Beta

Study guide: Exam P-1

The Beta Distribution X \sim \text{Beta}(\alpha, \beta) is a continuous distribution on (0, 1) parameterized by shape parameters \alpha > 0 and \beta > 0. It is commonly used to model rates, proportions, and probabilities.

f(x) = \frac{x^{\alpha - 1}(1-x)^{\beta - 1}}{B(\alpha,\, \beta)}

0 < x < 1

\text{where } B(\alpha, \beta) = \frac{\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha + \beta)}

• E[X] = \alpha/(\alpha + \beta) and \text{Var}(X) = \alpha\beta / [(\alpha+\beta)^2(\alpha+\beta+1)]
• When \alpha = \beta = 1 it reduces to the continuous uniform distribution on (0,1)
• For integer \alpha and \beta the normalizing constant is a binomial coefficient, 1/B(\alpha,\beta) = (\alpha+\beta-1)\dbinom{\alpha+\beta-2}{\alpha-1}, so the density is a plain polynomial and probabilities integrate directly — no special functions needed
• The distribution is skewed right when \alpha < \beta, left when \alpha > \beta, and symmetric when \alpha = \beta
• The k-th order statistic of n i.i.d. \text{Uniform}(0,1) variables is exactly \text{Beta}(k,\, n-k+1), which is where E[X_{(k)}] = k/(n+1) comes from

Expected Loss Ratio from a Beta Model An insurer models its loss ratio X \sim \text{Beta}(3, 2). Find the mean and variance of X.

Answer With \alpha = 3, \beta = 2: E[X] = \frac{3}{3+2} = \frac{3}{5} = 0.60 \text{Var}(X) = \frac{3 \cdot 2}{(5)^2 (6)} = \frac{6}{150} = 0.04 The expected loss ratio is 60% with a standard deviation of \sqrt{0.04} = 0.20.

Probability the Loss Ratio Exceeds a Threshold With X \sim \text{Beta}(3,2) modelling the loss ratio, find P(X > 0.8).

Answer Both parameters are integers, so write out the density explicitly:

\begin{align*}
B(3,2) &= \frac{\Gamma(3)\Gamma(2)}{\Gamma(5)} = \frac{2! \cdot 1!}{4!} = \frac{2}{24} = \frac{1}{12} \\
f(x) &= 12\,x^2(1-x), \quad 0 < x < 1
\end{align*}

Integrate over the tail:

\begin{align*}
P(X > 0.8) &= \int_{0.8}^{1} 12\left(x^2 - x^3\right) dx \\
           &= 12\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0.8}^{1} \\
           &= 12\left[\left(\tfrac{1}{3} - \tfrac{1}{4}\right) - \left(0.17067 - 0.10240\right)\right] \\
           &= 12\left[0.08333 - 0.06827\right] \\
           &\approx 0.181
\end{align*}

About an 18% chance the loss ratio exceeds 80%.

Beta as the Distribution of a Uniform Order Statistic Four claim-settlement delays are i.i.d. \text{Uniform}(0,1) years. Find the expected value of the second-smallest delay.

Answer The k-th order statistic of n i.i.d. uniforms is \text{Beta}(k, n-k+1). With k=2, n=4 that is \text{Beta}(2,3):

\begin{align*}
E[X_{(2)}] &= \frac{\alpha}{\alpha + \beta} \\
           &= \frac{2}{2+3} \\
           &= 0.4
\end{align*}

This matches the general uniform result k/(n+1) = 2/5 — the four delays split the year into five equal expected gaps.