Percentile

Study guides: Exam P-1, Exam FAM, Exam MAS-I, Exam 7

The Percentile (or quantile) $x_p$ of a Random Variable $X$ is the smallest value at which the CDF reaches $p$ , so at least a proportion $p$ of the distribution lies at or below it. When $F$ is continuous and strictly increasing, it is the value that solves $F(x_p) = p$ , found by inverting the CDF.

$x_p = \min\{x : F(x) \geq p\}$

$F(x_p) = p \quad (F \text{ continuous, strictly increasing})$

$x_p = F^{-1}(p)$

Here $p \in (0, 1)$ ; the $100p$ -th percentile is the quantile of order $p$ .
The 50th percentile ( $p = 0.5$ ) is the median.
Where $F$ jumps past $p$ — a discrete variable, or a payment with a probability mass at 0 — no $x$ solves $F(x) = p$ ; the percentile is the value at which the CDF first reaches or passes $p$ . For a Poisson with mean 3, $F(3) = 0.647$ and $F(4) = 0.815$ , so its 75th percentile is 4.
Percentiles describe tail behavior and underlie actuarial risk measures such as Value at Risk (VaR), the loss level a portfolio will not exceed with probability $p$ .

90th Percentile of an Exponential Distribution Claim sizes follow an Exponential Distribution with CDF $F(x) = 1 - e^{-x/10}$ for $x \geq 0$ . Find the 90th percentile.
Answer Set $F(x_{0.90}) = 0.90$ and solve:
 $\begin{align*} 1 - e^{-x_{0.90}/10} &= 0.90 \\ e^{-x_{0.90}/10} &= 0.10 \\ x_{0.90} &= -10 \ln(0.10) \\ &= 10 \ln(10) \approx 23.03 \end{align*}$ 
About 90% of claims fall below 23.03.

95th Percentile of a Normal Loss Annual losses follow a Normal Distribution $X \sim N(\mu = 100,\ \sigma^2 = 225)$ . Find the 95th percentile.
Answer The 95th percentile of the standard normal is $z_{0.95} = 1.645$ . Transform back with $\sigma = 15$ :
 $\begin{align*} x_{0.95} &= \mu + z_{0.95}\,\sigma \\ &= 100 + 1.645(15) \\ &= 124.68 \end{align*}$ 
There is a 95% chance the annual loss is below 124.68.

Percentiles of a Payment Under a Deductible Losses $X$ are exponential with mean $5,000, and a policy pays $Y = (X - 1{,}000)_+$ . Find the 10th and 90th percentiles of $Y$ .
Answer $Y = 0$ whenever $X \le 1{,}000$ , so the CDF of $Y$ jumps at zero: $F_Y(0) = P(X \le 1{,}000) = 1 - e^{-0.2} = 0.1813$ Since $F_Y(0) = 0.1813 \ge 0.10$ , the 10th percentile is $y_{0.10} = 0$ , even though $F_Y(y) = 0.10$ has no solution. Above the jump $F_Y$ is continuous and increasing, so the 90th percentile solves $F_Y(y) = 0.90$ :
 $\begin{align*} 1 - e^{-(y + 1{,}000)/5{,}000} &= 0.90 \\ y + 1{,}000 &= 5{,}000 \ln 10 \\ y_{0.90} &= 11{,}512.93 - 1{,}000 \\ &= 10{,}512.93 \end{align*}$ 
Every percentile with $p \le 0.1813$ is 0; above that, the continuous rule applies.