X and Y are independent random variables. A student wants to find E[(X+Y)²] using moment generating functions. Which combination of MGF properties makes this approach work?
AM_{X+Y}(t) = M_X(t) + M_Y(t) for independent variables, then differentiate twice at t=0
BThe n-th derivative of M at t=0 gives E[Z^n], combined with M_{X+Y}(t) = M_X(t)·M_Y(t) for independent variables
Ce^{t(X+Y)} = e^{tX} + e^{tY}, so expectations add directly
DThe normal approximation applies since X and Y are independent, giving a standard formula
Two properties combine: for independent X and Y, M_{X+Y}(t) = M_X(t)·M_Y(t) (product, not sum — this follows from E[e^{tX}·e^{tY}] = E[e^{tX}]·E[e^{tY}] by independence). Then M''_{X+Y}(0) gives E[(X+Y)²]. The sum-equals-product property is one of the most important and non-obvious facts about MGFs, and it's what makes them so useful for studying sums of independent variables.
Question 2 Multiple Choice
A student computes the mean and variance of distribution D and concludes: 'I've fully identified the distribution — any distribution with these parameters must be the same.' Is this claim sound?
AYes — the first two moments (mean and variance) uniquely determine any distribution
BNo — two moments are not sufficient to identify a distribution in general; the MGF uniqueness theorem requires the MGF to agree in a neighborhood of t=0, encoding all moments
CYes — mean and variance determine distributions up to location and scale, which is sufficient for identification
DNo — only the characteristic function can uniquely determine a distribution; the MGF is not sufficient
Two moments are far from sufficient to identify a distribution. There exist families of distinct distributions with identical means and variances. The MGF uniqueness theorem says that if two distributions have the same MGF in a neighborhood of t=0, they must be identical — but this requires the entire MGF (which encodes all moments), not just the first two. The characteristic function always exists and also characterizes distributions uniquely, but the issue here is that moment-matching with finitely many moments is insufficient.
Question 3 True / False
Differentiating M(t) = E[e^{tX}] twice and evaluating at t=0 gives E[X²], not the variance of X.
TTrue
FFalse
Answer: True
M''(0) = E[X²], which is the second raw moment. The variance is Var(X) = E[X²] − (E[X])² = M''(0) − (M'(0))². These are different quantities: E[X²] is the second moment; variance is the second central moment. Students often conflate 'second derivative of MGF' with 'variance,' but variance requires subtracting the square of the mean.
Question 4 True / False
If the MGF of random variable X equals the MGF of random variable Y for most t, then X and Y should be the same random variable defined on the same probability space.
TTrue
FFalse
Answer: False
Equal MGFs imply equal distributions (same probability law over outcomes), not that X and Y are literally the same random variable or defined on the same sample space. You can have two completely separate random experiments where X and Y follow the same distribution — their MGFs are equal, but they are distinct random variables. The uniqueness theorem is about distributions, not about the random variables themselves.
Question 5 Short Answer
Why is multiplying MGFs the key tool for studying sums of independent random variables, and what would be the harder alternative?
Think about your answer, then reveal below.
Model answer: For independent X and Y, M_{X+Y}(t) = M_X(t)·M_Y(t) because E[e^{t(X+Y)}] = E[e^{tX}·e^{tY}] = E[e^{tX}]·E[e^{tY}] by independence. This reduces finding the distribution of X+Y to algebraic multiplication of functions, then identifying the result. The harder alternative is direct convolution: computing the density or PMF of X+Y by integrating (or summing) over all ways X and Y can combine to give each total, which is an integral/sum and becomes very cumbersome for repeated sums or more complex distributions.
The payoff is especially clear in CLT proofs: showing the MGF of the standardized sum of n i.i.d. variables converges to e^{t²/2} (the standard normal's MGF) is much cleaner than working with convolutions directly. The product property turns a problem about distributions of sums into a problem about products of functions.