Converse assumed

The theorem runs one way. You used it in the other.

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 B

  $-n/\sum \ln X_i$

  Why this is wrong, and why it looks right

  False — this estimates $\theta$, not $1/\theta .-n/\sum \ln X_i$ is the reciprocal of the correct estimator and is in fact the MLE of $\theta$. Reciprocals do not commute with expectation, so an unbiased estimator of $1/\theta$ cannot simply be inverted to give one for $\theta$.

  From this question

  For n ≥ 2, let $X_1,\dots ,X_n$ be a random sample with density f(x|$\theta )=$ $\theta x^{\theta -1}$ on $0<x<1,\theta >0$ unknown. Which of the following is the UMVUE for $1/\theta$?

  See the full question

  Part B · december 2023 · free

• Option B

  G is a cyclic group.

  Why this is wrong, and why it looks right

  False — 'need not be' is the point. P has order $17^2$ and may be ${\mathbb{Z}}_{17}\times {\mathbb{Z}}_{17}$ rather than ${\mathbb{Z}}_{289}$, in which case $G\cong {\mathbb{Z}}_7\times {\mathbb{Z}}_{17}\times {\mathbb{Z}}_{17}$ has no element of order 2023. Abelian does not imply cyclic once a prime appears squared; ${\mathbb{Z}}_7\times {\mathbb{Z}}_{289}$ is cyclic but it is not the only option.

  From this question

  Let G be a group of order 2023. Which of the following statements are true?

  See the full question

  Part C · june 2023

• Option B

  If P(A | B) = 1 then P(B | A) = 1.

  Why this is wrong, and why it looks right

  False, and here the symmetry genuinely breaks. P(A | B) = 1 means P(A ∩ B) = P(B), that is B is contained in A up to null sets. Then P(B | A) = P(B)/P(A), which is less than 1 whenever P(B) < P(A). Containment runs one way only.

  From this question

  Let A, B be two events in a discrete probability space with P(A) > 0 and P(B) > 0. Which of the following are necessarily true?

  See the full question

  Part C · june 2023

• Option D

  If P(A | B) > P(B) then P(B | A) > P(A).

  Why this is wrong, and why it looks right

  False. Written out, the hypothesis is $P(A\cap B)>P(B{)}^2$ and the conclusion $P(A\cap B)>P(A{)}^2$, which are different statements. Take B ⊂ A with P(A) = 9/10 and P(B) = 1/10: then P(A | B) = 1 > 1/10 = P(B), so the hypothesis holds, but P(B | A) = 1/9 (verified) and SymPy confirms 1/9 − 9/10 < 0, so the conclusion fails. Compare with (C): the same shape of statement, but comparing against the other event's probability destroys the symmetry.

  From this question

  Let A, B be two events in a discrete probability space with P(A) > 0 and P(B) > 0. Which of the following are necessarily true?

  See the full question

  Part C · june 2023

• Option D

  If $\nabla u\cdot \nabla v=0$ at every point, then f is an entire function.

  Why this is wrong, and why it looks right

  False — it is (C) read backwards. Take f(z) = z̄, so u = x and v = −y: SymPy gives $\nabla u\cdot \nabla v=1\cdot 0+0\cdot (-1)=0$, yet f is nowhere holomorphic. Orthogonal gradients are one consequence of Cauchy–Riemann, not equivalent to them.

  From this question

  Let $f:\mathbb{C}\to \mathbb{C}$ be a real-differentiable function and define u(x, y) = Re f(x + iy), v(x, y) = Im f(x + iy). Let $\nabla u=(u_x,u_y)$ denote the gradient. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023

• Option A

  There is no continuous real-valued function g defined on any interval of $\mathbb{R}$ containing 0 such that f(x, g(x)) = 0.

  The analysis of this one comes with the PYQ pack.

  Part C · june 2023

• Option C

  a is not divisible by 48.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  rank A = 2.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option B

  rank A = 1.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option A

  If A is a real 3×2 matrix and B a real 2×3 matrix with BA $=I_2$, then AB $=I_3$.

  The analysis of this one comes with the PYQ pack.

  Part B · december 2023

• Option B

  There exists a holomorphic surjective map $f:{\Omega }_2\to {\Omega }_1$.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option D

  There exists a holomorphic injective map $f:{\Omega }_2\to {\Omega }_1$.

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option A

  Statement 1

  The analysis of this one comes with the PYQ pack.

  Part A · september 2022

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