Definition of Percentile (undefined notation)?

Question

From Loss Models, 4th ed., by Klugman et al.:

Definition 3.6 The $100p^{\text{th}}$ percentile of a random variable is any value $\pi_p$ such that $F\left(\pi_{p^{-}}\right) \leq p \leq F\left(\pi_{p}\right)$.

I have no idea what $\pi_{p^{-}}$ is, and cannot find it defined earlier in the text. Does anyone know what this notation is?

Note: I would prefer a rigorous definition, if available.

Answer 1

It means the distribution function as you approach from the left. This is because distribution functions are càdlàg, also known as "right continuous with left limit". But they don't have to be continuous from both sides. For example, a step function is allowed as long as a limit exists from the left and it is right continuous, even though the right and left limits are not equal. Therefore, the distribution's left limit will be less than or equal to its right limit--which is its value since it is right continuous.

Update

For example, given the following distribution: $$ F(x) = \begin{cases} 0,\quad&\textrm{for}\;x < 0\\ 0.25\quad&\textrm{for}\;0 \leq x< 0.5\\ 0.5\quad&\textrm{for}\;0.5 \leq x< 1.0\\ 0.75\quad&\textrm{for}\;1.0 \leq x < 1.5\\ 1\quad&\textrm{for}\;x\geq 1.5\\ \end{cases} $$

This is obviously a discrete distribution with: $$ P(x) = \begin{cases} 0.25\quad&\textrm{for}\;x = 0\\ 0.25\quad&\textrm{for}\;x = 0.5\\ 0.25\quad&\textrm{for}\;x = 1.0\\ 0.25\quad&\textrm{for}\;x = 1.5 \end{cases} $$ What is the 30th percentile? Well, it is any value in $(0.5, 1.0]$. So, for example, $0.6$ is the $30^{th}$ percentile (among other values) for this distribution since the left limit is $0.25$, the right limit is $0.5$ and $0.25 \leq 0.3 \leq 0.5$.

Answer 2

This notation is, in my opinion, a bit of a ham-handed attempt to define percentile for both continuous and discrete distributions (or some combination thereof).

What $\pi_{p^-}$ appears to mean is "some value in the sample space arbitrarily close to, but still strictly less than, $\pi_p$."

Consider a normal distribution, and let's use our conventional notion of percentile. The 50th percentile is quite clearly at zero; $e^{-x^2/2}$ is symmetric about $x=0$, so let $\pi_p = 0$.

$F(\cdot)$ denotes our cumulative distribution function, so $F(\pi_p) = \int_{-\infty}^{\pi_p} f(x)\ dx$ generally, but for a discrete distribution we would interpret it as a Lebesgue integral over a counting measure, but whatever. In this case, we have $F(0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 \exp \left(-\frac{x^2}{2}\right)\ dx$.

Now, the closest value that we have that is strictly below $F(0)$ would be $F(0^{-})$. We know that we can get arbitrarily close to $0$ from the left, and as we do so, $F$ evaluated at this point gets closer and closer to $F(0)$. So the percentile value is the only value that fits in here. You can formalize the argument using a limit if you'd like.

Now, we may want to look at discrete distributions. In such a case, we can't get arbitrarily close to $\pi_p$, but we can choose the closest thing smaller than $\pi_p$ that is still within the sample space. Therefore, we may define our percentile using the same definition.