Math Challenge - September 2021

[fresh_42]

Is there a prize for the highest grades?

Well, we would have to ask @Greg Bernhardt. Maybe we could create a badge for it.

Otherwise I wouldn't care if I got graded or not.

I have written a TeX-file and created a pdf listing all of my problems and solutions up to the end of the year. It has 530 pages. Additionally, we have had many problems from other members. This means: the counting has to be done by somebody else! I suggest an Excel sheet.

 

[Not anonymous]

15. For a partition $\pi$ of $N := \{1, 2, ..., 9\}$, let $\pi(x)$ be the number of elements in the part containing $x$. Prove that for any two partitions $\pi_1$ and $\pi_2$, there are two distinct numbers $x$ and $y$ in $N$ such that $\pi_j(x) = \pi_j(y)$ for $j=1,2$.

Spoiler: Question 15

Claim 1: Suppose a partition $\pi_1$ of $N = \{1,2,...,9\}$ has 4 distinct numbers, say $a_1, a_2, a_3, a_4$, such that $\pi_1(a_1) = \pi_1(a_2) = \pi_1(a_3) = \pi_1(a_4)$. Then there cannot be another partition $\pi_2$ of $N$ such that $\pi_1(x) = \pi_1(y)$ but $\pi_2(x) \neq \pi_2(y)$ for all distinct $x, y \in N$. In other words, for any other partition $\pi_2$ of $N$, we can find distinct $x, y \in N$ such that $\pi_j(x) = \pi_j(y)$ for $j=1,2$

Proof by contradiction: Suppose there exists a partition $\pi_2$ of the form claimed to not exist above. Then we must have $\pi_2(a_i) \neq \pi_2(a_j)$ for all $i, j \in \{1,2,3,4\}$ with $i \neq j$. Hence $\pi_2(a_1), \pi_2(a_2), \pi_2(a_3), \pi_2(a_4)$ must all be distinct, i.e. each of $a_1, a_2, a_3, a_4$ must belong to a different part in $\pi_2$ and this implies that $\pi_2$ has at least 4 parts of distinct cardinalities. Thus, the number of numbers in $\pi_2$ must be at least $1+2+3+4 = 10$, but this contradicts the fact that $\pi_2$ is a partition of $N$ and hence must have only 9 distinct numbers. This proves that any other partition of $\pi_2$ of $N$ must be such that there will exist at least 2 distinct numbers $x, y \in \{a_1, a_2, a_3, a_4\}$ such that $\pi_2(x) = \pi_2(y)$. And since by definition of $a_i$'s, $\pi_1(x) = \pi_1(y)$, we have found $x,y$ such that $\pi(x) = \pi(y)$ in both $\pi_1$ and $\pi_2$.

---

Now consider the following cases for $\pi_1$.

Case 1: $\pi_1$ has at least 1 part with a cardinality of 4 or more (i.e. the part is a subset of $N$ with at least 4 elements) or has at least 2 parts with the same cardinality and these equal-sized parts have 2 or more elements.

In this case, it is easy to see that there are at least 4 distinct numbers in $N$, say $a_1, a_2, a_3, a_4$, such that $\pi_1(a_1) = \pi_1(a_2) = \pi_1(a_3) = \pi_1(a_4)$. We simply take any 4 numbers from the largest cardinality part in $\pi_1$ as that part is stated to have 4 or more elements. This $\pi_1$ clearly satisfies the condition of claim 1.

Case 2: All parts in $\pi_1$ have a cardinality of 3 or less and at most one part in $\pi_1$ can have a cardinality of 2 or 3.

In this case, since at most one part can have a cardinality greater than 1 and the maximum allowed cardinality of a part is 3, $\pi_1$ must have at least 4 parts whose cardinality is 1 (single element parts). This is because the maximum number of elements contained in the parts of size more than one is $2+3=5$ (one part of size 2, one part of size 3) and the remaining $9-5=4$ numbers must be covered by single-element parts. Here again, $\pi_1$ clearly satisfies the condition of claim 1.

