Problem 75:  Random quadratics ($✓$ $✓$ $✓$) 1988 Paper II

The random variable $B$ is normally distributed with mean zero and unit variance. Find the probability that the quadratic equation

${X}^{2}+2BX+1=0$

has real roots.

Given that the two roots ${X}_{1}$ and ${X}_{2}$ are real, ﬁnd, giving your answers to three signiﬁcant ﬁgures:

(i)
the probability that both ${X}_{1}$ and ${X}_{2}$ are greater than $\frac{1}{5}$;
(ii)
the expected value of $|{X}_{1}+{X}_{2}|$.

It is quite difficult to ﬁnd statistics questions at this level that are not too difficult and are also not simple applications of standard methods. For example, ${\chi }^{2}$ tests are not really suitable, because the theory is sophisticated while the applications are usually rather straightforward. Most questions in the probability/statistics area tend therefore to concentrate on probability, and many of these have a bit of pure mathematics thrown in.

Here, the random variable is the coefficient of a quadratic equation, which is rather pleasing. But you have to handle the inequalities carefully. The difficulty is increased by the conditional element: for parts (i) and (ii) you are only interested in the case of real roots.

If you don’t have statistical tables handy, don’t bother to ﬁnd some: just leave the answers in terms of the probability, $\Phi \left(z\right)$, that a standard ($\mu =0$, $\sigma =1$) normally distributed random variable $Z$ satisﬁes $Z\le z\phantom{\rule{0.3em}{0ex}}$.

Solution to problem 75

The solution of the quadratic is

$X=-B±\sqrt{{B}^{2}-1}$

which has real roots if $|B|\ge 1$. Let $\Phi \left(z\right)$ be the probability that a standard normally distributed variable $Z$ satisﬁes $Z\le z$. Then the probability that $|B|\ge 1$ is (taking the two tails of the normal distribution)

$\Phi \left(-1\right)+\left(1-\Phi \left(1\right)\right)=2-2\phantom{\rule{0.3em}{0ex}}\Phi \left(1\right)=0.3174.$

(i) We need the smaller root to be greater than $\frac{1}{5}$. The smaller root is $-B-\sqrt{{B}^{2}-1}$. Now provided $\sqrt{{B}^{2}-1}$ is real, we have

However, if $B<-\frac{1}{5}$, then the condition that $\sqrt{{B}^{2}-1}$ is real, i.e. $|B|\ge 1$, implies the stronger condition $B\le -1$. The condition that both roots are real and greater than $\frac{1}{5}$ is therefore

$-\frac{13}{5}

and the probability that both roots are real and greater than $\frac{1}{5}$ is

$\Phi \left(-1\right)-\Phi \left(-2.6\right)=\Phi \left(2.6\right)-\Phi \left(1\right)=0.9953-0.8413=0.1540.$

The conditional probability that both roots are greater than $\frac{1}{5}$ given that they are real is

(ii) The sum of the roots is $-2B$, so we want the expectation of $|2B|$ given that $|B|\ge 1$, which is

$\frac{\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{-1}\left(-2x\right){e}^{-\frac{1}{2}{x}^{2}}\mathrm{d}x+\frac{1}{\sqrt{2\pi }}{\int }_{1}^{\infty }2x{e}^{-\frac{1}{2}{x}^{2}}\mathrm{d}x}{\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{-1}{e}^{-\frac{1}{2}{x}^{2}}\mathrm{d}x+\frac{1}{\sqrt{2\pi }}{\int }_{1}^{\infty }{e}^{-\frac{1}{2}{x}^{2}}\mathrm{d}x}=\frac{2×\frac{1}{\sqrt{2\pi }}×2{e}^{-\frac{1}{2}}}{2\left(\right1-\Phi \left(1\right)\left)\right}=3.05.$