Question bank — Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial

True — $X=X_1+\cdots +X_n$ jahan har $X_i\in \{0,1\}$; yahi decomposition exactly is liye hai ki $E[X]=np$ aur $\mathrm{Var}(X)=npq$.

True — sirf $0$ aur $1$ values hain, aur $0^2=0$, $1^2=1$; yahi trick hai jo $E[X^2]=E[X]=p$ deti hai aur isliye $\mathrm{Var}=p-p^2$.

False — woh equality sirf Poisson ka signature hai (dono $=\lambda$); Bernoulli mein $p$ vs $pq$ hai, Binomial mein $np$ vs $npq$ hai, etc.

False — linearity kisi bhi variables ke liye hold karta hai, dependent ho ya nahi; independence sirf variances add karne ke liye chahiye hoti hai.

Trials-counting convention mein False ($X\ge 1$, kyunki pehle success ke liye kam se kam ek trial chahiye); failures-counting convention mein True, jahan $0$ matlab "pehle hi trial par success, zero failures."

True — "1st success ka wait" same hai "trials to first success" se, aur $\left(\genfrac{}{}{0ex}{}{k-1}{0}\right)=1$ formula ko $q^{k-1}p$ mein collapse kar deta hai.

False — $k$ bina kisi upper bound ke $0,1,2,\dots$ par range karta hai; yeh Binomial ka $n\to \infty$ limit hai, to ceiling $n$ gaayab ho jaata hai.

True — trial 1 par success certain hai, to $X=1$ hamesha; $\mathrm{Var}=q/p^2=0/1=0$, ek degenerate point mass.

False — Geometric mein success position fixed hai (yeh last trial hai), to "choose" karne ko kuch nahi hai; koi binomial coefficient appear nahi hota.

True — $P(X>m+n\mid X>m)=P(X>n)$; pehle wait kiye gaye trials koi information nahi dete, yahi memorylessness hai, jo discrete laws mein sirf Geometric ke paas hai.

Wrong — $k$-va trial forced stopping success hai, to sirf $r-1$ successes pehle $k-1$ trials mein free hain: $\left(\genfrac{}{}{0ex}{}{k-1}{r-1}\right)$.

Wrong window — Poisson $\lambda$ ko interval ke saath scale karna chahiye: 2 minutes ke liye $\lambda =6$ use karo, jo $\frac{6^5{e}^{-6}}{5!}$ deta hai.

Covariance missing hai — generally $\mathrm{Var}(X+Y)=\mathrm{Var}X+\mathrm{Var}Y+2\mathrm{Cov}(X,Y)$; cross term sirf independence ke under vanish karta hai (Figure 2 relay picture mein yeh dikhata hai).

Overstated — mean $np$ sirf linearity se follow karta hai aur dependent trials ke liye bhi hold karta hai; independence sirf $\mathrm{Var}=npq$ ke liye chahiye.

Wrong — $npq=np(1-p)\to \lambda \cdot 1=\lambda$ kyunki $np\to \lambda$ fixed rehta hai aur $(1-p)\to 1$; variance $\lambda$ ke roop mein survive karta hai.

Wrong — yeh ek geometric series hai: ${\sum }_{k\ge 1}{q}^{k-1}p=p\cdot \frac{1}{1-q}=p/p=1$, ek perfectly valid distribution.

Signs flip ho gaye hain — formula hai $\mathrm{Var}=E[X^2]-(E[X]{)}^2=p-p^2=pq$, jo non-negative hai jaisa ki variance hona chahiye.

Woh factor $n$ chote trials mein saari "no success" probability collect karta hai; yeh standard exponential limit hai, jahan exactly $e^{-\lambda }$ Poisson formula mein enter karta hai (Figure 1 mein curve converge hoti dikhti hai).

Kyunki $r$-ve success ka wait waits ka relay hai: success 1 tak jaao, phir clock restart karo aur success 2 tak jaao, aur aage bhi — $r$ independent Geometric legs end-to-end, to unki average lengths simply stack hoti hain: $r\cdot \frac{1}{p}=\frac{r}{p}$ (Figure 2).

