Uniform Discrete

Study guide: Exam P-1

The Discrete Uniform Distribution X \sim \text{Unif}\{1, \ldots, n\} assigns equal probability to each of n possible integer values. It is the natural model for any experiment where all outcomes are equally likely, such as rolling a fair die.

P(X = k) = \frac{1}{n}, \quad k = 1, 2, \ldots, n

• E[X] = (n+1)/2 and \text{Var}(X) = (n^2 - 1)/12
• Over a general integer range \{a, a+1, \ldots, b\} there are n = b - a + 1 values, with E[X] = (a+b)/2 and \text{Var}(X) = (n^2-1)/12. The variance depends only on how many values there are, not where the range sits — shifting a distribution never changes its spread.
• Do not confuse it with the continuous uniform on (a,b), whose variance is (b-a)^2/12. For \{1,\ldots,6\} the discrete answer is 35/12 \approx 2.92; treating it as continuous on (1,6) gives 25/12 \approx 2.08.

Rolling a Fair Six-Sided Die A fair six-sided die is rolled. Find the expected value and variance of the outcome X.

Answer Here n = 6, so: E[X] = \frac{6+1}{2} = 3.5 \text{Var}(X) = \frac{6^2 - 1}{12} = \frac{35}{12} \approx 2.917

Deductible Applied to a Discrete Uniform Loss A loss X is equally likely to be any integer from 1 to 10 (in thousands). A deductible of 4 applies, so the insurer pays Y = (X-4)_+. Find E[Y].

Answer Each outcome has probability 1/10. Losses of 1 through 4 pay nothing; losses of 5 through 10 pay 1, 2, \ldots, 6:

\begin{align*}
E[Y] &= \frac{1}{10}\sum_{k=5}^{10}(k - 4) \\
     &= \frac{1}{10}(1+2+3+4+5+6) \\
     &= \frac{21}{10} \\
     &= 2.1
\end{align*}

The insurer expects to pay 2.1 thousand per loss, against a ground-up mean of E[X] = 5.5 — the deductible removes E[\min(X, 4)] = (1+2+3+4 \cdot 7)/10 = 3.4 on average, less than the full 4, because a loss of 1, 2 or 3 is withheld only up to its own amount.

A Range That Does Not Start at 1 Claim counts are equally likely to be any integer from 20 to 29. Find the mean and variance.

Answer There are n = 29 - 20 + 1 = 10 equally likely values:

\begin{align*}
E[X] &= \frac{20 + 29}{2} = 24.5 \\
\text{Var}(X) &= \frac{10^2 - 1}{12} = \frac{99}{12} = 8.25
\end{align*}

Applying E[X] = (n+1)/2 = 5.5 here would be the classic error: that formula assumes the range starts at 1.