Discrete random variables — PMF, CDF: Visual walkthrough

4.9.3 · D2 · Maths › Probability Theory & Statistics › Discrete random variables — PMF, CDF

Hum poore walkthrough mein ek hi example use karte hain: ek fair coin 3 baar uchhalo, heads gino. Baaki sab toh sirf background hai.

Step 1 — Raw outcomes se shuru karo (abhi koi number nahi)

KYA HAI. Kisi bhi "random variable" se pehle, hamare paas sirf un cheezoan ki list hoti hai jo physically ho sakti hain. Coin 3 baar uchhaalo: har toss Heads (H) ya Tails (T) hai. Kul $2 \times 2 \times 2 = 8$ equally likely sequences hain.

KYUN. Hume har symbol earn karna hoga. Letter $Ω$ (ek hollow "O", padho "omega") sirf ek naam hai us box ka jo saare 8 sequences ko hold karta hai — yeh sample space hai. Abhi kuch bhi mathematical nahi hai; yeh ek bag of tickets hai.

PICTURE. 8 tickets mein se har ek ko ek chhoti card ki tarah draw kiya gaya hai. Yeh sab equally likely hain, isliye har ek par $\frac{1}{8}$ probability hai.

Figure — Discrete random variables — PMF, CDF

Step 2 — Random variable $X$ har card ko ek number mein badalta hai

KYA HAI. Define karo $X =$ "card par Heads ki sankhya." Yeh hamara random variable hai. Yeh ek card padhta hai aur ek plain number output karta hai.

$X : Ω ⟶ {0, 1, 2, 3}$

$X$ — rule (ek function, yaani ek machine: ek input card, ek output number).
$Ω$ — Step 1 se input box.
${0, 1, 2, 3}$ — wahi numbers jo nikal sakte hain (3 tosses mein 4 heads nahi aa sakte).

YEH TOOL KYUN? Hum word " $H T H$ " par algebra nahi karna chahte — hume ek number chahiye taaki hum add, compare, aur total probabilities kar sakein. $X$ translator hai. Kyunki iska output list ${0, 1, 2, 3}$ finite hai (isliye countable), hum har individual number par probability rakh sakte hain aur simply add kar sakte hain — yahi discrete ka poora matlab hai.

PICTURE. Arrows 8 cards mein se har ek ko uske head-count tak le jaate hain. Dhyan do ki kai cards ek hi number par aa jaate hain — yeh piling-up hi unequal probabilities banata hai.

Step 3 — Count karo kitne cards har number par land karte hain → PMF

KYA HAI. Har output value $k$ ke liye, wahan land karne wale arrows ikkathe karo aur unke weights add karo. Woh total hi $k$ par probability mass hai, likha jaata hai $p_{X} (k)$ .

$p_{X} (k) = P (X = k) = count (how many cards give k heads) \times weight per card \frac{1}{8}$

Charon buckets solve karte hain:

$p_{X} (0) = \frac{1}{8}, p_{X} (1) = \frac{3}{8}, p_{X} (2) = \frac{3}{8}, p_{X} (3) = \frac{1}{8}$

$p_{X} (1) = \frac{3}{8}$ kyunki teen cards ( $H T T, T H T, T T H$ ) mein exactly ek head hai, har ek $\frac{1}{8}$ worth hai.
Subscript $X$ sirf yeh batata hai ki "yeh PMF variable $X$ ki hai."
$P (X = k)$ padho "probability ki output $k$ ke barabar hai."

KYUN. Yeh hamare bars ki heights hain. Aur inhein do rules follow karne hi honge — kyunki koi textbook keh raha hai isliye nahi, balki kyunki yeh ek hi 8 cards ke counts hain:

$p_{X} (k) \geq 0$ — cards ka count negative nahi ho sakta. (Probability axiom: $P \geq 0$ .)
$\sum_{k} p_{X} (k) = \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$ — har card kahin na kahin land karta hai, isliye saare 8 exactly ek baar count hote hain. (Countable additivity over disjoint buckets → $P (Ω) = 1$ .)

PICTURE. Har value par ek bar (stick); uski height mass hai. Charon heights milke puri 1 banti hain — total ink wahi hai jo hum 8 cards se lekar aaye the.

Step 4 — Left-to-right chalo aur accumulate karo → CDF janam leta hai

KYA HAI. Ab socho ki ek marker ko bahut door left ( $- \infty$ ) se right ki taraf khench rahe ho, aur jab bhi tum kisi bar ko cross karte ho toh uski mass ek bucket mein daal lo. Bucket mein jitna hai jab marker position $x$ par ho, wahi CDF hai:

$F_{X} (x) = P (X \leq x) = \sum_{x_{i} \leq x} p_{X} (x_{i})$

$F_{X} (x)$ — $x$ tak aur $x$ ko include karte hue collect kiya gaya running total.
$X \leq x$ — "output zyada se zyada $x$ hai"; $\leq$ (less-than-or-equal) matter karta hai, jaise Step 6 mein dikhta hai.
$\sum_{x_{i} \leq x}$ — har us value ki masses add karo jo $x$ par ya $x$ se left hain.

