Question bank — Moment generating function (MGF) — definition, use

Sach — $M_X(0)=E[e^{0\cdot X}]=E[1]=1$. Ye free sanity check hai; agar tumhari algebra kuch aur deti hai toh galti ho gayi.

Jhooth — $t$ ek bookkeeping dummy hai jo sirf power series banane ke liye use hota hai. Ye kabhi $X$ ka outcome represent nahi karta; hum derivatives near $t=0$ ki parwah karte hain, $X=t$ ki nahi.

Sach — ye uniqueness property hai. MGF, jab $0$ ke paas exist karta hai, ek fingerprint hai jo poora distribution pin kar deta hai.

Jhooth — mean aur variance sirf pehle do moments hain; MGF saare moments encode karta hai (skew, tails, …). Distributions do moments par match kar sakte hain lekin baaki jagah differ kar sakte hain.

Jhooth — iske liye $X\perp Y$ chahiye, kyunki tabhi $E[e^{tX}{e}^{tY}]$ split hokar $E[e^{tX}]E[e^{tY}]$ banta hai. Correlated variables ye split tod dete hain.

Jhooth — tumhe chahiye ki $E[e^{tX}]$ ==$0$ ke aas-paas ek interval mein finite== ho. Heavy-tailed variables jaise Cauchy ke liye $E[e^{tX}]=\infty$ sabhi $t\ne 0$ ke liye hota hai, toh koi MGF exist nahi karta.

Sach — unke existence ke intervals ke overlap par, $M_{X+Y}(t)=M_X(t)M_Y(t)$ finite numbers ka finite product hai, toh wahan finite hai.

Sach — kyunki $|e^{itX}|=1$ bounded hai, expectation hamesha finite hoti hai. Yehi wajah hai ki Characteristic Function woh bhaari, hamesha-available tool hai jab MGF fail ho jaate hain.

Jhooth — mean hai $M_X^{\prime }(0)$. Differentiate karke $0$ par evaluate karna hi ek single moment ko isolate karta hai; $1$ par evaluate karna sab ko mix kar deta hai.

Jhooth — tumhe sirf $0$ ke aas-paas ek interval par existence chahiye, aur $t<\lambda$ aise interval ko include karta hai ($0$ ke dono taraf). Uniqueness phir bhi hold karti hai.

Galti ye hai ki $E[e^X]$ ko $E[X]$ samjh liya — ye alag quantities hain. Moments aate hain $0$ par derivatives se: $E[X]=M_X^{\prime }(0)$, MGF ko kahin bhi evaluate karne se nahi.

Integral ${\int }_0^{\infty }{e}^{-(\lambda -t)x}dx$ tabhi converge karta hai jab exponent negative ho, yaani $\lambda -t>0$, toh domain hai $t<\lambda$. Usse aage MGF $+\infty$ hai.

Normal hona independence guarantee nahi karta. Product rule ke liye $X\perp Y$ chahiye; do correlated Normals ka sum phir bhi Normal hoga lekin unke sum ka MGF individual MGFs ka plain product nahi hoga.

Ruko — $M_X(0)$ zaroor $1$ ke barabar hona chahiye kyunki $E[1]=1$. $1$ se alag value prove karta hai ki upar koi algebra error hai; baaki par bharosa nahi kar sakte.

$t^n$ ka coefficient $E[X^n]/n!$ hai, $E[X^n]$ nahi. Moment khud recover karne ke liye $n!$ se multiply karna hoga (equivalently, $0$ par $n$-th derivative lena hoga).

Zarurat nahi — Binomial, $n$ independent Bernoullis ka sum hai, toh uska MGF parts multiply karke $(1-p+p{e}^t{)}^n$ hai. Ye sum ko poori tarah bypass kar deta hai.

Constant exponent mein baitha hai, multiplier ke roop mein nahi: $M_{aX}(t)=E[e^{taX}]=M_X(at)$. $X$ ko scale karna MGF ke argument ko $at$ par rescale karta hai, poori function ko multiply nahi karta.

Iska Taylor Series $\sum \frac{t^n{X}^n}{n!}$ mein $X^n$, $t^n$ slot par baitha hai, toh $E$ lene par $E[X^n]$ $t$ ki har power par aa jaata hai. $t$ ki har power ek clean "tag" ka kaam karti hai exactly ek moment ke liye.

Dono same idea hain — $0$ par $n$-th derivative hi $t^n$ coefficient ka $n!$ guna hai. Differentiate karna sirf ek mechanical tarika hai us ek coefficient ko extract karne ka bina poori series expand kiye.

Kyunki $e^{t(X+Y)}=e^{tX}{e}^{tY}$, aur independence $E[e^{tX}{e}^{tY}]=E[e^{tX}]E[e^{tY}]$ allow karta hai. Toh independent variables ka addition unke MGFs ka multiplication ban jaata hai — convolution arithmetic mein badal jaati hai.

$n$ i.i.d. variables ka sum ek MGF ko $n$-th power tak uthane jaisa ban jaata hai, jo ek manageable cheez hai. Ye dikhana ki ye power $e^{t^2/2}$ (standard Normal ka MGF) ki taraf converge karti hai, uniqueness ke zariye distribution mein convergence prove kar deta hai.

$t=0$ par ye trivially sabke liye $1$ hai, toh akele kuch nahi kehta. Sirf ek two-sided neighborhood par finiteness guarantee karti hai ki derivatives (moments) exist karte hain aur uniqueness fingerprint hold karta hai.

$Y=aX+b$ likhne par $e^{t(aX+b)}=e^{bt}\cdot e^{(at)X}$ milta hai. Shift $b$ ek constant hai jo $e^{bt}$ ke roop mein factor out hota hai; scale $a$, $X$ ke saath rehta hai, isliye MGF ke argument ko $at$ par rescale karta hai.

Ye $\log M_X(t)$ hai, aur log lena independent sums ke liye product rule ko cumulant functions ka plain sum mein badal deta hai — toh ek sum ke cumulants bas add ho jaate hain.

$M_X(t)=E[e^{tc}]=e^{tc}$. Iske $0$ par derivatives $E[X^n]=c^n$ dete hain, bilkul sahi, aur $\mathrm{Var}=0$ expected hai.

Haan — $t=0<\lambda$ hamesha, aur $M_X(0)=\lambda /(\lambda -0)=1$. Domain $t<\lambda$ aaram se $0$ ke aas-paas ek interval contain karta hai.

Haan — Lognormal ke har order ke finite moments hain, lekin $E[e^{tX}]=\infty$ har $t>0$ ke liye, toh $0$ ke aas-paas koi interval kaam nahi karta. Finite moments MGF guarantee nahi karte.

Tab $X$ aur $Y$ maximally dependent hain, independent nahi, toh $M_{2X}(t)=M_X(2t)\ne M_X(t{)}^2$ generally. Product rule chupchap fail ho jaata hai.

Uniqueness se, $X$ constant $0$ hai: iska MGF $E[e^{t\cdot 0}]=1$ hai. Ek flat MGF jo $1$ ke barabar hai matlab hai ki zeroth ke baad ke saare moments zero hain.

Nahi — $\mathrm{Var}(X+b)=\mathrm{Var}(X)$, aur $M_{X+b}(t)=e^{bt}{M}_X(t)$ sirf mean ($M^{\prime }(0)$) shift karta hai, $M^{\prime \prime }(0)-(M^{\prime }(0){)}^2$ ko unchanged chhodta hai. MGF sahi tarah se variance ki shift-invariance encode karta hai.