Continuous random variables — PDF, CDF, percentiles: Question bank

4.9.7 · D5 · Maths › Probability Theory & Statistics › Continuous random variables — PDF, CDF, percentiles

Quick symbol reminder taaki kuch bhi bina naam ke use na ho:

$f (x)$ = probability density function (PDF) — " $x$ par probability kitni ghani hai." Iska height density hai, iska area probability hai.
$F (x) = P (X \leq x)$ = cumulative distribution function (CDF) — " $x$ se neeche kitni probability hai," yani ekdam left se $x$ tak $f$ ka scooped area.
$x_{p}$ = == $p$ -th percentile==, woh value jo $F (x_{p}) = p$ solve karta hai.
Support = woh set of $x$ jahan $f (x) > 0$ ho; bahar $f = 0$ hota hai.

True or false — justify karo

PDF value

f (x)

1 se bada ho sakta hai.

True.

f

ek density hai, probability nahi; sirf uska area total 1 hona chahiye. Jaise

f (x) = 2 x

[0, 1]

2

tak pahunchta hai, aur

f (x) = 5

[0, 0.2]

theek hai kyunki uska area

5 \times 0.2 = 1

hai.

Ek continuous RV ke liye, har constant

c

ke liye

P (X = c) = 0

hota hai.

True. Ek single point zero-width ka interval hai, isliye

\int_{c}^{c} f d t = 0

. Probability areas mein rehti hai, aur ek line segment ka koi area nahi hota.

Kyunki

P (X = c) = 0

hai, event

X = c

impossible hai.

False. Zero probability ka matlab impossible nahi hota —

X

har trial mein kisi exact value par zaroor land karta hai. "Impossible" (empty event) aur "probability zero" sirf discrete variables ke liye ek jaise hote hain, continuous ke liye nahi.

Continuous RV ke liye

P (a \leq X \leq b) = P (a < X < b)

hota hai.

True. Endpoints

a

aur

b

dono zero probability carry karte hain, isliye unhe include ya exclude karne se koi fark nahi padta.

\leq

<

ka distinction yahan irrelevant hai.

CDF

F (x)

kahin par 1 se bada ho sakta hai.

False.

F (x) = P (X \leq x)

ek probability hai, isliye

0 \leq F (x) \leq 1

hamesha. Yeh

0

1

tak badhta hai aur kabhi overshoot nahi karta — yeh total-area-equals-1 axiom hai jo accumulation ke through dikh raha hai.

F

hamesha non-decreasing hota hai.

True.

x

ko right mein move karne par

f

ke neeche aur area add hota hai, aur

f \geq 0

hone ki wajah se woh added area kabhi negative nahi ho sakta. Isliye

F

flat ho sakta hai (jahan

f = 0

ho) par kabhi girta nahi.

Agar

F

kisi interval par flat hai, toh wahan

f = 0

hai.

True. FTC ke anusaar

f (x) = F^{'} (x)

hai, aur ek flat stretch ki slope zero hoti hai, isliye wahan density zero hai — us interval par koi probability accumulate nahi ho rahi.

Har non-negative function ek valid PDF hoti hai.

False. Non-negativity zaruri hai par kaafi nahi; total area bhi exactly 1 hona chahiye. Jaise

f (x) = 1

[0, 2]

non-negative hai par uska area

2 \neq = 1

hai.

Continuous RV ke liye median hamesha mean ke barabar hoti hai.

False. Yeh sirf tab coincide karte hain jab density apne centre ke baare mein symmetric ho. Skewed density (jaise exponential) mein mean long tail ki taraf khichta hai jabki median wahin rehti hai jahan half area hota hai.

Agar do RVs ka median same ho, toh unka distribution bhi same hoga.

False. Median sirf ek point pin karta hai jahan

F = 0.5

ho; infinitely many different shapes of

f

woh single crossing share kar sakte hain jabki baaki jagah alag hon.

X

ke continuous RV hone ke liye

f

continuous honi chahiye.