KYUN sum, integrate nahi? Kyunki mass isolated points par hai, kisi region mein spread nahi hai. Points ke beech integrate karne ko kuch hai hi nahi. Sticks ko add karna natural operation hai — yahi "discrete" ka faida hai.

PICTURE. Marker move karta hai; bucket ka level flat stretches mein badh-ta hai jo sudden lifts se break hoti hain. Values ke beech mein kuch scoop nahi hota (flat); value cross karne par uski poori stick scoop ho jaati hai (ek jump).

Kuch points evaluate karte hain:

$F_{X} (0) = \frac{1}{8}, F_{X} (1) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8}, F_{X} (2) = \frac{4}{8} + \frac{3}{8} = \frac{7}{8}, F_{X} (3) = \frac{7}{8} + \frac{1}{8} = 1$

Step 5 — Graph staircase kyun hai (har property isko dekh kar samajhao)

KYA HAI. $F_{X} (x)$ ko saare real $x$ ke liye plot karo, sirf chaar values ke liye nahi. Hume ek step function milta hai: horizontal treads aur vertical risers.

KYUN har feature forced hai:

Non-decreasing. Right move karne par sirf non-negative sticks add ho sakti hain; bucket kabhi nahi drain hota. Isliye staircase kabhi bas chadhti hai ya level rehti hai — kabhi girti nahi.
Values ke beech flat. Maano $1$ aur $2$ ke beech koi mass scoop nahi hoti, isliye total unchanged rehta hai — $\frac{4}{8}$ height par ek horizontal tread.
Limits. Bahut left, kisi bar se pehle, bucket empty hai: $F_{X} (- \infty) = 0$ . Bahut right, last bar ke baad, sab kuch collect ho chuka hai: $F_{X} (+ \infty) = 1$ . Top tread exactly $1$ hai kyunki Step 3 mein masses $1$ tak sum hui thi.

PICTURE. Number line par poora staircase, jisme har riser ki height par uski swallow ki gayi mass label hai.

Step 6 — Tricky corner: jump ka endpoint kaun own karta hai?

KYA HAI. Ek hi riser ko zoom karo, $x = 1$ par. Question: exactly $x = 1$ par, bucket low tread $\frac{1}{8}$ par hai ya high tread $\frac{4}{8}$ par? Answer: high wala. Kyunki $F_{X} (x) = P (X \leq x)$ mein $\leq$ use hota hai, point $x = 1$ khud include hota hai, isliye mass $p_{X} (1)$ already scoop ho jaati hai jis instant marker $1$ par pahunchta hai.

KYUN MATTER KARTA HAI. Yahi right-continuity hai: jab tum $1$ ko right se approach karte ho toh $\frac{4}{8}$ milta hai aur wahan rehta hai — $1$ par value right side se agree karti hai. Lekin left se approach karne par sirf $\frac{1}{8}$ milta hai. Hum us left-limit ko $F_{X} (1^{-})$ kehte hain (" $1$ se bilkul left"). Dono sides ka difference exactly mass hai:

$p_{X} (x_{i}) = F_{X} (x_{i}) - F_{X} (x_{i}^{-}) (riser height = mass)$

$F_{X} (x_{i})$ — riser ka top (mass included, closed dot ●).
$F_{X} (x_{i}^{-})$ — bottom, bilkul left wali value (mass abhi include nahi, open dot ○).
Unka difference PMF wapas deta hai — CDF secretly apne jumps mein har mass store karta hai.

PICTURE. Ek riser, filled dot top par (value yahan belong karti hai) aur hollow dot bottom par (value left tread se belong nahi karti), aur ek bracket jo jump $= p_{X} (1) = \frac{3}{8}$ measure karta hai.

Step 7 — Interval probabilities: staircase se ek range padho

KYA HAI. Maano hum $P (0 < X \leq 2)$ chahte hain — "0 se zyada, zyada se zyada 2 heads." Do tread heights padho aur subtract karo:

$P (a < X \leq b) = F_{X} (b) - F_{X} (a)$ $P (0 < X \leq 2) = F_{X} (2) - F_{X} (0) = \frac{7}{8} - \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$

KYUN YEH EXACT BRACKETING. $F_{X} (2)$ sab kuch $\leq 2$ count karta hai; $F_{X} (0)$ sab kuch $\leq 0$ count karta hai. Subtract karne se shared part cancel ho jaata hai aur values ${1, 2}$ bachti hain — left end $0$ bahar nikal jaata hai, right end $2$ reh jaata hai. Yeh exactly $(0, 2]$ bracket hai: left par open, right par closed.

