Puzzle editor Anirban DasGupta returns with these puzzles in our contest model. Each correct answer receives 3 points, each incorrect answer receives -2 points, and each item left unanswered receives -1 point. You can answer just one of the two problems, 55.1 and 55.2, but it will be fantastic if you attempt both.

Puzzle 55.1

Let $X_1,\cdots ,X_n$ be iid $C(\mu ,1)$, the Cauchy distribution with median $\mu$ and scale parameter $1$.

(a) Prove rigorously that the number of roots $T_n$ of the likelihood equation is an odd integer between 1 and $2n-1$.

(b) Find with precise reasoning $\underset{n\to \mathrm{\infty }}{lim}P(T_n>1)$.

Puzzle 55.2, the contest problem. For each question, just say True or False, without the need to provide a proof. But answers with some explanations are especially welcome. Here are the items.
(a) A quadratic $a{x}^2+bx+c$ is called a Gaussian quadratic if $a,b,c$ are i.i.d. standard normal. If $X_1,X_2$ denote the two roots of a Gaussian quadratic, then the expectation of $\frac{1}{X_1}+\frac{1}{X_2}$ does not exist.

(b) In the standard linear model $E(\mathbf{Y}\mathbf{)}\mathbf{=}\mathbf{X}\beta$, no quadratic function ${\mathbf{Y}}^{\prime }\mathbf{A}\mathbf{Y}$ can be an unbiased estimate of a linear function ${\mathbf{c}}^{\prime }\mathbf{b}\mathbf{e}\mathbf{t}\mathbf{a}$.

(c) There exists a location parameter density on the real line such that the average of the three sample quartiles is asymptotically the most efficient among all convex combinations of the three sample quartiles.

(d) Suppose $f:[0,1]\to \mathbb{R}$ is a strictly increasing continuous function. Then there is a minimum value of $D(r)={\int }_0^1|f(x)-r|dx$, and the minimum is attained at a unique real number $r_0$.

(e) Let $X_n$ denote an $n\times n$ matrix all of whose elements are $\pm 1$. If $D_n$ denotes the supremum of the determinant of all such matrices $X_n$, then $(D_n{)}^{1/n}$ has a finite limit superior.

(f) Sixteen equally good soccer teams are going to play in a tournament, in which teams are paired up in random, and a team to lose a match is eliminated from the tournament. The probability that teams $1$ and $2$ will meet each other at some point in the tournament is more than $10\mathrm{\%}$.

Solution to Puzzle 54

Congratulations to IMS student member Yutong Wang (London School of Economics and Political Science), whose answers to Puzzle 54 were correct!

Guest puzzle editor Stanislav Volkov explains:

LEMMA 1:
Suppose that $X_i$, $i=1,\dots ,n$ are Bernoulli($p$) random variables and denote $S_n=X_1+\cdots +X_n$.
Then $\text{Var}(S_n)\le n^{2}p(1-p)$ and this upper bound can be achieved.

PROOF:
Since $\text{Var}(X_i)=p(1-p)$ and $\text{Cov}(X_i,X_j)\le \sqrt{\text{Var}(X_i)\text{Var}(X_j)}=p(1-p)$ by the Cauchy-Schwarz inequality, we have
$\text{Var}(S_n)=\sum _{i=1}^n\text{Var}(X_i)+2\sum _{1\le i<j\le n}\text{Cov}(X_i,X_j)\le np(1-p)+n(n-1)\cdot p(1-p)=n^{2}p(1-p).$
On the other hand, this variance can be achieved by setting all $X_i\equiv X_1$.

LEMMA 2:
Suppose that $X_i$, $i=1,\dots ,n$ are Bernoulli($p$) random variables and denote $S_n=X_1+\cdots +X_n$. Let $m=\lfloor np\rfloor$, and let $a=np-m$ be the fractional part of $np$ (which can be $0$). Then $\text{Var}(S_n)\ge a(1-a)$, and this lower bound can be achieved.

