Expectation, Variance and Standard Deviation

For a discrete random variable X with probability mass function $P(X=x_i)=p_i$:

• Expected value / mean: $E(X) = \mu = \sum_i x_i\,p_i$ — multiply each value by its probability and add.
• The probabilities must satisfy $\sum_i p_i = 1$ and $p_i \ge 0$; this is always the first thing to check (and how you find an unknown $k$).
• Expectation is linear: $E(aX+b) = aE(X)+b$, and for independent parts $E(X+Y)=E(X)+E(Y)$ (so the mean of the sum on two dice is $3.5+3.5=7$).

To find the mean from a pmf:

• If a probability is unknown, first use $\sum P(x) = 1$ to solve for it.
• Then apply $E(X) = \sum x\,P(x)$ row by row.
• Linearity shortcut: for the sum of two independent variables, $E(X+Y) = E(X)+E(Y)$. Each fair die has mean $\tfrac{1+2+\cdots+6}{6} = 3.5$, so the expected sum of two dice is $3.5+3.5 = 7$ — no need to list all 36 outcomes.

The three core formulas of this subtopic:

• E(X²): $E(X^2) = \sum x^2\,P(x)$ — square each value, weight by its probability.
• Variance: $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$ (equivalently $\sum(x-\mu)^2 P(x)$, but the E(X²) form is faster).
• Standard deviation: $\sigma = \sqrt{\mathrm{Var}(X)}$.
• From a CDF: if F(x) is given, recover the pmf by $P(x_i) = F(x_i) - F(x_{i-1})$ (with $P(x_1)=F(x_1)$) before computing E(X) or E(X²).

For a payoff g(X):

• Expected payoff: $E(g(X)) = \sum g(x)\,P(x)$ — weight each cash outcome (a gain positive, a loss negative) by its probability.
• Typical 3-coin game: $P(\text{all heads or all tails}) = \tfrac{2}{8} = \tfrac14$ and $P(\text{one or two heads}) = \tfrac{6}{8} = \tfrac34$.
• Variance of a winning amount uses the same E(X²) − [E(X)]² machinery on the payoff values: list the winnings with their probabilities, then apply the variance formula.
• $E = 0$ ⟹ fair game; $E > 0$ ⟹ expected gain; $E < 0$ ⟹ expected loss.

For the discrete uniform distribution on $\{1,2,\ldots,n\}$, $P(x)=\tfrac1n$:

• Mean: $E(X) = \dfrac{n+1}{2}$ (the middle value).
• Variance: $\mathrm{Var}(X) = \dfrac{n^2-1}{12}$.
• Handy ratio: $\dfrac{\mathrm{Var}(X)}{E(X)} = \dfrac{(n^2-1)/12}{(n+1)/2} = \dfrac{n-1}{6}$ — the fastest route to 'find n' questions.
• A weighted pmf $P(x) = \dfrac{2x}{n(n+1)}$ on $1,\ldots,n$ is a different distribution: $E(X) = \sum x\cdot\dfrac{2x}{n(n+1)} = \dfrac{2}{n(n+1)}\cdot\dfrac{n(n+1)(2n+1)}{6} = \dfrac{2n+1}{3}$.

The two-equation setup:

• Equation 1 (normalization): $\sum P(x) = 1$.
• Equation 2 (mean or an extra relation): $E(X) = \sum x\,P(x) = \text{given}$, or a stated link such as $P(X=3)=2P(X=1)$.
• Solve the linear system for the unknown probabilities.
• Range problems: when the probabilities depend on a parameter p, impose $0 \le P(x) \le 1$ on EVERY row to get an interval for p; the mean (a linear function of p) attains its extreme values at the endpoints of that interval.

Standard-distribution means that appear here:

• Geometric (trials until first success), success probability p: mean $= \dfrac{1}{p}$. Rolling an n-faced die until a number $< n$ shows has $p = \dfrac{n-1}{n}$, so mean $= \dfrac{n}{n-1}$.
• Hypergeometric (n drawn without replacement from N containing K successes): $E(X) = \dfrac{nK}{N}$.
• Small-pmf via combinations: for 'X = number of queens in 2 cards' build $P(X=k) = \dfrac{\binom{4}{k}\binom{48}{2-k}}{\binom{52}{2}}$, then $E(X)=\sum kP$, $E(X^2)=\sum k^2 P$.
• Larger-of-two: X = larger of two numbers drawn from $\{1,\ldots,m\}$ has $P(X=k) = \dfrac{k-1}{\binom{m}{2}}$; then $E(X)=\sum k P(X=k)$.