Statistics Graduate Level Quiz: Mgf And Characteristic Functions

10 questions · exam conditions

0:00

Mgf And Characteristic FunctionsQuestion 1 of 10

A nonnegative random variable $X$ has characteristic function $\phi_X(t)=(1-2it)^{-3}.$ What is the third raw moment $E[X^3]$ ?

E[X^3]=48.

E[X^3]=480.

E[X^3]=216.

E[X^3]=60.

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Mgf And Characteristic Functions

Practice Mgf And Characteristic Functions in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Mgf And Characteristic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A nonnegative random variable $X$ has characteristic function $\phi_X(t)=(1-2it)^{-3}.$ What is the third raw moment $E[X^3]$ ?

$E[X^3]=48.$
$E[X^3]=480.$ (correct answer)
$E[X^3]=216.$
$E[X^3]=60.$

Explanation: When you see a characteristic function on a statistics exam, your instinct should immediately be: identify the distribution, then use moment formulas. The characteristic function

\phi_X(t) = (1-2it)^{-3}

matches the gamma distribution

\text{Gamma}(k, \theta)

with shape

k = 3

and scale

\theta = 2

, since the gamma CF is

(1 - i\theta t)^{-k}

. For a

\text{Gamma}(k, \theta)

distribution, the

n

-th raw moment is

E[X^n] = \theta^n \cdot \frac{\Gamma(k+n)}{\Gamma(k)}

. For the third raw moment with

k=3

and

\theta=2

E[X^3] = 2^3 \cdot \frac{\Gamma(6)}{\Gamma(3)} = 8 \cdot \frac{120}{2} = 8 \cdot 60 = 480.

So B) 480 is correct. You can verify this using the CF derivative method:

E[X^n] = i^{-n} \phi_X^{(n)}(0)

. Differentiating three times and evaluating at

t=0

yields the same result. Choice A) 48 likely comes from forgetting the

\Gamma

ratio and computing

8 \times 6 = 48

, confusing

\Gamma(6)/\Gamma(3)

with just

k=6

. Choice C) 216 is

6^3

, suggesting the student used

E[X]^3 = (k\theta)^3 = 6^3

instead of

E[X^3]

— a classic moment vs. moment-of-moments confusion. Choice D) 60 is simply

\Gamma(6)/2 = 60

, dropping the

\theta^3

factor entirely. Your study tip: memorize the gamma moment formula

E[X^n] = \theta^n \Gamma(k+n)/\Gamma(k)

and always check whether the CF's scale parameter needs to be raised to the

n

-th power.

Question 2

The joint characteristic function of $X$ and $Y$ is $\phi_{X,Y}(s,t)=\exp\left\{-\frac{1}{2}(s^2+4st+9t^2)\right\}.$ Which pair of random variables is independent?

$X$ and $Y$
$X+Y$ and $2X-Y$
$X$ and $Y-2X$ (correct answer)
$X-2Y$ and $2X+Y$

Explanation: When working with joint characteristic functions, the key tool for testing independence is this: two random variables

U

and

V

are independent if and only if their joint characteristic function factors as

\phi_{U,V}(s,t) = \phi_U(s)\cdot\phi_V(t).

For jointly normal variables, this happens precisely when

U

and

V

are uncorrelated. First, read off the covariance structure from

\phi_{X,Y}(s,t)=\exp\left\{-\tfrac{1}{2}(s^2+4st+9t^2)\right\}.

The exponent reveals the covariance matrix:

\text{Var}(X)=1,\quad\text{Cov}(X,Y)=2,\quad\text{Var}(Y)=9.

\sigma_X=1

\sigma_Y=3

, and

\text{Cov}(X,Y)=2

. For choice C, compute

\text{Cov}(X,\, Y-2X) = \text{Cov}(X,Y) - 2\text{Var}(X) = 2 - 2(1) = 0.