Masses se check karo: strictly $0$ se upar aur zyada se zyada $2$ wali values $1$ aur $2$ hain, isliye $p_{X} (1) + p_{X} (2) = \frac{3}{8} + \frac{3}{8} = \frac{6}{8} = \frac{3}{4}$ . ✓

PICTURE. Staircase jisme do treads $F_{X} (0)$ aur $F_{X} (2)$ highlight hain aur unka difference shaded hai — literally do levels ke beech capture ki gayi vertical height.

Ek picture mein poora summary

Ek hi canvas par sab kuch: 8 cards 4 sticks mein fold ho jaate hain (PMF); woh sticks, left-to-right scoop hote hue, staircase mein stack ho jaati hain (CDF); har riser ki height uski stick ke barabar hai; top tread $1$ hai; aur koi bhi interval do tread heights ka difference hai.

Recall Feynman retelling — poora walkthrough plain words mein

Ek game khelo: coin teen baar uchhhalo, heads gino. Har possible flip-sequence apne card par likho — 8 hain, sab equally likely hain, isliye har card ek-aathwan liter "probability water" worth hai. Ab cards ko 0, 1, 2, 3 label wale buckets mein sort karo is hisaab se ki kitne heads dikhte hain. Bucket 0 mein 1 card jaata hai, bucket 1 mein 3, bucket 2 mein 3, bucket 3 mein 1. Har bucket mein paani ki height PMF hai — aur kyunki humne exactly 1 liter pour kiya tha aur ek bhi drop nahi hua, heights ka sum 1 hota hai. Ab ek khali jug lo aur bucket 0 se right ki taraf chalo, raaste mein har bucket ka paani jug mein daalte jao. Jug ka level CDF hai: buckets ke beech chalte waqt flat rehta hai, phir jab bhi koi bucket cross karo jump kar leta hai. Kyunki " $\leq$ " ka matlab hai ki bucket us instant count ho jaata hai jab tum pahunchte ho, jump right wali value ka hota hai — isliye hum top par solid dot aur neeche hollow dot draw karte hain. Har jump ki size exactly us bucket ka paani hai, isliye staircase secretly poori PMF yaad rakhta hai. Aakhir mein, kisi range mein land hone ka chance nikalane ke liye, sirf dono ends par jug ka level padho aur subtract karo — beech mein jo paani mila wahi tumhara answer hai. Flat treads, sudden risers, poore ek liter par top out: woh staircase hi poori kahaani hai.

Active recall

Poori derivation jin teen drawings se guzarti hai, woh kaunsi hain, order mein?

8 equally-likely outcome cards, chaar PMF sticks (heights), aur CDF staircase.

Discrete RV ke liye hum point masses add kyun kar sakte hain, integrate kyun nahi?

Values countable/isolated points hain, isliye mass individual points par baith-ti hai aur beech mein integrate karne ko kuch nahi.

Staircase par ek riser ki height kiske barabar hai?

Us value par probability mass:

p_{X} (x_{i}) = F_{X} (x_{i}) - F_{X} (x_{i}^{-})

Picture dekh kar batao CDF non-decreasing kyun hai?

Right chalne par sirf non-negative sticks bucket mein scoop hoti hain; kuch kabhi remove nahi hota, isliye level kabhi nahi girta.

Exactly

x = 1

par

F_{X} (1)

low ya high tread hai, aur kyun?

High tread, kyunki

P (X \leq x)

mein

\leq

use hota hai, isliye

1

par mass already included hai (right-continuous; top par solid dot).

3 coin tosses ke liye

P (0 < X \leq 2)

compute karo aur batao kaunse endpoints in/out hain.

F_{X} (2) - F_{X} (0) = \frac{7}{8} - \frac{1}{8} = \frac{3}{4}

;

0

exclude hai,

2

include hai (values

1, 2

Connections

Parent topic — PMF, CDF — yeh page uska visual derivation hai.
Probability Axioms — "heights $\geq 0$ " aur "heights sum to 1" axioms hain jo pictures mein draw hue hain.
Continuous random variables — PDF, CDF — sticks ki jagah smooth density; staircase ek smooth curve ban jaata hai.
Expectation and Variance of Discrete RVs — $E [X]$ nikalane ke liye har value ko uski stick height se weight karo.
Binomial Distribution — yeh 3-coin-toss masses hi $Binomial (3, \frac{1}{2})$ hain.
Poisson Distribution, Conditional Probability — same stick-and-staircase machinery, naye masses.