Problem 75: Random quadratics ($\u2713$ $\u2713$ $\u2713$) 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 $|{X}_{1}+{X}_{2}|$.

Comments

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

$$X=-B\pm \sqrt{{B}^{2}-1}$$which has real roots if $\left|B\right|\ge 1$. Let $\Phi \left(z\right)$ be the probability that a standard normally distributed variable $Z$ satisﬁes $Z\le z$. Then the probability that $\left|B\right|\ge 1$ 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

$$\begin{array}{rcll}-B-\sqrt{{B}^{2}-1}>\frac{1}{5}& \iff & B+\frac{1}{5}<-\sqrt{{B}^{2}-1}& \text{}\\ & \iff & {\left(B+\frac{1}{5}\right)}^{2}>{B}^{2}-1\phantom{\rule{2em}{0ex}}\text{and}\left(B+\frac{1}{5}\right)0& \text{}\\ & \iff & \frac{2}{5}B+\frac{1}{25}-1\phantom{\rule{2em}{0ex}}\text{and}B-\frac{1}{5}& \text{}\\ & \iff & -\frac{1}{5}B-\frac{13}{5}.& \text{}\end{array}$$However, if $B<-\frac{1}{5}$, then the condition that $\sqrt{{B}^{2}-1}$ is real, i.e. $\left|B\right|\ge 1$, implies the stronger condition $B\le -1$. The condition that both roots are real and greater than $\frac{1}{5}$ is therefore

$$-\frac{13}{5}<B\le -1$$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

$$\begin{array}{rcll}\mathrm{P}\left(\right.\text{bothroots}\frac{1}{5}\text{}|\text{bothreal}\left)\right.& =& \frac{\mathrm{P}\left(\right.\text{bothroots}\frac{1}{5}\text{andbothrootsreal}\left)\right.}{\mathrm{P}\left(\right.\text{bothrootsreal}\left)\right.}& \text{}\\ & =& \frac{0.1540}{0.3174}=0.485.& \text{}\end{array}$$

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

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