Median

Study guide: Exam P-1

The Median $m$ of a Random Variable $X$ is its 50th percentile: the smallest value $m$ with $F(m) \ge 0.5$ . For a continuous distribution that is the point where $F(m) = 0.5$ , splitting the probability in half.

$F(m) = 0.5$

$m = F^{-1}(0.5)$

For a continuous $X$ with a strictly increasing CDF the median is unique: solve $F(m) = 0.5$ , or equivalently $\int_{-\infty}^{m} f(x)\,dx = 0.5$ .
For a discrete $X$ the CDF usually jumps past 0.5 rather than landing on it, so $F(m) = 0.5$ has no solution; the median is then the smallest $m$ with $F(m) \ge 0.5$ — the same rule as for every other Percentile.
Median versus mean. For a normal distribution the median equals the mean; for a skewed one they differ, the mean pulled toward the heavier tail. Claim severities are right-skewed — a few large claims pull the mean up but leave the median alone — so typically mode < median < mean. For the exponential, $m = \theta \ln 2 \approx 0.693\theta$ against a mean of $\theta$ ; for the lognormal, $m = e^{\mu}$ against $e^{\mu + \sigma^2/2}$ .
Deductibles pass straight through. The payment $(X - d)_+$ is a continuous, non-decreasing function of the loss, so its median is the loss median less the deductible, floored at zero: $\max(m - d,\ 0)$ . The same holds for every percentile. The mean has no such property: $E[(X - d)_+] \ne E[X] - d$ in general.
The sample median is the middle order statistic, or the average of the two middle values when $n$ is even.

Median Fire Damage Ratio The fraction $X$ of a building's value destroyed in a fire has density $f(x) = 2(1 - x)$ for $0 < x < 1$ . Find the median, and compare it with the mean and the mode.
Answer Integrate for the CDF, then set it to one half:
 $\begin{align*} F(x) &= \int_0^x 2(1 - t)\,dt \\ &= 1 - (1 - x)^2 \\ (1 - m)^2 &= 0.5 \\ m &= 1 - \sqrt{0.5} \\ &= 0.293 \end{align*}$ 
The mean is $\int_0^1 2x(1 - x)\,dx = \tfrac{1}{3} \approx 0.333$ , and the density is largest at $x = 0$ , so the mode is $0$ . The ordering $0 < 0.293 < 0.333$ is the right-skew pattern: half of fires destroy less than 29% of the building, even though the average loss is a third of its value.

Median Claim Count The number of claims $N$ on a policy has $P(N = 0, 1, 2, 3) = 0.35, 0.25, 0.20, 0.20$ . Find the median and the mean.
Answer Accumulate until the CDF first reaches 0.5: $F(0) = 0.35$ , then $F(1) = 0.60$ . So the median is $1$ .
 $\begin{align*} E[N] &= 1(0.25) + 2(0.20) + 3(0.20) \\ &= 1.25 \end{align*}$ 
No value of $N$ has $F = 0.5$ exactly — a discrete median is found where the CDF crosses one half, not where it equals it.

Median Payment Under a Deductible Losses are exponential with mean $5,000, and a policy pays $Y = (X - 1{,}000)_+$ . Find the median and the mean of $Y$ .
Answer The median loss is $5{,}000 \ln 2 = 3{,}465.74$ . Because $F(1{,}000) = 1 - e^{-0.2} = 0.181 < 0.5$ , the non-decreasing map $x \mapsto (x - 1{,}000)_+$ carries the median straight through:
 $\begin{align*} m_Y &= 3{,}465.74 - 1{,}000 \\ &= 2{,}465.74 \end{align*}$ 
The mean does not shift that way. For the exponential, $E[(X - d)_+] = \theta e^{-d/\theta}$ :
 $\begin{align*} E[Y] &= 5{,}000\,e^{-0.2} \\ &= 4{,}093.65 \end{align*}$ 
That is not $5{,}000 - 1{,}000 = 4{,}000$ . The median payment is $2,466 and the mean payment $4,094: subtracting the deductible carries the median through, but not the mean.