Exercises — Discrete random variables — PMF, CDF

KYA check karte hain: sirf do rules.

1. Har entry $\ge 0$ hai? Haan — $0.1,0.4,0.3,0.2$ sab non-negative hain. ✓
2. Kya inка sum 1 hai? $0.1+0.4+0.3+0.2=1.0$. ✓

Dono rules hold karte hain, toh haan, yeh ek legal PMF hai. Kyunki $5$ ek listed value nahi hai, wahan koi mass nahi baithta: $p_X(5)=0$.

KYA karte hain: $\le x$ wali saari values ka mass sum karo.

• $F_X(1)=p_X(0)+p_X(1)=0.1+0.4=0.5$. (Values $0,1$ hain $\le 1$.)
• $F_X(2.5)=p_X(0)+p_X(1)+p_X(2)=0.1+0.4+0.3=0.8$. (Values $0,1,2$ hain $\le 2.5$; value $3$ nahi hai.)

Dhyan do ki $F$ flat hai $2$ aur $3$ ke beech: $3$ tak pahunchne se pehle collect karne ke liye kuch naya nahi hai.

KYUN normalization: total mass rule $\sum p_X=1$ woh equation hai jo unknown $c$ ko pin down karti hai. $c(1^2+2^2+3^2)=c(1+4+9)=14c=1\Rightarrow c=\frac{1}{14}.$ Toh $p_X(1)=\frac{1}{14},p_X(2)=\frac{4}{14},p_X(3)=\frac{9}{14}$.

$P(X\ge 2)$: values $2$ aur $3$ ka mass add karo: $P(X\ge 2)=\frac{4}{14}+\frac{9}{14}=\frac{13}{14}.$ Complement se cross-check: $1-p_X(1)=1-\frac{1}{14}=\frac{13}{14}$. ✓

KYUN do answers: bracket type decoration nahi hai — points positive mass carry karte hain, toh endpoint include ya exclude karna count change karta hai.

$P(2<X\le 5)$ $2$ ko exclude karta hai, $5$ ko include karta hai → values $\{3,4,5\}$: $F_X(5)-F_X(2)=\frac{5}{6}-\frac{2}{6}=\frac{3}{6}=\frac{1}{2}.$

$P(2\le X\le 5)$ $2$ ko include karta hai → values $\{2,3,4,5\}$: $F_X(5)-F_X(2^{-})=F_X(5)-F_X(1)=\frac{5}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}.$ Extra $\frac{1}{6}$ exactly $p_X(2)$ hai, woh mass jo humne rakhna choose kiya.

KYUN jumps = mass: parent note ke according, $p_X(x_i)=F_X(x_i)-F_X(x_i^{-})$ — har vertical step ki height exactly wahan baithne wali probability hai. Flat pieces ka matlab koi mass nahi.

Staircase $x=0,1,2,4$ par jump karti hai:

• $0$ par: $0.2-0=0.2$
• $1$ par: $0.5-0.2=0.3$
• $2$ par: $0.9-0.5=0.4$
• $4$ par: $1-0.9=0.1$

Toh RV $\{0,1,2,4\}$ par live karta hai (nahi $3$ par — $F$ puri $[2,4)$ mein flat hai, toh wahan koi mass nahi) aur $p_X(0)=0.2,p_X(1)=0.3,p_X(2)=0.4,p_X(4)=\mathrm{0.1.}$ Sanity check: $0.2+0.3+0.4+0.1=1$. ✓

• $P(X=2)$ = $2$ par jump = $0.4$.
• $P(1\le X<2)$: values $\ge 1$ aur $<2$ → sirf $\{1\}$, toh $=p_X(1)=0.3$. (CDF se: $F_X(2^{-})-F_X(1^{-})=0.5-0.2=0.3$.)
• $P(X>1)$ = $1-P(X\le 1)=1-F_X(1)=1-0.5=0.5$. (Values $2$ aur $4$: $0.4+0.1=0.5$. ✓)

