Moments and tails

A moment assumed to exist, or tail behaviour assumed to be tame.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option A

  2

  Why this is wrong, and why it looks right

  This is $E[X]=\lambda=2$ itself. It never squares anything, so it can't be $E[X^{2}]$.

  From this question

  Let X ~ Poisson(2). Then $E[X^{2}]$ equals

  Part B · free

• Option B

  4

  Why this is wrong, and why it looks right

  This is $(E[X])^{2}=2^{2}=4$, which equals $E[X^{2}]$ only if Var(X)=0. Poisson(2) has Var(X)=2>0, so $E[X^{2}]=Var(X)+(E[X])^{2}=2+4=6$, not 4.

  From this question

  Let X ~ Poisson(2). Then $E[X^{2}]$ equals

  Part B · free

• Option D

  8

  Why this is wrong, and why it looks right

  $8=\lambda^{2}+2\lambda=4+4$ double-counts the mean term. The correct combination uses the factorial moment $E[X(X-1)]=\lambda^{2}: E[X^{2}]=E[X(X-1)]+E[X]=\lambda^{2}+\lambda=4+2=6$.

  From this question

  Let X ~ Poisson(2). Then $E[X^{2}]$ equals

  Part B · free

• Option D

  The strong law of large numbers requires finite variance.

  Why this is wrong, and why it looks right

  False — option 1 already shows the SLLN holds under E|$X_{1}$|$<\infty$ alone. The variance is needed only for the rate/distributional statements about the CLT, not for almost-sure convergence of the average itself.

  From this question

  Let $X_{1}, X_{2}, \dots$ be i.i.d. with E|$X_{1}$| $< \infty, E X_{1} = \mu$, and ${\bar{S}}_{n} = (X_{1} + \cdots + X_{n})/n$. Which of the following are true?

  Part C · free

• Option B

  p > 1

  Why this is wrong, and why it looks right

  p > 1 keeps the integral fine near 0, but it fails at infinity: e.g. p = 2 makes the integrand behave like $x^{p-2}$ $= x^{0} = 1$ for large x, and $\int^\infty 1$ dx diverges. This range gets the binding endpoint exactly backwards.

  From this question

  For which values of p does $\int_{0}^\infty$ $x^{p-1}/(1$ + x) dx converge?

  Part C · free

• Option C

  p > 0

  Why this is wrong, and why it looks right

  p > 0 is necessary for convergence at x = 0 but says nothing about $x \to \infty$. Taking p = 2, which satisfies p > 0, gives integrand ~ $x^{p-2}$ $= x^{0}$ for large x, so the tail $\int^\infty 1$ dx diverges.

  From this question

  For which values of p does $\int_{0}^\infty$ $x^{p-1}/(1$ + x) dx converge?

  Part C · free

• Option A

  $E(X^{4}) = 3$

  Why this is wrong, and why it looks right

  False. X is standard Cauchy, and the Cauchy distribution has no finite moments of any positive order. SymPy samples the fourth-moment integrand $x^{4}/(\pi(1 + x^{2}))$ and finds it growing without bound — it behaves like $x^{2}/\pi$ for large x, so the integral diverges. Not even E(X) exists, let alone $E(X^{4})$.

  From this question

  Suppose U ~ Uniform(0,1) and $X = \tan(\pi(U -$ ½)). Then which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option B

  converges absolutely

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  The distribution of X is symmetric about its mean.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  Var(S) = 2.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  0.50

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  0.90

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  0.95

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  $f \in L^{4}(\mathbb{R})$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  $f \in L^\infty(\mathbb{R})$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  There exists an $\alpha \in (0, \infty)$ such that $lim_{x\to\alpha^{-}}$ |y(x)| $= \infty$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option A

  $(13/5)e^{-2}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option B

  $(1/5)(e^{-2}$ + $2e^{-1})$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option C

  E|$X_{n} - X$| → 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  $X_{n} \to X$ in $L^{2}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option A

  $X_{1}$ and $(X_{1} + X_{2})/\sqrt{2}$ have the same distribution.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  $(1/n)\sum(i=1..n)X_{i}$ converges to 0 in probability, as $n \to \infty$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  E(|$X_{1}$|$^(3/4)) = \infty$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  Finite mean implies finite variance.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  $(T_{n} - n)/\sqrt{3n}$ converges in distribution to a random variable Z, where Z ~ N(0, 1)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  $(T_{n} - n)/\sqrt{V}_{n}$ converges in distribution to a random variable Z, where Z ~ N(0, 1)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• Every IVP has a unique solution
• Convergence in probability ⇒ almost sure convergence
• Uncorrelated ⇒ independent
• A continuous right-hand side gives a unique solution
• A locally unique solution exists for all time
• Every random variable has a moment generating function

Other mistakes that cost marks