False. "Continuous RV" ka matlab hai ki

X

values ka continuum leta hai (isliye

F

continuous hai bina jumps ke). Density

f

khud jumps le sakti hai — jaise uniform density ek flat block hai jiske edges par sudden drops hain. Dekho Uniform distribution.

Galti dhundho

f (x) = 2 x

ke liye

f (0.7) = 1.4

hai, isliye

X = 0.7

ki probability

1.4

hai."

Galti: height se probability padhna.

1.4

ek density hai, aur yeh 1 se bhi zyada hai, jo ek probability kabhi nahi ho sakti.

P (X = 0.7) = 0

hai; probability area hoti hai, height nahi.

"90th percentile nikalne ke liye main

f (x_{p}) = 0.9

solve karta hoon."

Galti: $F$ ki jagah $f$ use karna. Percentile ek accumulated quantity hoti hai, isliye

F (x_{p}) = 0.9

solve karo. Kisi point par density value seedha yeh nahi batati ki uske neeche kitni probability hai.

f (x) = 2 x

[0, 1]

ke liye

F (x) = x^{2}

har jagah hai."

Galti: support bhool jaana.

F (x) = x^{2}

sirf

[0, 1]

par hai; support se neeche

F = 0

aur uske upar

F = 1

hai. CDF ko ekdam left mein 0 aur ekdam right mein 1 tak pahunchna chahiye.

P (1 \leq X \leq 3) = \int_{1}^{3} f = f (3) - f (1)

Galti: galat function subtract karna. Interval probability

F (b) - F (a)

hai, yaani CDF (antiderivative) ka difference, density

f

ka nahi. Tum accumulated areas subtract karte ho, heights nahi.

x \geq 0

par

f (x) = λ e^{- λ x}

ke liye CDF

F (x) = e^{- λ x}

hai."

Galti: sign aur boundary. Integrate karne par

F (x) = [- e^{- λ t}]_{0}^{x} = 1 - e^{- λ x}

milta hai. Proposed answer 1 se 0 tak ghatta hai, jo non-decreasing rule violate karta hai — yeh flipped sign ka clear signal hai. Dekho Exponential distribution.

"Kyunki

\int f = 1

hai,

f

ka maximum at most 1 hoga."

Galti: total area aur peak height ko confuse karna. Ek tall, narrow spike ki height

≫ 1

ho sakti hai phir bhi area 1 enclose kar sakti hai. Koi cheez density ki height cap nahi karti; sirf total area fixed hota hai.

"100th percentile woh sabse badi value hai jo

X

le sakta hai, isliye

F (x_{1.00}) = 1

use dega."

Galti: percentiles ko $0 < p < 1$ chahiye. Unbounded support wale distributions (jaise exponential) ke liye,

F (x) = 1

ka koi finite solution nahi hota — "100th percentile"

+ \infty

hota hai. Percentiles strictly

(0, 1)

ke andar define hoti hain.

Why questions

Continuous variables ke liye hum probability ki jagah density kyun use karte hain?

Kyunki infinitely many exact values hoti hain, har single value zero probability carry karti hai; density hume batati hai ki total probability of 1 kaise spread hai taaki intervals ko area ke through positive probability milti rahe.

PDF ka exactly 1 tak integrate karna kyun zaroori hai, sirf kuch finite mein nahi?

Total area woh probability hai ki

X

koi value lega, jo certainty hai — probability 1. Finite-but-not-1 area ka matlab hoga ki probabilities certainty tak nahi jodti.

f (x) = F^{'} (x)

kyun hai?

Fundamental Theorem of Calculus ke anusaar, accumulated area ka derivative woh rate hai jis par area add ho raha hai — aur wahi rate exactly density hai. Dekho Fundamental Theorem of Calculus.

Continuous RV ke liye jahan

f > 0

ho wahan

F

invertible kyun hoti hai?

Jahan

f > 0

ho,

F

strictly increasing hoti hai, isliye har probability level

p

exactly ek

x

par hit hoti hai. Wahi unique crossing hai jo

x_{p} = F^{- 1} (p)

ko well-defined banata hai. Dekho Quantile function and inverse-transform sampling.