KYUN binomial: independent trials, har Head same $p=0.6$ ke saath; $X$ successes count karta hai. $3$ flips mein exactly $k$ Heads paane ke tarike $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ hain. $p_X(k)=\left(\genfrac{}{}{0ex}{}{3}{k}\right)(0.6{)}^k(0.4{)}^{3-k}.$

• $p_X(0)=\left(\genfrac{}{}{0ex}{}{3}{0}\right)(0.6{)}^0(0.4{)}^3=1\cdot 1\cdot 0.064=0.064$
• $p_X(1)=\left(\genfrac{}{}{0ex}{}{3}{1}\right)(0.6{)}^1(0.4{)}^2=3\cdot 0.6\cdot 0.16=0.288$
• $p_X(2)=\left(\genfrac{}{}{0ex}{}{3}{2}\right)(0.6{)}^2(0.4{)}^1=3\cdot 0.36\cdot 0.4=0.432$
• $p_X(3)=\left(\genfrac{}{}{0ex}{}{3}{3}\right)(0.6{)}^3(0.4{)}^0=1\cdot 0.216\cdot 1=0.216$

Total check: $0.064+0.288+0.432+0.216=1.000$. ✓

$F_X(1)=p_X(0)+p_X(1)=0.064+0.288=\mathrm{0.352.}$

KYUN Poisson: yeh "kitne independent rare events ek fixed window mein land karte hain" ke liye standard PMF hai, jo ek parameter $\lambda =$ mean count se governed hai. $P(X\le 2)=p_X(0)+p_X(1)+p_X(2)=e^{-2}\left(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}\right)=e^{-2}(1+2+2)=5e^{-2}.$ Numerically $5e^{-2}\approx 5\times 0.13534=0.6767$.

(a) Infinite tail ke saath Normalization. Values countably infinite hain, toh hum ek geometric series sum karte hain: ${\sum }_{k=1}^{\infty }c{\left(\frac{1}{2}\right)}^k=c\cdot \frac{1/2}{1-1/2}=c\cdot 1=c=1.$ (${\sum }_{k=1}^{\infty }{r}^k=\frac{r}{1-r}$ for $|r|<1$ use kiya, yahan $r=\frac{1}{2}$.) Toh $c=1$ aur $p_X(k)=(\frac{1}{2}{)}^k$.

(b) $P(X\le 2)=p_X(1)+p_X(2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$.

(c) Definition se $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$.

• $A=\{X\le 2\}$, $B=\{X\ge 2\}$. Unka intersection exactly $\{X=2\}$ hai, toh $P(A\cap B)=p_X(2)=\frac{1}{4}$.
• $P(B)=P(X\ge 2)=1-p_X(1)=1-\frac{1}{2}=\frac{1}{2}$. $P(X\le 2\mid X\ge 2)=\frac{1/4}{1/2}=\frac{1}{2}.$

Proof. Events $\{X\le a\}$ aur $\{X>a\}$ disjoint hain (koi value dono nahi ho sakti) aur unka union sab kuch hai jo $X$ kar sakta hai. Probability axioms se (dekho Probability Axioms), disjoint events jo $\Omega$ cover karte hain unki probabilities 1 sum karti hain: $P(X\le a)+P(X>a)=1\Rightarrow P(X>a)=1-P(X\le a)=1-F_X(a).$

Apply. Die, $\{1,\dots ,6\}$ par uniform: $F_X(4)=\frac{4}{6}=\frac{2}{3}$, toh $P(X>4)=1-\frac{2}{3}=\frac{1}{3}.$ (Directly check karo: values $5,6$ dete hain $\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$. ✓)

PMF read/verify karo (rules) → CDF tak sum karo (staircase) → correct endpoints ke saath intervals read karo → mass recover karne ke liye jumps invert karo → named models se PMFs banao (Binomial, Poisson, geometric) → axioms se complement/conditional identities prove karo.