Since

X

and

Y-2X

are jointly normal (linear combinations of normals) and uncorrelated, they are independent — confirming C is correct. For A,

\text{Cov}(X,Y)=2\neq 0

, so

X

and

Y

are dependent. For B, compute

\text{Cov}(X+Y,\,2X-Y) = 2\text{Var}(X) + \text{Cov}(X,Y) - \text{Cov}(Y,X) - \text{Var}(Y)...

working it out:

2(1)+(2)-(2)-9 = -7\neq 0.

For D,

\text{Cov}(X-2Y,\,2X+Y) = 2(1)+\text{Cov}(X,Y)-4\text{Cov}(Y,X)-2(9) = 2+2-8-18=-22\neq 0.

Your go-to strategy: extract the covariance matrix directly from the quadratic form in the exponent, then check whether your target linear combinations have zero covariance. For jointly Gaussian variables, zero covariance equals independence.

Question 3

A random variable $X$ has characteristic function $\phi_X(t)=0.3+0.7e^{2it-t^2/2}.$ Which pair gives the mean and variance of $X$ ?

$E[X]=1.4,\quad \operatorname{Var}(X)=3.50.$
$E[X]=1.4,\quad \operatorname{Var}(X)=0.70.$
$E[X]=2.0,\quad \operatorname{Var}(X)=1.00.$
$E[X]=1.4,\quad \operatorname{Var}(X)=1.54.$ (correct answer)

Explanation: When you encounter a characteristic function, remember that moments are extracted via derivatives:

E[X] = \frac{\phi_X'(0)}{i}

and

E[X^2] = \frac{\phi_X''(0)}{i^2}

, with

\operatorname{Var}(X) = E[X^2] - (E[X])^2

. Here,

\phi_X(t) = 0.3 + 0.7e^{2it - t^2/2}

. Taking the first derivative:

\phi_X'(t) = 0.7(2i - t)e^{2it - t^2/2}

t=0

\phi_X'(0) = 0.7(2i) = 1.4i

, so

E[X] = \frac{1.4i}{i} = 1.4

. For the second derivative, differentiating again:

\phi_X''(t) = 0.7\left[(-1) + (2i-t)^2\right]e^{2it - t^2/2}

t=0

\phi_X''(0) = 0.7[{-1} + (2i)^2] = 0.7(-1 - 4) = -3.5

, so

E[X^2] = \frac{-3.5}{i^2} = \frac{-3.5}{-1} = 3.5

. Therefore

\operatorname{Var}(X) = 3.5 - (1.4)^2 = 3.5 - 1.96 = 1.54

, confirming answer D. Choice A correctly computes

E[X^2] = 3.5

but forgets to subtract

(E[X])^2

, reporting it as the variance directly. Choice B gets the mean right but uses only the Gaussian component's variance (1) scaled by 0.7, ignoring the full moment calculation. Choice C mistakes the Gaussian mean parameter (2) for

E[X]

and its variance parameter (1) for

\operatorname{Var}(X)

, ignoring the mixing weight entirely. Study tip: Always subtract

(E[X])^2

when computing variance from characteristic functions — forgetting this step is the most common trap, and it's exactly what choice A exploits.

Question 4

Let $X_1$ and $X_2$ be independent copies of a random variable having characteristic function $\phi_X(t)=e^{-2|t|}.$ Define $Y=(X_1+X_2)/2.$ Which statement about the distribution of $Y$ is correct?

$Y$ has the same distribution as $X_1$ . (correct answer)
$Y$ has the same distribution as $2X_1$ .
$Y$ has the same distribution as $X_1/2$ .
$Y$ is normally distributed with variance $4$ .

