Discrete Random Variables, PMF and CDF

A probability mass function $P(X=x)$ of a discrete random variable must satisfy TWO axioms:

• Each probability is valid: $0 \le P(X=x_i) \le 1$ for every value $x_i$.
• The total mass is one: $\displaystyle\sum_i P(X=x_i) = 1$ — summed over ALL values the variable can take.

These two rules are the engine of the whole subtopic: every 'find the constant' question is just ΣP = 1 solved for the unknown, and every 'is this a valid distribution?' check is these two axioms.

When the pmf entries are linear in $k$ (e.g. $2k, k, 2k, 4k, k$, or a piecewise rule $P=kx$/$P=k(5-x)$):

• Sum all entries and set the total equal to 1: $\sum P(X=x)=1$.
• Solve the resulting LINEAR equation for $k$ — a single step.
• Substitute back to read off any required probability or range.

If a fixed number appears (e.g. $P(0)=0.1$ with the rest in $k$), include it in the sum: $0.1 + (\text{terms in }k) = 1$.

Translate the inequality into exactly which values to add:

• $P(X < a)$: all values strictly below $a$.
• $P(a \le X < b)$: from $a$ up to but NOT including $b$.
• $P(X \ge a) = 1 - P(X < a)$ — use the complement to avoid adding a long tail.
• $P(X > a) = 1 - P(X \le a)$.

The complement rule is the workhorse whenever the 'up to' side has fewer cells than the 'from' side.

The two standard quadratics and their admissible roots:

• $6k^2 + 5k - 1 = 0 \Rightarrow (6k-1)(k+1)=0 \Rightarrow k = \tfrac16$ (reject $k=-1$).
• $10k^2 + 9k - 1 = 0 \Rightarrow (10k-1)(k+1)=0 \Rightarrow k = \tfrac1{10}$ (reject $k=-1$).

Always reject the negative root — a negative $k$ would make some $P(X=x)$ negative. After finding k, evaluate the required range, remembering the $k^2$ terms: e.g. $P(X\ge 6)=2k^2+(7k^2+k)=9k^2+k$.

For $P(X=x)=k\,r^x$ on $x=0,1,\ldots,n$:

• Finite geometric sum: $\displaystyle\sum_{x=0}^{n} r^x = \dfrac{r^{n+1}-1}{r-1}$ (for $r\ne 1$).
• Set $k \cdot (\text{that sum}) = 1$ and solve for $k$.

For $r=2,\ n=4$: $2^0+2^1+2^2+2^3+2^4 = 1+2+4+8+16 = 31$, so $31k=1$ and $k=\tfrac1{31}$.

For the infinite pmf $P(X=x)=k(x+1)r^x,\ x=0,1,2,\ldots$:

• Key AGP sum (memorise): $\displaystyle\sum_{x=0}^{\infty}(x+1)r^{x} = \frac{1}{(1-r)^{2}}$ for $|r|<1$.
• Set $k\cdot\dfrac{1}{(1-r)^2} = 1$, so $k = (1-r)^2$.
• For $r=\tfrac12$: $k=(1-\tfrac12)^2=\tfrac14$. For $r=\tfrac15$: $k=(1-\tfrac15)^2=(\tfrac45)^2=\tfrac{16}{25}$.

Then any $P(X=x)$ follows by direct substitution — e.g. $P(X=0)=k(1)(r^0)=k$.

Identify the values $X$ takes, then find each $P(X=x)$ by the right counting rule:

• With-replacement draws (independent trials): binomial — $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$. Two cards with replacement, jack has $p=\tfrac4{52}=\tfrac1{13}$: $P(0)=(\tfrac{12}{13})^2=\tfrac{144}{169},\ P(1)=2\cdot\tfrac1{13}\cdot\tfrac{12}{13}=\tfrac{24}{169},\ P(2)=\tfrac1{169}$.
• Without-replacement draws: hypergeometric — $P(X=r)=\dfrac{\binom{D}{r}\binom{N-D}{n-r}}{\binom{N}{n}}$. 4 defective + 16 good, draw 3: $P(0)=\tfrac{\binom{16}{3}}{\binom{20}{3}}=\tfrac{28}{57}$, and so on.
• Counting outcomes (equally likely): three fair coins, $X=$ heads: $P(X=k)=\binom{3}{k}(\tfrac12)^3$, giving $\tfrac18,\tfrac38,\tfrac38,\tfrac18$.
• Sequential 'until' experiments: multiply along each branch — a coin tossed until a head or 4 tails gives $\tfrac12,\tfrac14,\tfrac18,\tfrac18$ for $X=1,2,3,4$ (the last cell pools TTTH and TTTT).

For a discrete random variable with values $x_1<x_2<\cdots$:

• Definition: $F(x)=P(X\le x)=\displaystyle\sum_{x_i \le x}P(X=x_i)$; it is non-decreasing and reaches 1 at the top value.
• Recover the pmf (differencing): $P(X=x_i)=F(x_i)-F(x_{i-1})$, with $P(X=x_1)=F(x_1)$.
• Tail from the CDF: $P(X>a)=1-F(a)$; $P(X\le a)=F(a)$.

The continuous analogues of the pmf rules:

• Normalisation (find the constant): $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$, integrated over the support only.
• CDF: $F(x)=\displaystyle\int_{-\infty}^{x} f(t)\,dt$; $F$ rises from 0 to 1, and $F'(x)=f(x)$.
• Interval probability: $P(a<X<b)=\displaystyle\int_a^b f(x)\,dx = F(b)-F(a)$. For continuous $X$, $<$ and $\le$ give the same value.
• Two-condition pdf: if the pdf has TWO unknowns, use $\int f=1$ AND a given point value (like $f(2)=2$) to solve the pair.