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

True — $M_X(0)=E[e^{0\cdot X}]=E[1]=1$. It is the free sanity check; if your algebra gives anything else you slipped.

False — $t$ is a bookkeeping dummy used only to build the power series. It never stands for an outcome of $X$; we care about derivatives near $t=0$, not $X=t$.

True — this is the uniqueness property. The MGF, when it exists near $0$, is a fingerprint that pins down the whole distribution.

False — mean and variance are just the first two moments; the MGF encodes all moments (skew, tails, …). Distributions can match on two moments yet differ everywhere else.

False — it needs $X\perp Y$, because only then does $E[e^{tX}{e}^{tY}]$ split into $E[e^{tX}]E[e^{tY}]$. Correlated variables break the split.

False — you need $E[e^{tX}]$ ==finite in an interval around $0$==. Heavy-tailed variables like the Cauchy have $E[e^{tX}]=\infty$ for all $t\ne 0$, so no MGF exists.

True — on the overlap of their intervals of existence, $M_{X+Y}(t)=M_X(t)M_Y(t)$ is a finite product of finite numbers, so it is finite there.

True — because $|e^{itX}|=1$ is bounded, the expectation is always finite. That is exactly why the Characteristic Function is the heavier, always-available tool when MGFs fail.

False — the mean is $M_X^{\prime }(0)$. Differentiating and evaluating at $0$ is what isolates a single moment; evaluating at $1$ mixes them all.

False — you only need existence on an interval around $0$, and $t<\lambda$ includes such an interval (both sides of $0$). Uniqueness still holds.

The error is confusing $E[e^X]$ with $E[X]$ — they are different quantities. Moments come from derivatives at $0$: $E[X]=M_X^{\prime }(0)$, not from evaluating the MGF anywhere.

The integral ${\int }_0^{\infty }{e}^{-(\lambda -t)x}dx$ only converges when the exponent is negative, i.e. $\lambda -t>0$, so the domain is $t<\lambda$. Beyond that the MGF is $+\infty$.

Being Normal does not grant independence. The product rule needs $X\perp Y$; two correlated Normals still add to a Normal but their sum's MGF is not the plain product of the individual MGFs.

Stop — $M_X(0)$ must equal $1$ since $E[1]=1$. A value other than $1$ proves an algebra error upstream; the rest cannot be trusted.

The coefficient of $t^n$ is $E[X^n]/n!$, not $E[X^n]$. You must multiply by $n!$ (equivalently, take the $n$-th derivative at $0$) to recover the moment itself.

No need — a Binomial is a sum of $n$ independent Bernoullis, so its MGF is $(1-p+p{e}^t{)}^n$ by multiplying the parts. This sidesteps the sum entirely.

The constant sits in the exponent, not as a multiplier: $M_{aX}(t)=E[e^{taX}]=M_X(at)$. Scaling $X$ rescales the argument of the MGF, it does not multiply the whole function.

Its Taylor Series $\sum \frac{t^n{X}^n}{n!}$ has $X^n$ sitting on the $t^n$ slot, so taking $E$ places $E[X^n]$ on each power of $t$. Each power of $t$ acts as a clean "tag" for exactly one moment.

They are the same idea — the $n$-th derivative at $0$ is $n!$ times the $t^n$ coefficient. Differentiating is just the mechanical way to extract that one coefficient without expanding the whole series by hand.

Because $e^{t(X+Y)}=e^{tX}{e}^{tY}$, and independence lets $E[e^{tX}{e}^{tY}]=E[e^{tX}]E[e^{tY}]$. So addition of independent variables becomes multiplication of their MGFs — convolution made into arithmetic.

Summing $n$ i.i.d. variables becomes raising one MGF to the $n$-th power, a manageable object. Showing that this power converges to $e^{t^2/2}$ (the standard Normal's MGF) proves convergence in distribution via uniqueness.

At $t=0$ it is trivially $1$ for everyone, so that alone says nothing. Only finiteness on a two-sided neighborhood guarantees the derivatives (moments) exist and that the uniqueness fingerprint holds.

Writing $Y=aX+b$ gives $e^{t(aX+b)}=e^{bt}\cdot e^{(at)X}$. The shift $b$ is a constant that factors out as $e^{bt}$; the scale $a$ lives with $X$, so it rescales the MGF's argument to $at$.

It is $\log M_X(t)$, and taking a log turns the product rule for independent sums into a plain sum of cumulant functions — so cumulants of a sum just add.

$M_X(t)=E[e^{tc}]=e^{tc}$. Its derivatives at $0$ give $E[X^n]=c^n$, exactly right, and $\mathrm{Var}=0$ as expected.

Yes — $t=0<\lambda$ always, and $M_X(0)=\lambda /(\lambda -0)=1$. The domain $t<\lambda$ comfortably contains an interval around $0$.

Yes — the Lognormal has finite moments of every order, but $E[e^{tX}]=\infty$ for every $t>0$, so no interval around $0$ works. Finite moments do not guarantee an MGF.

Then $X$ and $Y$ are maximally dependent, not independent, so $M_{2X}(t)=M_X(2t)\ne M_X(t{)}^2$ in general. The product rule silently fails.

By uniqueness, $X$ is the constant $0$: its MGF is $E[e^{t\cdot 0}]=1$. A flat MGF equal to $1$ means every moment past the zeroth vanishes.

No — $\mathrm{Var}(X+b)=\mathrm{Var}(X)$, and $M_{X+b}(t)=e^{bt}{M}_X(t)$ only shifts the mean ($M^{\prime }(0)$), leaving $M^{\prime \prime }(0)-(M^{\prime }(0){)}^2$ unchanged. The MGF correctly encodes shift-invariance of variance.