Explanation: When you see a characteristic function of the form

e^{-c|t|}

, you should immediately recognize the Cauchy distribution family. The key property to exploit here is the stability of the Cauchy distribution: linear combinations of Cauchy random variables remain Cauchy, and the scale parameter transforms predictably through characteristic functions. Recall that if

X

has characteristic function

\phi_X(t)

, then a scaled version

aX

has characteristic function

\phi_X(at)

. For independent

X_1, X_2

, the characteristic function of

X_1 + X_2

\phi_X(t)^2

. Putting this together,

Y = (X_1 + X_2)/2

has characteristic function:

\phi_Y(t) = \phi_X(t/2)^2 = \left(e^{-2|t/2|}\right)^2 = e^{-2|t|}

This is identical to

\phi_X(t)

, so by the uniqueness theorem for characteristic functions, $Y$ has the same distribution as $X_1$ — confirming answer A. For the wrong answers: B would require

\phi_Y(t) = e^{-4|t|}

, which corresponds to scaling

X_1

by 2 (doubling the Cauchy scale parameter) — that's not what we computed. C would require

\phi_Y(t) = e^{-|t|}

, which would arise if you naively divided the scale by 2 without accounting for the squaring from independence — a common algebraic error. D is fundamentally wrong because a Cauchy distribution has no finite variance and is certainly not normal; the heavy-tailed

e^{-|t|}

form (linear in

|t|

) is a hallmark of the Cauchy, not the Gaussian (which is

e^{-\sigma^2 t^2/2}

). Study tip: When a characteristic function is exponential in

|t|

, think Cauchy stability — always compute

\phi_Y(t)

explicitly and apply the uniqueness theorem rather than guessing distributional form from intuition.

Question 5

Suppose $X$ and $Y$ are claimed to be independent, $S=X+Y$ , and the characteristic functions of $X$ and $S$ are reported as $\phi_X(t)=e^{-2t^2}$ and $\phi_S(t)=e^{-t^2}$ , respectively. Which conclusion is correct?

$Y$ must be normal with mean zero and variance $-2$ , interpreted through formal subtraction of variances.
$Y$ must be normal with mean zero and variance $2$ because the exponents differ by $t^2$ .
No probability distribution for an independent $Y$ is compatible with the two reported characteristic functions. (correct answer)
$Y$ may be normal with variance $2$ provided its covariance with $X$ is sufficiently negative.

Explanation: Characteristic functions are uniquely tied to probability distributions, and one of their most powerful properties is the multiplication rule for independent random variables: if

X \perp Y

, then

\phi_{X+Y}(t) = \phi_X(t) \cdot \phi_Y(t)

. This question tests whether you can apply that rule and recognize when reported characteristic functions are internally inconsistent. If

X

and

Y

are truly independent and

S = X + Y

, then

\phi_Y(t)

must equal

\phi_S(t)/\phi_X(t)

. Plugging in the given functions:

\phi_Y(t) = e^{-t^2}/e^{-2t^2} = e^{t^2}

. Here's the problem —

e^{t^2}

is not a valid characteristic function. A valid characteristic function must satisfy

|\phi(t)| \leq 1

for all

t

, and must be positive semi-definite. The function

e^{t^2}

grows without bound and violates these conditions. Therefore, no legitimate probability distribution for

Y

can reconcile the two given characteristic functions, making C correct. Choice A fails because you cannot have a normal distribution with negative variance — variance must be non-negative by definition, and no such distribution exists. Choice B sounds plausible because it correctly identifies the exponent difference, but it draws the wrong conclusion: a normal distribution with variance 2 would have characteristic function

e^{-t^2}

, not

e^{+t^2}

, which is invalid. Choice D introduces covariance as a fix, but independence already implies zero covariance — you cannot adjust covariance while maintaining independence. Study tip: Whenever you're given characteristic functions for a claimed relationship, immediately check consistency using the multiplication rule. If the implied

\phi_Y(t)

explodes or violates

|\phi(t)| \leq \phi(0) = 1

, the setup is mathematically impossible.

