# 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:

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 $$0, we have $$\frac{x}{c}>1$$. Similarly, $$-x 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.

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*}