Normal quantile relation

Question

If $X$ is a random variable having a standard normal distribution and $z(p)$ is the point having probability $p$ to the right of it. Is there any relation between $z(p)$ and $z(p/2)$?

Answer 1

There is, of course, a formal relationship between $z(p)$ and $z(p/2)$ such as the one stated in the answer by Xi'an but note that, as also pointed out in ping's comment on that answer, Xi'an's answer is merely a restatement of the question asked. The question asked by the OP is of some interest to those interested in digital communications because the bit error rate (BER) $p$ in some systems is related to the signal-to-noise ratio (SNR) as $$\text{BER} = Q(\text{SNR})$$ where $Q(x) = 1-\Phi(x)$ is the complementary standard normal CDF. Communications engineers spend a great deal of time and effort in achieving as large an SNR as they can since increasing the SNR results in minimizing BER, and they might be assumed to have some interest in finding out how much more they need to increase SNR in order to halve the BER.

With that as preliminaries explaining why the answers to the OP's question might be of some interest to a few readers of stats.SE, let us consider $Q(z) = \int_z^\infty \phi(x) \mathrm dx$ where $\phi(x)$ is the standard normal density function. $Q(z)$ is a convex function on the positive real line having value $Q(0)= \frac 12$, and decreasing to limiting value $0$ as $z \to \infty$. The derivative of $Q(z)$ is $-\phi(z)$ and the tangent to $Q(z)$ lies below the curve. The tangent line at the point $(z(p),p)$ has equation $$y = p -\phi(z(p))[z-z(p)]$$ and thus reaches value $p/2$ at \begin{align} z &=z(p) + \frac{p}{2\phi(z(p))}\\ &= z(p) + 0.5\frac{Q(z(p))}{\phi(z(p))} \end{align} where the fraction on the right can be recognized as Mills' ratio evaluated at $z(p)$. Since the $Q(z)$ curve is above the tangent, $z(p/2)$ is larger than the above value of $z$, that is, $$z(p/2) - z(p) > 0.5\frac{Q(z(p))}{\phi(z(p))}. \tag{1}$$ Similarly, the tangent to the $Q(z)$ curve at $(z(p/2), p/2)$ has equation $$y = p/2 -\phi(z(p/2))[z-z(p/2)]$$ and thus has value $p$ at \begin{align} z &=z(p/2) - \frac{p/2}{\phi(z(p/2))}\\ &= z(p/2) - \frac{Q(z(p/2))}{\phi(z(p/2))} \end{align} Since the $Q(z)$ curve is above the tangent, $z(p)$ is larger than the above value of $z$, that is, $$z(p/2) - z(p) < \frac{Q(z(p/2))}{\phi(z(p/2))}\tag{2}$$ where we can identify the fraction as Mill's ratio evaluated at $z(p/2)$. Thus, we have the bounds $$ 0.5\frac{Q(z(p))}{\phi(z(p))} < z(p/2) - z(p) < \frac{Q(z(p/2))}{\phi(z(p/2))}.\tag{3}$$ Now, it is well known that for $z>0$, $Q(z) < z^{-1} \phi(z)$ which bound is not very useful for small $z$ but is tight for large $z$. Thus, for large $z$ (corresponding to small $p$, we can modify $(3)$ to $$\frac{1}{2z(p)} < z(p/2) - z(p) < \frac{1}{z(p/2))}.\tag{4}$$

Alternatively, note that Eqs. $(26.2.22)$ and $(26.2.23)$ of Abramowitz and Stegun's Handbook of Mathematical Functions tell us that for $p < \frac 12$, $z(p)$ can be approximated as $t - h(t)$ where $h(t)$ is a rational function of $$t = \sqrt{\frac{1}{\ln\left(\frac{1}{p^2}\right)}} = \frac{1}{\sqrt{-2\ln(p)}}.$$ This could be used to derive an approximation for $z(p/2)$ in terms of $z(p)$.