Question 6

Suppose $X$ is lognormal. Its positive integer moments are all finite, but its MGF is infinite for every $t>0$ . Let $Y$ be another random variable satisfying $E[Y^k]=E[X^k]$ for every positive integer $k$ . Which conclusion is valid?

The equality of all integer moments forces $X$ and $Y$ to have the same distribution, even without finite MGFs.
The moment equalities need not determine the distribution, but equality of the characteristic functions for all real arguments would. (correct answer)
Because every integer moment of $X$ exists, its MGF must be finite on some open interval containing zero.
The absence of a positive-side MGF means that even the characteristic function cannot uniquely determine the distribution of $X$ .

Explanation: This question tests your understanding of the moment problem — specifically, when do moments uniquely determine a distribution? The key distinction is between the moment determinacy problem and the role of MGFs versus characteristic functions. The lognormal distribution is the classic example of a moment-indeterminate distribution: even though all positive integer moments

E[X^k]

are finite, there exist infinitely many other distributions sharing the exact same integer moments. This was shown by Heyde (1963). So

Y

need not equal

X

in distribution, even with

E[Y^k] = E[X^k]

for all

k \in \mathbb{Z}^+

. However, the characteristic function

\varphi(t) = E[e^{itX}]

always exists for all real

t

(since

|e^{itX}| = 1

), and by the Lévy uniqueness theorem, it uniquely determines the distribution. So B is correct: moment equality is insufficient, but characteristic function equality would force distributional equality. A is wrong because it asserts that integer moment equality implies distributional equality — precisely the claim the lognormal counterexample refutes. C contains a critical misconception: finiteness of all integer moments does not imply a finite MGF on any interval around zero. The MGF requires finiteness of

E[e^{tX}]

, which involves all moments growing fast enough — lognormal moments grow too rapidly (faster than any exponential), making the MGF infinite for every

t > 0

. D is wrong because it conflates the MGF with the characteristic function. The characteristic function always exists and always uniquely determines the distribution, regardless of MGF behavior. Study tip: Remember the triad — moments can fail to determine a distribution (moment problem), MGFs may not exist, but characteristic functions always exist and always uniquely characterize distributions.

Question 7

A random variable $X$ has moment generating function $M_X(t)=e^{2t}(1-3t)^{-4},\quad t<1/3.$ Define $Y=(X-14)/6.$ Which expression is the moment generating function of $Y$ , including its maximal interval of finiteness containing zero?

$M_Y(t)=e^{-2t}(1-t/2)^{-4},\quad t<2.$ (correct answer)
$M_Y(t)=e^{2t}(1-t/2)^{-4},\quad t<2.$
$M_Y(t)=e^{-2t}(1-18t)^{-4},\quad t<1/18.$
$M_Y(t)=e^{-7t/3}(1-t/2)^{-4},\quad t<2.$

Explanation: When you need to find the MGF of a linear transformation

Y = aX + b

, use the fundamental property:

M_Y(t) = e^{bt} M_X(at)

. Here,

Y = (X - 14)/6 = \frac{1}{6}X - \frac{14}{6}

, so

a = 1/6

and

b = -14/6 = -7/3

. Applying the formula, substitute

at = t/6

into

M_X

M_Y(t) = e^{-7t/3} \cdot M_X(t/6) = e^{-7t/3} \cdot e^{2(t/6)}(1 - 3(t/6))^{-4}

= e^{-7t/3} \cdot e^{t/3}(1 - t/2)^{-4} = e^{-2t}(1 - t/2)^{-4}

For the domain, the original MGF requires

t/6 < 1/3

, which gives

t < 2

. So the correct answer is A. B has the wrong sign on the exponential — it uses

+2t

instead of

-2t

, likely from forgetting the negative sign in

b = -7/3

or misapplying the shift. C correctly captures the negative exponential idea but fails to scale the argument properly; substituting

t/6

into

3t

gives

t/2

, not

18t

— this error comes from mistakenly multiplying instead of dividing. D is a tempting trap: it correctly computes

e^{-7t/3}

but stops before combining the exponential terms, forgetting to merge

e^{-7t/3} \cdot e^{t/3} = e^{-2t}

. Study tip: Always simplify exponential factors fully after substitution — combining

e^{bt} \cdot e^{a \mu t}

into a single exponential is the step most students skip, and it's exactly where the wrong answers hide.

