Use transforms only on their valid domain

Try

A Bernoulli variable X with success probability 1/4 has E(e^(tX))=3/4+(1/4)e^t. Differentiate at t=0 to recover its mean. Predict what a second derivative produces: it gives E(X²), not variance directly.

Predict

An exponential transformation can magnify a tail dramatically. Predict whether a variable can have every polynomial moment finite while E(e^(tX)) diverges for all positive t. The lognormal distribution supplies such a counterexample, so moment existence and MGF existence are distinct requirements.

Experiment

For independent Bernoulli variables with probabilities 1/4 and 1/2, the sum MGF is [3/4+e^t/4][1/2+e^t/2]. Expanding gives 3/8+(1/2)e^t+(1/8)e^(2t). Read off probabilities 3/8,1/2,1/8 for sums 0,1,2. These match direct convolution.

For an exponential variable with rate 3, integrate M(t)=∫₀∞3e^(−(3−t)x)dx=3/(3−t), valid only for t<3. At t≥3 the integral diverges. Mark that domain beside the formula before differentiating or comparing transforms.

Understand

The moment-generating function is M_X(t)=E(e^(tX)) wherever that expectation is finite. It always equals one at zero, but that isolated value is not enough for the usual uniqueness and differentiation theorems. Finiteness on an open interval containing zero supplies a standard sufficient condition for recovering all moments by derivatives and identifying the distribution uniquely.

Under those conditions M′(0)=E(X) and M″(0)=E(X²). Variance is M″(0)−[M′(0)]². Justifying differentiation under the expectation requires control supplied by the local exponential integrability, rather than an unrestricted exchange of limit and integral.

For independent variables, M_(X+Y)(t)=M_X(t)M_Y(t) wherever the relevant expectations are finite. This follows from e^(t(X+Y))=e^(tX)e^(tY) and independence. A product identity for a few derivatives, such as matching mean and variance, does not prove independence or identify an entire distribution.

For a Poisson variable, summing the exponential series gives M(t)=exp(λ(e^t−1)) for every real t. Multiplying independent Poisson MGFs adds their parameters, identifying a Poisson sum by uniqueness. The distributional conclusion is stronger than simply adding expectations.

For a normal variable with mean μ and variance σ², completing the square gives M(t)=exp(μt+σ²t²/2). This formula supports independent-normal sum calculations, while a statement about marginal normality alone does not establish independence or joint normality.

A lognormal X=e^Z, with a nondegenerate normal Z, has all positive integer moments finite because E(X^k)=E(e^(kZ)). Yet for positive t the growth of exp(t e^Z) outruns the normal tail, so its MGF diverges. More generally, even all moments need not uniquely determine a law without additional conditions. Characteristic functions provide a broader transform theory, but are beyond this introductory MGF lesson.

A transform calculation should therefore record three separate facts: the expression obtained, the real values of t for which it is finite, and the theorem permitting the intended conclusion. Matching a formula outside its convergence domain is not an identification argument.

Apply

Derive the exponential MGF with its convergence range and calculate both derivatives. Then explain where independence and a uniqueness theorem enter the Poisson-sum argument. A successful symbolic simplification is not enough unless the transform exists on a domain where identification is justified.

Continue

Finally use inequalities to bound probabilities and prove a weak law, then distinguish concentration from the central limit theorem’s standardized shape statement.