Humein $\sum k{q}^{k-1}$ chahiye, aur $\sum q^k=\frac{1}{1-q}$ ko term-by-term differentiate karne se exactly $\sum k{q}^{k-1}=\frac{1}{(1-q{)}^2}$ milta hai — derivative "pull down" karta hai woh $k$ jo humein chahiye.

Stopping rule hai "$r$-va success reach karo"; definition ke mutabiq tum ruk jaate ho jis pal $r$-va success aata hai, to final trial wohi success hai — warna tum pehle ruk gaye hote.

Agar ek success time ka $p$ fraction hota hai (jaise die ke six ke liye $1/6$), to tum $1/p$ trials mein ek success expect karte ho — six ke liye $6$ rolls wait karna everyday intuition se match karta hai.

Trials independent hain, to saare covariances zero hain aur variances cleanly add ho jaate hain; independence drop karo aur cross-covariance terms appear ho jaayenge.

"Trials to first success" success ko khud count karta hai (mean $1/p$); "failures before success" ek trial drop karta hai, sab kuch ek se shift karta hai aur mean $q/p$ deta hai.

Yeh ek single Bernoulli$(p)$ mein reduce ho jaata hai: $\left(\genfrac{}{}{0ex}{}{1}{k}\right)p^k{q}^{1-k}$ with $k\in \{0,1\}$ — woh atom jis par poori family build hai.

$X=0$ certainty ke saath (koi success kabhi possible nahi); mean $np=0$ aur variance $npq=0$, $0$ par ek degenerate spike.

$X=n$ certainty ke saath (har trial succeed karta hai); mean $np=n$ aur variance $npq=n\cdot 1\cdot 0=0$, $n$ par ek degenerate spike — $p=0$ case ka mirror image.

Pehla success kabhi nahi aata — $P(X<\infty )=0$, total probability infinity tak leak ho jaati hai, to $X$ ek proper finite-valued variable nahi hai aur mean $1/p$ infinity tak blow up karta hai. Yahi woh boundary hai jiske baare mein formula $1/p$ warn kar raha hai.

Zero success chance ke saath $r$-va success kisi bhi finite $r\ge 1$ ke liye kabhi nahi aata; $X$ probability $1$ ke saath infinite hai (ek defective distribution), aur mean $r/p$ diverge karta hai — wahi leak hai jo Geometric mein hai, $r$-fold amplified.

"Opportunities" ki koi fixed number nahi hai — limit mein $n\to \infty$ hai, to koi bhi non-negative integer $k$, chahe kitna bhi bada ho, positive probability rakhta hai.

$X=0$ certainty ke saath: $P(X=0)=0^0{e}^0/0!=1$ aur $P(X=k)=0$ for $k\ge 1$; mean $=$ variance $=\lambda =0$, $0$ par ek degenerate spike (koi events kabhi nahi hote).

$k=r$ — tum $r$ se kam trials mein $r$ successes nahi pa sakte, to support exactly $r$ par start hota hai ($\left(\genfrac{}{}{0ex}{}{r-1}{r-1}\right)=1$ match karta hai).

$k=r$ par ek point mass: har trial succeed karta hai, to $r$-va success exactly trial $r$ par milta hai; mean $r/p\to r$ aur variance $rq/p^2\to 0$.

Trivially, tumhe zero successes chahiye, to tum kisi bhi trial se pehle ruk jaate ho: $X=0$ certainty ke saath. Yeh koi information carry nahi karta lekin "$r$ independent Geometric waits" picture confirm karta hai — zero waits matlab zero trials.

$X=1$ par ek point mass: success almost certainly turant hoti hai, to $P(X=1)=p\to 1$ aur mean $1/p\to 1$.

Haan — bade window par dono mean aur variance $2\lambda$ ban jaate hain, phir bhi equal; interval scale karne se $\lambda$ scale hota hai lekin Poisson identity intact rehti hai.