Question 8

For each positive integer $n$ , let $X_n$ be normally distributed with mean zero and variance $n$ . Its characteristic function is $\phi_n(t)=e^{-nt^2/2}.$ The pointwise limit is one at $t=0$ and zero at every $t\ne 0$ . Which conclusion is correct?

$X_n$ converges in distribution to the constant zero because the limiting characteristic function equals one at zero.
$X_n$ converges in distribution to a probability law whose characteristic function vanishes at every nonzero argument.
$X_n$ has no weak limit on the real line because the pointwise limit is discontinuous at zero. (correct answer)
$X_n$ converges to a centered normal law with infinite variance, interpreted as a proper limiting distribution.

Explanation: Whenever you see a question about convergence in distribution, your first instinct should be to invoke Lévy's continuity theorem: a sequence of distributions converges weakly to a proper probability law if and only if its characteristic functions converge pointwise to a function that is continuous at zero. This continuity condition is not optional — it's the decisive criterion. Here,

\phi_n(t) = e^{-nt^2/2}

converges pointwise to the function

f(t) = \mathbf{1}_{t=0}

, which equals 1 at

t=0

and 0 everywhere else. This limiting function is discontinuous at zero (the left- and right-hand limits as

t \to 0

equal 0, not 1). Because the pointwise limit fails the continuity-at-zero requirement, Lévy's theorem tells you that no proper probability distribution on

\mathbb{R}

has this as its characteristic function. The sequence

X_n

therefore has no weak limit on the real line — intuitively, the mass is "escaping to infinity" as

n \to \infty

. This confirms C is correct. A is wrong because convergence in distribution requires the limiting characteristic function to be continuous at zero and be a valid characteristic function everywhere — the value at zero alone tells you nothing. B is wrong because no proper probability law on

\mathbb{R}

can have a characteristic function that vanishes at all nonzero

t

; such a function isn't continuous at zero, so it violates Bochner's theorem. D is wrong because "infinite variance normal" is not a proper distribution on

\mathbb{R}

; the concept is informal and has no rigorous probability measure to support it. Study tip: Memorize Lévy's continuity theorem precisely — the continuity at zero condition is the most commonly tested and most commonly forgotten part.

Question 9

Let $N$ have a Poisson distribution with mean $3$ . Conditional on $N$ , let $Z_1,\ldots,Z_N$ be independent exponential random variables with rate $2$ , independent of $N$ , and define $S=\sum_{j=1}^{N}Z_j$ , with the empty sum equal to zero. Which pair gives both the MGF of $S$ and the probability that $S=0$ ?

$M_S(t)=\exp\left\{3\left(\frac{2}{2-t}-1\right)\right\},\quad \Pr(S=0)=e^{-3}.$ (correct answer)
$M_S(t)=\left(\frac{2}{2-t}\right)^3,\quad \Pr(S=0)=0.$
$M_S(t)=\exp\left\{3\cdot\frac{2}{2-t}\right\},\quad \Pr(S=0)=e^{-3}.$
$M_S(t)=\exp\left\{\frac{6t}{2-t}\right\},\quad \Pr(S=0)=1-e^{-3}.$

Explanation: When you encounter a compound random variable like

S = \sum_{j=1}^N Z_j

where

N

is itself random, the key tool is the law of total expectation applied to MGFs: condition on

N

first, then average over it. Given

N = n

S

is a sum of

n

independent Exp(2) random variables, each with MGF

\frac{2}{2-t}

. So the conditional MGF is

\left(\frac{2}{2-t}\right)^n

. Now average over

N \sim \text{Poisson}(3)

