S1.5 - Discrete random variables

A discrete random variable assigns a numerical value to each outcome of a random process and can take only separate, countable values. Write the random variable with a capital letter, such as $X$, and a possible value with a lower-case letter, such as $x$.

If three T-shirts are selected and $X$ is their median selling price, then repeated selections may produce many ordered combinations, but $X$ can still take only the listed price values. Different combinations that give the same median belong to the same event $X=x$.

The probability function gives the probability of each possible value, $p(x)=P(X=x)$. Every probability is between 0 and 1, and the probabilities over the complete support of $X$ add to 1.

To construct a distribution, list every possible value, combine the probabilities of all disjoint routes that produce that value, and check the final probabilities total 1. To recover a probability from a cumulative distribution, subtract adjacent cumulative totals: here $P(X=1)=F(1)-F(0)=0.70-0.20=0.50$.

Translate an event before reading the table. For example, $P(3X-3<X+2)=P(X<2.5)=P(X\le2)=1$ for this integer-valued support. A non-integer cutoff is handled by including precisely the supported values that satisfy the inequality.

The expectation $E(X)$ is the probability-weighted long-run mean of a discrete random variable. Variance measures its expected squared spread about that mean.

For $X=0,1,2$ with probabilities $0.20,0.50,0.30$,$$E(X)=0(0.20)+1(0.50)+2(0.30)=1.10,$$$$E(X^2)=0^2(0.20)+1^2(0.50)+2^2(0.30)=1.70.$$Therefore $\operatorname{Var}(X)=1.70-1.10^2=0.49$, and the standard deviation is $\sqrt{0.49}=0.70$.

If $Y=5-3X$, then $E(Y)=5-3(1.10)=1.70$ and $\operatorname{Var}(Y)=(-3)^2(0.49)=4.41$. Adding a constant shifts every value without changing spread; multiplying by $a$ multiplies every deviation by $a$, so variance is multiplied by $a^2$.

A discrete uniform distribution has a finite set of possible values that are all equally likely. If there are $n$ values, each has probability $1/n$.

Suppose $B$ is equally likely to be $1,3,5$ or $7$. Then$$E(B)=\frac{1+3+5+7}{4}=4,$$$$E(B^2)=\frac{1^2+3^2+5^2+7^2}{4}=21,$$so $\operatorname{Var}(B)=21-4^2=5$. Symmetry also shows immediately that the mean is the midpoint, 4.

For equally likely consecutive integers $a,a+1,\ldots,b$, the number of values is $n=b-a+1$ and the mean is $(a+b)/2$. The variance may be found from the same $E(X^2)-[E(X)]^2$ method, which remains valid for any explicitly listed uniform support.