What is Discrete random variable; law?

Example 22.2 (The classical laws)

• Bernoulli $\mathcal{B}(p)$: $X \in \{0, 1\}$, $\P(X = 1) = p$. Indicator of an event.
• Binomial $\mathcal{B}(n, p)$: $\P(X = k) = \binom nk p^k(1-p)^{n-k}$, $0 \leq k \leq n$: number of successes in $n$ independent Bernoulli trials (High School volume; re-proved below via sums of independent variables).
• Geometric $\mathcal{G}(p)$: $\P(X = k) = (1-p)^{k-1}p$, $k \in \N^*$: rank of the first success (Example 21.5).
• Poisson $\mathcal{P}(\lambda)$: $\P(X = k) = e^{-\lambda}\frac{\lambda^k}{k!}$, $k \in \N$ — a probability measure by the exponential series. The law of rare events (Chapter 23).

Example 22.10 (Transfer in action)

For $X \sim \mathcal P(\lambda)$, compute $\E\bigl(\frac1{1+X}\bigr)$ — the law of $\frac1{1+X}$ itself is awkward, but transfer never asks for it:

Two lessons. Computationally: recognizing a shifted exponential series is the whole work — transfer reduces expectations of $f(X)$ to series manipulation. Structurally: the naive plug-in value would be $\frac1{1 + \E X} = \frac1{1 + \lambda}$, and the true answer is larger,

exactly as Jensen’s inequality for the convex function $t \mapsto \frac1{1+t}$ demands. Expectations of convex images sit above the naive plug-in value, and transfer plus a series check makes the abstract inequality concrete.

Example 22.12 (Products, with and without independence)

Roll two fair dice. If $Y$ is the second die (independent of the first), $\E(XY) = \E(X)\E(Y) = 3.5^2 = 12.25$. If instead $Y = X$ (the “product” of a die with itself),

same marginal laws in both scenarios, different joint laws, different product expectations. The moral, worth engraving: $\E(XY)$ is a functional of the pair, not of the two marginals — and the gap $\E(X^2) - \E(X)^2 \approx 2.92$ is, by König–Huygens, precisely the variance $\frac{35}{12}$ of the die.