M_S(t) = E\!\left[E[e^{tS}\mid N]\right] = E\!\left[\left(\frac{2}{2-t}\right)^N\right] = G_N\!\left(\frac{2}{2-t}\right)

where

G_N(z) = e^{3(z-1)}

is the PGF of Poisson(3). Substituting gives

M_S(t) = \exp\!\left\{3\!\left(\frac{2}{2-t}-1\right)\right\}

. For

\Pr(S=0)

: this happens only when

N=0

, so

\Pr(S=0) = \Pr(N=0) = e^{-3}

. That's answer A. B is wrong on two counts: the MGF

\left(\frac{2}{2-t}\right)^3

treats

N

as fixed at 3 rather than random, and it implies

S

is always positive, forcing

\Pr(S=0)=0

— ignoring the positive probability that

N=0

. C incorrectly omits the

-1

in the exponent, failing to subtract the normalization term when applying the PGF formula. D uses an algebraically incorrect simplification of the exponent and gives the wrong probability

1 - e^{-3}

, which is actually

\Pr(N \geq 1)

. Study tip: Memorize that for

N \sim \text{Poisson}(\lambda)

, the compound sum MGF is

\exp\{\lambda(M_Z(t)-1)\}

. The

-1

inside is essential — it ensures

M_S(0)=1

Question 10

Under a probability measure $P$ , a random variable $X$ has MGF $M_P(t)=(1-t)^{-2}$ for $t<1$ . A new probability measure $Q$ is defined by exponential tilting with parameter $1/2$ , so that $dQ/dP=e^{X/2}/M_P(1/2).$ Which is the MGF of $X$ under $Q$ , with its correct domain?

$M_Q(t)=(1-2t)^{-2},\quad t<1.$
$M_Q(t)=(1-t/2)^{-2},\quad t<2.$
$M_Q(t)=4(1/2-t)^{-2},\quad t<1/2.$
$M_Q(t)=(1-2t)^{-2},\quad t<1/2.$ (correct answer)

Explanation: Exponential tilting (also called an Esscher transform) is a fundamental tool in mathematical finance and risk theory. When you see a Radon-Nikodym derivative of the form

dQ/dP = e^{\theta X}/M_P(\theta)

, your job is to compute the MGF under

Q

by absorbing the tilt into the moment-generating machinery. The MGF of

X

under

Q

is:

M_Q(t) = E^Q[e^{tX}] = E^P\!\left[e^{tX}\cdot\frac{dQ}{dP}\right] = \frac{E^P[e^{(t+\theta)X}]}{M_P(\theta)} = \frac{M_P(t+\theta)}{M_P(\theta)}.

With

\theta = 1/2

and

M_P(t)=(1-t)^{-2}

, first compute

M_P(1/2) = (1-1/2)^{-2} = 4

. Then:

M_Q(t) = \frac{(1-(t+1/2))^{-2}}{4} = \frac{(1/2-t)^{-2}}{4}.

Note that

(1/2-t)^{-2} = 4(1-2t)^{-2}

, so

M_Q(t) = \frac{4(1-2t)^{-2}}{4} = (1-2t)^{-2}

. The domain requires

t + 1/2 < 1

, i.e.,

t < 1/2

. This confirms D. Choice A gives the right functional form

(1-2t)^{-2}

but uses the wrong domain

t<1

, ignoring that tilting shifts the convergence strip. Choice B results from an algebra error — using

\theta/2

instead of

\theta

in the shift, producing a weaker tilt. Choice C correctly computes the intermediate expression

4(1/2-t)^{-2}

but fails to simplify it, leaving a distractor that looks numerically distinct. Your study tip: always track two things under exponential tilting — the shifted functional form and the shifted domain. The new domain is the old domain translated by

-\theta

, a step that eliminates both A and C as viable options.