Mode

Study guide: Exam P-1

The Mode of a Random Variable $X$ is the value at which its PMF or PDF $f(x)$ is largest — the single most likely value of a discrete variable, or the peak of a continuous density.

$\text{mode} = \arg\max_{x}\, f(x)$

Interior peak (continuous): solve $f'(x) = 0$ and confirm it is a maximum. Maximising $\ln f(x)$ gives the same point and is usually easier for gamma- or lognormal-type densities, where the product becomes a sum.
Endpoint peak: a density that only decreases, such as the exponential, has its mode at the left end of the support. Calculus finds no critical point there, so always check the endpoints.
Discrete: compare neighbouring probabilities. For the Poisson, $p(k)/p(k-1) = \lambda/k$ , so probabilities rise while $k \le \lambda$ and the mode is $\lfloor \lambda \rfloor$ (both $\lambda - 1$ and $\lambda$ when $\lambda$ is an integer). For the binomial the same argument gives $\lfloor (n+1)p \rfloor$ (and $(n+1)p - 1$ as well when $(n+1)p$ is an integer).
A mode need not be unique. A uniform distribution has every point of its support as a mode, and a mixture of two claim populations can be bimodal.
Standard results: normal $\mu$ ; Gamma $(\alpha - 1)\theta$ for $\alpha \ge 1$ ; lognormal $e^{\mu - \sigma^2}$ . For a right-skewed severity, mode < median < mean is typical, and the lognormal always obeys it: $e^{\mu - \sigma^2} < e^{\mu} < e^{\mu + \sigma^2/2}$ .
The mode ignores the tail entirely. It describes the most common claim, not the cost of claims — pricing is done on the mean.

Most Likely Claim Size under a Gamma Severity Claim sizes have density $f(x) = \dfrac{x\,e^{-x/2{,}000}}{2{,}000^2}$ for $x > 0$ (a gamma with $\alpha = 2$ , $\theta = 2{,}000$ ). Find the mode and compare it with the mean.
Answer Maximise the log-density:
 $\begin{align*} \ln f(x) &= \ln x - \frac{x}{2{,}000} - 2\ln 2{,}000 \\ \frac{d}{dx}\ln f(x) &= \frac{1}{x} - \frac{1}{2{,}000} \\ 0 &= \frac{1}{x} - \frac{1}{2{,}000} \\ x &= 2{,}000 \end{align*}$ 
The second derivative $-1/x^2$ is negative, so this is a maximum: the mode is $2,000, matching $(\alpha - 1)\theta$ . The mean is $\alpha\theta = \$4{,}000$ , and the median, found numerically from $F(m) = 0.5$ , is about $3,357. The most common claim is half the average one — a premium set at the mode would collect half the expected cost.

Most Likely Claim Count for a Fleet A fleet's annual claim count is Poisson with $\lambda = 2.6$ . Find the mode and its probability.
Answer The ratio $p(k)/p(k-1) = 2.6/k$ is $2.6$ at $k = 1$ and $1.3$ at $k = 2$ , both above 1, then $0.87$ at $k = 3$ . Probabilities rise up to $k = 2$ and fall after, so the mode is $\lfloor 2.6 \rfloor = 2$ .
 $\begin{align*} p(2) &= \frac{e^{-2.6}\,(2.6)^2}{2} \\ &= 0.07427 \times 3.38 \\ &= 0.2510 \end{align*}$ 
Its neighbours are $p(1) = 0.1931$ and $p(3) = 0.2176$ . Two claims is the single likeliest outcome, yet it happens only a quarter of the time, and the mean of $2.6$ is not itself a possible count.