Random Variables

A random variable attaches a number to each outcome of an experiment. Once you have its distribution, two summary numbers — the expectation and the variance — describe almost everything you need.

Discrete random variables

A discrete random variable takes separate, countable values: the score on a die, the number of defective items in a batch.

Notation: capital $X$ for the variable, lower-case $x$ for a particular value, so $P(X = x)$ is the probability that $X$ takes the value $x$.

Probability distributions

A probability distribution lists every value with its probability. Two conditions must hold:

$$0 \le P(X = x) \le 1 \qquad \text{and} \qquad \sum P(X = x) = 1$$

The probabilities total $1$ ✓

Finding an unknown constant. If $X$ takes the values $1, 2, 3$ with $P(X = x) = kx$, then

$$k(1) + k(2) + k(3) = 1 \quad \Rightarrow \quad 6k = 1 \quad \Rightarrow \quad k = \tfrac16$$

Expectation

$$E(X) = \sum x\,P(X = x)$$

the long-run average value if the experiment were repeated indefinitely.

For the distribution above:

$$E(X) = 1(0.1) + 2(0.2) + 3(0.4) + 4(0.3) = 0.1 + 0.4 + 1.2 + 1.2 = 2.9$$

For a fair die:

$$E(X) = \frac{1+2+3+4+5+6}{6} = 3.5$$

For a function of $X$, weight the function's values by the same probabilities:

$$E\big(g(X)\big) = \sum g(x)\,P(X = x)$$

$$E(X^2) = 1(0.1) + 4(0.2) + 9(0.4) + 16(0.3) = 0.1 + 0.8 + 3.6 + 4.8 = 9.3$$

Variance

$$\operatorname{Var}(X) = E(X^2) - \big[E(X)\big]^2$$

For the distribution above:

$$\operatorname{Var}(X) = 9.3 - 8.41 = 0.89$$

The standard deviation is $\sqrt{0.89} = 0.943$.

For a fair die:

$$E(X^2) = \frac{1+4+9+16+25+36}{6} = \frac{91}{6} \qquad \operatorname{Var}(X) = \frac{91}{6} - \frac{49}{4} = \frac{35}{12} \approx 2.92$$

Linear transformations

$$E(aX + b) = aE(X) + b$$ $$\operatorname{Var}(aX + b) = a^2\operatorname{Var}(X)$$

With $E(X) = 2.9$ and $\operatorname{Var}(X) = 0.89$:

$$E(3X + 2) = 3(2.9) + 2 = 10.7$$ $$\operatorname{Var}(3X + 2) = 9(0.89) = 8.01$$

The cumulative distribution function

$$F(x) = P(X \le x)$$

a running total of the probabilities.

$F$ never decreases, and its final value is always 1.

Individual probabilities can be recovered by subtraction:

$$P(X = x) = F(x) - F(x-1)$$

so $P(X = 3) = 0.7 - 0.3 = 0.4$ ✓