Discrete random variables — PMF, CDF: Worked examples

4.9.3 · D3 · Maths › Probability Theory & Statistics › Discrete random variables — PMF, CDF

Scenario matrix

Kuch bhi solve karne se pehle, poora battlefield map karte hain. Ek PMF/CDF problem in cells mein se kisi ek mein hi land kar sakti hai:

#	Case class	Kya alag / dangerous hai	Example
A	Finite uniform support	sab masses equal hain — pure counting	Ex 1
B	Non-uniform finite support	masses alag hain; har ek ko track karna hoga	Ex 2
C	Endpoint bracketing ( $<$ vs $\leq$ )	points mass carry karte hain, toh kaunsa bracket matter karta hai	Ex 3
D	Parameter ke liye solve karna	unknown constant pin karne ke liye $\sum p = 1$ use karo	Ex 4
E	Degenerate RV	saari mass ek point pe; PMF/CDF collapse ho jaate hain	Ex 5
F	CDF staircase se PMF recover karna	heights nahi, jumps padho	Ex 6
G	Countably infinite support	infinite sum phir bhi 1 total hona chahiye (geometric series)	Ex 7
H	Real-world word problem	English $\to$ RV $\to$ probability mein translate karo	Ex 8
I	Exam twist — $P (a \leq X \leq b)$ aur left limits	subtle inclusive-both-ends formula	Ex 9

Ab hum har cell hit karenge.

Example 1 — Finite uniform (cell A)

Figure — Discrete random variables — PMF, CDF

Staircase dekho: yeh $(2, 3)$ pe flat hai, toh us gap mein kaheen bhi — including $2.7$ — same height $\frac{1}{2}$ milegi.

Example 2 — Non-uniform finite (cell B)

Example 3 — Endpoint bracketing (cell C)

Worked example Statement

Example 2 ke coin RV use karke, $P (0 < X \leq 2)$ aur $P (0 \leq X \leq 2)$ compute karo. Kya yeh equal hain?

Forecast: kya yeh dono answers match karte hain, ya $0$ pe bracket kuch change kar deta hai?

Strict-left version. $P (0 < X \leq 2) = F_{X} (2) - F_{X} (0) = 1 - \frac{1}{4} = \frac{3}{4}$ . Yeh step kyun? $F_{X} (2)$ sab kuch $\leq 2$ count karta hai; $F_{X} (0)$ subtract karne se $0$ pe mass remove ho jaati hai. Kyunki $0$ real mass carry karta hai ( $\frac{1}{4}$ ), use remove karna cosmetic nahi hai.
Inclusive-both version. $P (0 \leq X \leq 2) = F_{X} (2) = 1$ . Yeh step kyun? Ab $0$ included hai, toh hum uska mass subtract nahi karte. Sab kuch $\leq 2$ hai, toh hum saara $1$ collect karte hain.

Verify: dono answers ka difference exactly $p_{X} (0) = \frac{1}{4}$ hai, aur $\frac{3}{4} + \frac{1}{4} = 1$ ✓. Left end pe bracket usi point mass ke exactly barabar matter karta hai.

Example 4 — Parameter ke liye solve karna (cell D)

Example 5 — Degenerate RV (cell E)

Worked example Statement

$X = 7$ hamesha (ek "constant random variable"). Iska PMF aur CDF likho, aur CDF properties check karo.

Forecast: staircase kaisa dikhega jab sirf ek step ho?

PMF. $p_{X} (7) = 1$ aur $p_{X} (x) = 0$ har doosre $x$ ke liye. Yeh step kyun? Saari probability mass single reachable value pe baithti hai.
CDF. $F_{X} (x) = 0$ for $x < 7$ aur $F_{X} (x) = 1$ for $x \geq 7$ . Yeh step kyun? Rightward chalte hue $7$ tak kuch collect nahi hota, phir seedha full pe jump.
Property audit. Non-decreasing ✓ ( $0 \to 1$ jaata hai), right-continuous ✓ (jump $x = 7$ ko belong karta hai kyunki $\leq$ use include karta hai), limits $0$ aur $1$ ✓.

Verify: jump size $= F_{X} (7) - F_{X} (7^{-}) = 1 - 0 = 1 = p_{X} (7)$ ✓ — single jump se PMF recover ho gayi. Yeh degenerate case Expectation and Variance of Discrete RVs ka seed hai: $E [X] = 7$ , variance $= 0$ .

Example 6 — CDF se PMF recover karna (cell F)

Example 7 — Countably infinite support (cell G)

Worked example Statement

$X$ = pehla head aane tak coin flips ki sankhya, fair coin ke saath. Toh $X \in {1, 2, 3, \dots}$ (infinite!). PMF nikalo, confirm karo ki yeh $1$ sum karta hai, aur $P (X \leq 3)$ compute karo.

Forecast: positive masses ki ek infinite list — kya yeh sach mein exactly $1$ add ho sakti hai?

PMF. Flip $k$ pe pehla head paane ke liye, tumhe $k - 1$ tails phir ek head chahiye: $p_{X} (k) = (\frac{1}{2})^{k - 1} \cdot \frac{1}{2} = (\frac{1}{2})^{k}$ . Yeh step kyun? Independent flips multiply hote hain; yeh geometric pattern hai.
Geometric series se sum check. $\sum_{k = 1}^{\infty} (\frac{1}{2})^{k} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$ . Yeh step kyun? Ek geometric series $\sum a r^{k}$ jahan $∣ r ∣ < 1$ ho, $\frac{a}{1 - r}$ pe converge karta hai. Yeh tool humein chahiye kyunki hum haath se infinitely many terms add nahi kar sakte — closed form yeh kaam karta hai. Probability Axioms se countable additivity countably infinite union pe bhi apply hoti hai.
CDF. $F_{X} (3) = P (X \leq 3) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$ . Yeh step kyun? Pehle teen point masses sum karo.

