Continuous Univariate Distributions
Study guide: Exam P-1
A Continuous Univariate Distribution describes the probability law of a single random variable X that can take any value in a continuous interval or union of continuous intervals.
- The probability density function (PDF)
f(x)must satisfyf(x) \geq 0and\int_{-\infty}^{\infty} f(x)\,dx = 1 - Individual point probabilities are zero; all probabilities are computed as areas under
f:
P(a \leq X \leq b) = \int_a^b f(x)\, dx
- The CDF and the density are two views of the same object:
F(x) = \int_{-\infty}^{x} f(t)\,dtandf(x) = F'(x). For anything phrased as a tail probability, work from the survival functionS(x) = 1 - F(x)instead.
Choosing the right one
Read the question for the story the quantity follows, then check the support matches:
-
Every value in a range equally likely → Uniform Continuous Distribution
(a,b)on(a,b), withE[X] = (a+b)/2 -
Waiting time between events, memoryless → Exponential Distribution
(\theta)on(0,\infty), withE[X] = \theta -
Waiting time to the
\alpha-th event, or a skewed severity → Gamma(\alpha,\theta)on(0,\infty), withE[X] = \alpha\theta -
A proportion, rate, or probability → Beta
(\alpha,\beta)on(0,1), withE[X] = \alpha/(\alpha+\beta) -
A sum or average of many variables → Normal Distribution
(\mu,\sigma^2)on\mathbb{R}, withE[X] = \mu -
A positive, heavily right-skewed loss whose log is normal → Lognormal Distribution
(\mu,\sigma^2)on(0,\infty), withE[X] = e^{\mu+\sigma^2/2} -
The word memoryless is decisive: it appears only for the exponential (and its discrete counterpart, the geometric).
-
For the lognormal,
\muand\sigmaare the parameters of\ln X, not the mean and standard deviation ofX. Every lognormal probability reduces to a normal one by taking logs of both sides:P(X > c) = P(\ln X > \ln c). -
Apart from a uniform with
a < 0, the normal is the only one on this list that can go negative, and the only one with no lower bound at all — a red flag if the quantity is a loss. -
Insurance provisions (deductibles, limits, coinsurance) transform whichever severity distribution is chosen; see Transformations of Random Variables.
Finding a Probability from a PDF A random variable
Xhas PDFf(x) = 3x^2for0 < x < 1. FindP(0.5 < X < 1).Answer
P(0.5 < X < 1) = \int_{0.5}^{1} 3x^2\, dx = \left[x^3\right]_{0.5}^{1} = 1 - (0.5)^3 = 1 - 0.125 = 0.875