Hum

P (X = x)

ko discrete variables ki tarah sab

x

par add kyun nahi kar sakte?

Har term 0 hai aur uncountably many terms hain; sum meaningless ho jaata hai. Density ka integration sum ki jagah leta hai, jisse zero-probability points ka uncountable collection phir bhi positive interval probabilities build karta hai. Dekho Discrete random variables — PMF.

Normal distribution percentiles ke liye neat formula ki jagah

Φ

kyun use karta hai?

Uska CDF

Φ (x) = \int_{- \infty}^{x} f

ka koi elementary closed form nahi hai, isliye

Φ^{- 1}

(aur z-scores) tables ya numerics se nikalne padte hain, algebra se nahi. Dekho Normal distribution.

Expectation

\int x F (x) d x

ki jagah

\int x f (x) d x

kyun use karta hai?

Expectation har value

x

ko wahan ki probability ki density se weight karta hai, aur

f

wahi density hai;

F

accumulated probability hai, jo galat weighting hai. Dekho Expectation and Variance of continuous RVs.

Edge cases

Support ke ekdam left aur ekdam right mein

F (x)

kya hota hai?

F (- \infty) = 0

(abhi koi area scoop nahi hua) aur

F (+ \infty) = 1

(sara area collect ho gaya). Yeh limits total-area-equals-1 axiom se force hoti hain.

Jab

b \to a

ho toh

P (a \leq X \leq b)

ka kya hota hai?

Yeh

F (a) - F (a) = 0

tak shrink ho jaata hai. Yahi wajah hai ki ek single point (zero-width interval) ki probability zero hoti hai.

Kya ek valid PDF apni stated range ke kisi hisse par zero ho sakti hai?

Haan. Density kisi interval ke andar gaps par vanish ho sakti hai; woh gaps sirf koi probability contribute nahi karte. Sirf poori line par total area 1 hona chahiye.

Exponential

f (x) = λ e^{- λ x}

ke liye

P (X < 0)

kya hai?

Bilkul 0, kyunki support

x \geq 0

hai aur negative

x

ke liye

f = 0

hai. Integrate karne se pehle hamesha check karo ki density kahan defined hai.

Kya

F

uniform distribution ke edges par continuous hai jabki wahan

f

jump karta hai?

Haan —

F

smoothly ramp karta hai (ek straight line) bina kisi jump ke;

F

ki slope jump karti hai (

f

ke jumps se match karti hai), par

F

khud continuous rehta hai kyunki koi single point mass carry nahi karta.

Us distribution ka median kya hai jिसकी density

μ

ke baare mein symmetric hai?

Exactly

μ

: symmetry centre ke dono taraf equal area rakhti hai, isliye

F (μ) = 0.5

. Yahan median aur mean coincide karte hain (jaise Normal distribution mein).

Agar

f (x) > 0

sirf ek single point par ho, toh kya woh valid PDF hai?

Nahi. Ek single point ki width zero hai, isliye uska area 0 hai, 1 nahi — tum required total probability ko zero-width set par kisi bhi finite density se nahi pack kar sakte.

Recall Ek-line self-test

Upar har answer cover karo aur justification khud derive karo, sirf verdict nahi — agar tum sirf "true/false" bol sakte ho toh trap seekha nahi hai.

Connections

Continuous random variables — PDF, CDF, percentiles (parent note)
Discrete random variables — PMF (jahan single points sach mein probability carry karte hain)
Expectation and Variance of continuous RVs (kyun $\int x f d x$ , $F$ nahi)
Normal distribution (median = mean by symmetry; $Φ$ ka koi closed form nahi)
Exponential distribution (skew ki wajah se median ≠ mean; uske CDF mein sign trap)
Uniform distribution ( $f$ jump karta hai par $F$ continuous rehta hai)
Fundamental Theorem of Calculus (kyun $f = F^{'}$ )
Quantile function and inverse-transform sampling (kyun $F$ invertible hai)