Verify: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8} = 0.875$ ✓, aur full infinite sum $= 1$ ✓. Related named PMF: Poisson Distribution ek aur infinite-support model hai.

Example 8 — Word problem (cell H)

Worked example Statement

Ek shop rainy hour mein $0$ , $1$ , ya $2$ umbrellas probabilities $0.5, 0.3, 0.2$ ke saath bechti hai. $X$ = ek ghante mein biki umbrellas. $F_{X} (1)$ aur probability nikalo ki kam se kam ek umbrella biki.

Forecast: $F_{X} (1)$ aur $P (X \geq 1)$ mein se kaunsa bada hai?

Translate karo. English directly PMF deti hai: $p_{X} (0) = 0.5, p_{X} (1) = 0.3, p_{X} (2) = 0.2$ . Yeh step kyun? RV ko naam dene se story numbers mein badal jaati hai jinhein hum sum kar sakte hain.
CDF at $1$ . $F_{X} (1) = p_{X} (0) + p_{X} (1) = 0.5 + 0.3 = 0.8$ . Yeh step kyun? " $\leq 1$ " $0$ aur $1$ masses collect karta hai.
Kam se kam ek. $P (X \geq 1) = 1 - p_{X} (0) = 1 - 0.5 = 0.5$ . Yeh step kyun? Complement: "kuch nahi bika" ke alawa sab kuch. $0.3 + 0.2$ add karne se sasta, aur match karta hai.

Verify: $p_{X} (1) + p_{X} (2) = 0.3 + 0.2 = 0.5$ ✓ complement answer se match karta hai. Aur $F_{X} (1) = 0.8 > P (X \geq 1) = 0.5$ (forecast: $F_{X} (1)$ bada, correct). Units: probabilities, sab $[0, 1]$ mein ✓.

Example 9 — Exam twist: inclusive both ends & left limits (cell I)

Worked example Statement

Example 6 ke CDF use karke, (a) $P (1 < X \leq 3)$ , (b) $P (1 \leq X \leq 3)$ , aur (c) $P (X = 3)$ sirf CDF se compute karo.

Forecast: kya (a) aur (b) alag honge, aur kitne se?

(a) strict left. $P (1 < X \leq 3) = F_{X} (3) - F_{X} (1) = 0.7 - 0.2 = 0.5$ . Yeh step kyun? $1$ pe mass exclude karta hai, $3$ pe mass keep karta hai. Yeh exactly $p_{X} (3) = 0.5$ hai.
(b) inclusive both ends. $P (1 \leq X \leq 3) = F_{X} (3) - F_{X} (1^{-}) = 0.7 - 0 = 0.7$ . Yeh step kyun? Left endpoint include karne ke liye hum CDF ko $1$ ke bilkul left se subtract karte hain, yaani $F_{X} (1^{-}) = 0$ , toh hum $p_{X} (1)$ nahi kho dete.
(c) point mass. $P (X = 3) = F_{X} (3) - F_{X} (3^{-}) = 0.7 - 0.2 = 0.5$ . Yeh step kyun? Ek point ki probability uska jump hai: value pe $F$ minus uske bilkul left $F$ .

Verify: (b) $-$ (a) $= 0.7 - 0.5 = 0.2 = p_{X} (1)$ ✓ — inclusive version ne exactly left-endpoint mass add ki. (c) $= 0.5$ , step 1 ke $p_{X} (3)$ se match karta hai ✓. Conditional Probability dekho ki yeh interval events conditional PMFs mein kaise feed karte hain.

Active recall

Recall

$P (a \leq X \leq b)$ kaunsa formula deta hai aur kyun? $P (a \leq X \leq b) = F_{X} (b) - F_{X} (a^{-})$ ::: $a^{-}$ use karne se ( $a$ ke bilkul left se) mass $p_{X} (a)$ andar rehti hai, unlike $F_{X} (a)$ jo use remove kar deta.

Recall CDF staircase se point mass

$p_{X} (x_{i})$ kaise nikaalte hain? Measure the jump height ::: $p_{X} (x_{i}) = F_{X} (x_{i}) - F_{X} (x_{i}^{-})$ .

Recall

$p_{X} (k) = (1/2)^{k}$ , $k \geq 1$ ke liye, total $1$ kyun hota hai? Geometric series ::: $\sum_{k \geq 1} (1/2)^{k} = \frac{1/2}{1 - 1/2} = 1$ .

Recall Degenerate RV

$X = 7$ ka variance kya hai? Zero ::: saari mass ek point pe hone ka matlab koi spread nahi, toh variance $= 0$ .

Related core notes: Discrete random variables — PMF, CDF · Probability Axioms · Continuous random variables — PDF, CDF · Expectation and Variance of Discrete RVs.

Worked examples — Discrete random variables — PMF, CDF

Scenario matrix

Example 1 — Finite uniform (cell A)

Example 2 — Non-uniform finite (cell B)

Example 3 — Endpoint bracketing (cell C)

Example 4 — Parameter ke liye solve karna (cell D)

Example 5 — Degenerate RV (cell E)

Example 6 — CDF se PMF recover karna (cell F)

Example 7 — Countably infinite support (cell G)

Example 8 — Word problem (cell H)

Example 9 — Exam twist: inclusive both ends & left limits (cell I)

Active recall

Case map