PROOF:
(a) First, let us show that this is indeed the lowest possible value. For simplicity, assume that $0<a<1$ (the case when $np$ is an integer can be handled similarly).

We claim that for any integer-valued random variable $S$, with $\mathbb{E}(S)=np\in (m,m+1)$, the variance cannot be less than $a(1-a)$, and moreover this minimum is achieved when $S\in \{m,m+1\}$ with probability one; in this case $P(S_n=m)=1-a$, $P(S_n=m+1)=a$, and the variance is indeed $a(1-a)$.

Suppose the contrary and the minimum is achieved for some $S$ such that $\mathbb{P}(S\notin \{m,m+1\})>0$. Then there exists either $i\le m-1$ such that $\mathbb{P}(S=i)>0$, or $j\ge m+2$ such that $\mathbb{P}(S=j)>0$. Because of symmetry, it suffices only to study the first case.

Indeed, assume that there is an $i\le m-1$ such that $\mathbb{P}(S=i)>0$. We shall show that the variance of $S$ can be made strictly smaller. Since $\mathbb{E}S>m$, there must be an integer $j\ge m+1$ such that $\mathbb{P}(S=j)>0$. Let $ϵ=min(\mathbb{P}(S=i),\mathbb{P}(S=j))>0$. Note also that $\mathbb{P}(S=m)<1$. Take a very small $\delta >0$ and create a new variable $S^{\prime }$ by changing the probabilities only at three points: $i,m,j$ by setting
$$\begin{array}{rl}\mathbb{P}(S^{\prime }=i)& =\mathbb{P}(S=i)-x,\\ \mathbb{P}(S^{\prime }=m)& =\mathbb{P}(S=m)+\delta ,\\ \mathbb{P}(S^{\prime }=j)& =\mathbb{P}(S=j)-y.\end{array}$$
To ensure that the probabilities sum up to $1$ and to keep $\mathbb{E}S=\mathbb{E}{S}^{\prime }$ unchanged, we need
$$\begin{array}{rl}-x+\delta -y& =0,\\ -ix+m\delta -jy& =0.\end{array}$$
Solving for $x$ and $y$ gives
$$x=\delta \frac{j-m}{j-i}>0,x=\delta \frac{m-i}{j-i}>0$$
which is possible as long as $\delta$ is small enough.
However, with this change, the variance of $S$ will decrease:
$$\begin{array}{rl}\text{Var}(S^{\prime })-\text{Var}(S)& =\mathbb{E}(S^{\prime 2})-\mathbb{E}(S^2)=–i^{2}x+m^2\delta -j^{2}y=-(j–m)(m–i)\delta <0\end{array}$$
so $\text{Var}(S^{\prime })<\text{Var}(S)$ contradicting the assumption that $S$ has the smallest variance. By the same argument, we can show that if $\mathbb{P}(S=j)>0$ for some $j\ge m+2$, then $S$ does not have the smallest variance.
(b) Now let us show how this minimum can be achieved. Let $C$ be the circumference of a circle centred at the origin, and let $U$ be a point uniformly chosen on $C$. Consider $n$ arcs on $C$, $I_1,I_2,\dots ,I_n$, such that $I_k$ contains all the points located between angles $2\pi (k-1)p$ and $2\pi kp$. Let $X_k=1$ if $I_k$ contains $U$ and set $X_k=0$ otherwise. It is easy to check that $\mathbb{P}(X_k=1)=p$ for all $k=1,2,\dots ,n$.

On the other hand, it is easy to check that $S_n\in \{m,m+1\}$ As a result, $S_n-m$ is a Bernoulli($a$) random variable, resulting in the statement of the claim.

Now let $Y_i=2X_i-1$. Then $T_n=2S_n-n$ and thus by Lemma 1 $\text{Var}(T_n)=4\text{Var}(S_n)$ which has the largest achievable value at $4n^{2}p(1-p)$. By the same token, from Lemma 2, $\text{Var}(T_n)=4\text{Var}(S_n)$ which has the lowest achievable value at $4a(1-a)$ where $a=np-\lfloor np\rfloor$.