Which of the following is FALSE about moment generating functions?
AThe kth derivative of M(t) evaluated at t = 0 equals E[Xᵏ]
BIf the MGF exists on an open interval around 0, it uniquely determines the probability distribution
CThe MGF always exists for any random variable, since e^{tX} is always a real number
DConvergence of MGFs to M(t) implies convergence of the corresponding distributions
The MGF M(t) = E[e^{tX}] requires computing an expectation, which may be infinite. For heavy-tailed distributions like the Cauchy, E[e^{tX}] = ∞ for any t ≠ 0 — the MGF does not exist. The other three statements are all true when the MGF exists on an open interval around 0. This is precisely why the characteristic function φ(t) = E[e^{itX}] is preferred in rigorous probability theory: |e^{itX}| = 1 always, so the characteristic function always exists.
Question 2 Multiple Choice
A sequence of standardized random variables has MGFs that converge pointwise to exp(t²/2). This fact is most directly useful for proving:
AThat the random variables have finite variance equal to 1
BThat the distributions converge to a standard normal distribution
CThat the characteristic functions of the sequence do not exist
DThat the random variables are independent
The MGF of a standard normal N(0,1) is exp(t²/2). The MGF continuity theorem states that if the MGFs of a sequence converge pointwise to the MGF of some distribution X in a neighborhood of 0, then the distributions converge weakly to the distribution of X. So convergence of MGFs to exp(t²/2) directly implies convergence in distribution to N(0,1). This is one proof route for the Central Limit Theorem — show that the MGF of the standardized sum of iid variables converges to exp(t²/2).
Question 3 True / False
The moment generating function M(t) = E[e^{tX}] typically exists for any random variable X, because e^{tX} is a well-defined real number for most value of X.
TTrue
FFalse
Answer: False
e^{tX} is indeed a real number for each specific value of X, but the MGF requires taking the *expectation* E[e^{tX}], which is an integral (or sum) over all possible values. For heavy-tailed distributions, this integral diverges — the MGF is infinite. The Cauchy distribution is the standard example: E[e^{tX}] = ∞ for all t ≠ 0. This is why the characteristic function E[e^{itX}], which uses complex exponentials satisfying |e^{itX}| = 1, is more general.
Question 4 True / False
Two random variables X and Y with the same moment generating function (wherever it exists on an open interval around 0) must have identical probability distributions.
TTrue
FFalse
Answer: True
This uniqueness property is what makes the MGF so powerful for proving limit theorems. It is analogous to the fact that a smooth function is determined by all its Taylor coefficients at a point — the MGF encodes all moments, and when those are sufficient to determine the distribution (which requires the MGF to exist on an open interval, ensuring the Taylor series converges), the distribution is fully pinned down. This is why showing that MGFs converge is equivalent to showing that distributions converge.
Question 5 Short Answer
Explain why differentiating the MGF M(t) = E[e^{tX}] exactly k times and evaluating at t = 0 recovers the kth moment E[Xᵏ].
Think about your answer, then reveal below.
Model answer: By the Taylor expansion, e^{tX} = 1 + tX + t²X²/2! + t³X³/3! + ⋯ Taking expectations term by term gives M(t) = 1 + tE[X] + t²E[X²]/2! + t³E[X³]/3! + ⋯ This is a power series in t with coefficient E[Xᵏ]/k! in front of tᵏ. Differentiating a power series k times and evaluating at t = 0 isolates the coefficient of tᵏ and multiplies by k!, yielding E[Xᵏ].
The Taylor series argument is the cleanest way to see why the MGF 'generates' moments. The function e^{tX} is just a device for packaging all powers of X into one expression, weighted by powers of t. Taking the expectation distributes this over all moments. The derivative operation then acts as a 'selector' that extracts one moment at a time by evaluating at t = 0, where all higher-power terms vanish. The key requirement is that the Taylor series of M(t) converges — guaranteed when M(t) exists in an open interval around 0.