I shared above to the older question has several good answers discussing both approaches.

Changing the sampling interval to $(-1, +1)$ makes things a bit difficult from the theoretical perspective. Experimentally, it is rather simple. This is the code I wrote to simulate the experiment for $\mathcal{U}(0,1)$. Changing it from (0, theta) to (-1, +1) gives me an average probability of $62.7\%$ with a standard deviation of $0.3\%$

I plotted the simulated PDF and CDF. In that order, they are:

PDFCDF

So I’m aiming to find a CDF that looks like the second image.

Theoretical Approach

The approach that I find easy to understand is outlined in this answer. Proceeding in a similar manner, we have

$$
f_A(a) = \begin{cases}
\frac{1}{2}, &-1\leq a\leq+1\\
0, &\text{ otherwise}
\end{cases}
$$

The PDFs are similar for $B$ and $C$.

The CDF for $A$ is

$$
F_A(a) = \begin{cases}
\frac{a + 1}{2}, &-1\leq a\geq +1\\
0,&a<-1\\
1,&a>+1
\end{cases}
$$

Let us assume $X=AC$. I proceed to calculate the CDF for $X$ (for $x>0$) as:

$$
\begin{align}
F_X(x) &= P(X\leq x)\\
&= P(AC\leq x)\\
&= \int_{c=-1}^{+1}P(Ac\leq x)f_C(c)dc\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P(Ac\leq x)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
\end{align}
$$

We take a quick detour to make some observations. First, when $0<c<x$, we have $\frac{x}{c}>1$. Similarly, $-x<c<0$ implies $\frac{x}{c}<-1$. Also, $A$ is constrained to the interval $[-1, +1]$. Also, we’re only interested when $x\geq 0$ because $B^2\geq 0$.

Continuing, the calculation

$$
\begin{align}
F_X(x) &= \frac{1}{2}\left(\int_{c=-1}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=-x}^{0}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=0}^{x}P\left(A\leq \frac{x}{c}\right)dc + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}P\left(A\leq \frac{x}{c}\right)dc + 0 + 1 + \int_{c=x}^{+1}P\left(A\leq \frac{x}{c}\right)dc\right)\\
&= \frac{1}{2}\left(\int_{c=-1}^{-x}\frac{x+c}{2c}dc + 0 + 1 + \int_{c=x}^{+1}\frac{x+c}{2c}dc\right)\\
&= \frac{1}{2}\left(\frac{1}{2}(-x+x(\log(-x)-\log(-1)+1) + 0 + 1 + \frac{1}{2}(-x+x(-\log(x)-\log(1)+1)\right)\\
&= \frac{1}{2}\left(2 + \frac{1}{2}(-x+x(\log(x)) -x + x(-\log(x))\right)\\
&= 1 – x
\end{align}
$$

I don’t think this is correct.

My Specific Questions

  1. What mistake am I making? Can I even obtain the CDF through integration?
  2. Is there an easier way? I used this approach because I was able to understand it well. There are shorter approaches possible (as is evident with the $\mathcal{U}(0,1)$ case) but perhaps I need to read more before I can comprehend them. Any pointers in the right direction would be helpful.

Answer

I would probably start by breaking into cases based on $A$ and $C$.

Conditioned on $A$ and $C$ having different signs, there are always real roots (because $4AC\leq 0$, so that $B^2-4AC\geq0$). The probability that $A$ and $C$ have different signs is $\frac{1}{2}$.

Conditioned on $A\geq0$ and $C\geq 0$, you return to the problem solved in the link above. Why? Because $B^2$ has the same distribution whether you have $B$ uniformly distributed on $(0,1)$ or on $(-1,1)$. At the link, they computed this probability as $\frac{5+3\log4}{36}\approx0.2544134$. The conditioning event here has probability $\frac{1}{4}$.

Finally, if we condition on $A<0$ and $C<0$, we actually end up with the same probability, as $4AC$ has the same distribution in this case as in the case where $A\geq0$ and $C\geq 0$. So, this is an additional $\frac{5+3\log 4}{36}\approx0.2544134$ conditional probability, and the conditioning event has probability $\frac{1}{4}$.

So, all told, the probability should be
$$
\begin{align*}
P(B^2-4AC\geq0)&=1\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{5+3\log4}{36}+\frac{1}{4}\cdot\frac{5+3\log 4}{36}\\
&=\frac{1}{2}+\frac{5+3\log4}{72}\\
&\approx0.6272…
\end{align*}
$$

Attribution
Source : Link , Question Author : Hungry Blue Dev , Answer Author : Nick Peterson

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