Two random variables X and Y have been shown to have the same MGF on an open interval around zero. What can you conclude?
AX and Y have the same mean and variance, but their full distributions may still differ
BX and Y have identical distributions
CX and Y are independent of each other
DX and Y are identically distributed only if they also have the same support
The uniqueness theorem for MGFs states: if two random variables have the same MGF on an open interval containing zero, they have the same distribution — not just the same moments, but identical distributions. This is what gives MGFs their real power beyond moment computation: you can prove distributional equalities by comparing MGFs instead of density functions. Options A and D are wrong because matching all moments (encoded by the MGF) determines the full distribution, not just the first two moments.
Question 2 Multiple Choice
How does the MGF formula M(t) = E[e^{tX}] encode the moments of a random variable X?
AThe moments are encoded in the base of the exponential — varying the base extracts different moments
BEach moment E[X^n] appears as the coefficient of t^n/n! in the Taylor expansion of M(t) around t = 0
CThe MGF encodes only the mean and variance; higher moments require separate calculations
DMoments are recovered by integrating M(t) over intervals centered at zero
Expanding e^{tX} as a Taylor series gives 1 + tX + t²X²/2! + t³X³/3! + ⋯ Taking expectations term by term yields M(t) = 1 + tE[X] + t²E[X²]/2! + ⋯ The coefficient of t^n/n! is exactly E[X^n]. This is why differentiating M(t) n times and evaluating at t = 0 gives M^(n)(0) = E[X^n] — the derivative operation extracts the coefficient by canceling the factorial. All moments are simultaneously encoded in M(t), not just the first two.
Question 3 True / False
For any random variable X, the moment generating function M(t) = E[e^{tX}] typically exists and uniquely determines the distribution.
TTrue
FFalse
Answer: False
The MGF does not always exist. For heavy-tailed distributions — most famously the Cauchy distribution — E[e^{tX}] is infinite for all t ≠ 0, so the MGF fails to exist. When the MGF doesn't exist, one must use the characteristic function φ(t) = E[e^{itX}], which always exists since |e^{itX}| = 1. The characteristic function has analogous uniqueness and moment properties but requires complex analysis. The existence caveat is important: MGF results apply only when the MGF exists on an open interval around zero.
Question 4 True / False
If X and Y are independent random variables, the MGF of X + Y equals the product of their individual MGFs.
TTrue
FFalse
Answer: True
M_{X+Y}(t) = E[e^{t(X+Y)}] = E[e^{tX} · e^{tY}]. By independence, E[e^{tX} · e^{tY}] = E[e^{tX}] · E[e^{tY}] = M_X(t) · M_Y(t). This multiplicative property is what allows MGF proofs of important results like the Central Limit Theorem: as you take sums of independent copies, their product MGFs converge to the MGF of the normal distribution, and by uniqueness the sums converge in distribution to normal. Without independence, the factorization fails.
Question 5 Short Answer
Explain why the n-th derivative of M(t) evaluated at t = 0 gives E[X^n], connecting this to the Taylor series expansion of e^{tX}.
Think about your answer, then reveal below.
Model answer: Start from the Taylor series: e^{tX} = Σ (tX)^n/n! = 1 + tX + t²X²/2! + t³X³/3! + ⋯ Taking expectations: M(t) = E[e^{tX}] = 1 + tE[X] + t²E[X²]/2! + t³E[X³]/3! + ⋯ This is a power series in t whose coefficient of t^n/n! is E[X^n]. When you differentiate M(t) n times with respect to t, you apply the power rule n times to each term: the t^n/n! term becomes 1, and all lower-degree terms vanish. Evaluating at t = 0 kills all remaining terms involving positive powers of t, leaving only E[X^n]. Differentiation is precisely the operation that extracts Taylor coefficients.
This connection between differentiation and moment extraction is the computational heart of MGF theory. It transforms the problem of computing moments (which might require difficult integrals) into the problem of differentiating a single function — often much easier for standard distributions like the Poisson, binomial, or normal.