Cases (1) and (2) cover all possible ways to form a partition $\pi_1$ of $N$. Since in both cases $\pi_1$ meets the condition stated in claim 1, it follows that it is not possible to have another partition $\pi_2$ of $N$ such that $\pi_1(x) \neq \pi_2(y)$ for all distinct $x, y \in N$. In other words, for any other partition $\pi_2$ of $N$, we can find distinct $x, y \in N$ such that $\pi(x) = \pi(y)$ in both $\pi_1$ and $\pi_2$. Hence proved.

 

[nuuskur]

The well known Maschke's theorem for Problem 4 :)

 

[fresh_42]

The well known Maschke's theorem for Problem 4 :)

But Maschke is easy enough to prove, fulfills the requirement to learn something, and last but not least doesn't provide a counterexample for $\operatorname{char}\mathbb{F}\,\mid\,|G| .$

 

[nuuskur]

IIRc the result is the group algebra is semisimple if and only if \mathrm{char}F \not\mid |G|.

 

[fresh_42]

IIRc the result is the group algebra is semisimple if and only if \mathrm{char}F \not\mid |G|.

Sure. Otherwise, there wouldn't be a counterexample.

 

[Office_Shredder]

It's funny, number 3 is just a continuous version of number 2 (if your random variables take discrete values with probabilities that can be expressed as integer ratios as each other, then #3 turns into #2 I think) but somehow, #3 seemed obviously true to me and #2 seemed like a crazy result when I first read it.Anyway, here's a proof for #3.

Edit: ahh my dog posted this when I started typing. I'll remove this when I finish writing the proof.

Spoiler

It suffices to show for $r>1$, that $P(r-1 < |X-Y| \leq r) \leq 2P(|X-Y| \leq 1)$, since then
$$P(|X-Y| \leq r) = P(| X-Y| \leq 1) + \sum_{k=2}^r P(r-1 < |X-Y| \leq r) \leq P(| X-Y| \leq 1) + 2(r-1)P(| X-Y| \leq 1).$$

 

[fresh_42]

It's funny, number 3 is just a continuous version of number 2 (if your random variables take discrete values with probabilities that can be expressed as integer ratios as each other, then #3 turns into #2 I think) but somehow, #3 seemed obviously true to me and #2 seemed like a crazy result when I first read it.Anyway, here's a proof for #3.

Edit: ahh my dog posted this when I started typing. I'll remove this when I finish writing the proof.

Spoiler

It suffices to show for $r>1$, that $P(r-1 < |X-Y| \leq r) \leq 2P(|X-Y| \leq 1)$, since then
$$P(|X-Y| \leq r) = P(| X-Y| \leq 1) + \sum_{k=2}^r P(r-1 < |X-Y| \leq r) \leq P(| X-Y| \leq 1) + 2(r-1)P(| X-Y| \leq 1).$$

Yes, and it's a bit short. I post my solution which is a bit more detailed.

Fix an integer $m,$ and let $S:=(x_1,\ldots,x_m)$ be a random sequence of $m$ elements, where each $x_i$ is chosen, randomly and independently, according to the distribution of $X.$ By the previous statement (problem #2)
$$
|S_r| <(2r-1)|S_1|.
$$
Therefore, the expectation of $|S_r|$ is smaller than that of $(2r-1)|S_1|.$ However, by the linearity of expectation it follows that the expectation of $|S_b|$ is precisely $m+m(m-1)p_b$ for every $b>0.$ Thus
$$
m+m(m-1)p_r< (2r-1)(m+(m-1)p_1),
$$
implying that for every integer $m,$
$$
p_r< (2r-1)p_1+\dfrac{2r-2}{m-1} \stackrel{m\to \infty }{\longrightarrow } (2r-1)p_1
$$

It can be proven that even the strict inequality
$$\operatorname{prob}\left(|X-Y|\leq 2\right) < 3\cdot \operatorname{prob}\left(|X-Y|\leq 1\right)$$
holds, in which case we speak of the 1-2-3 theorem.

 

[julian]

Problem #8:

Spoiler

We have

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & = P \left( \sum_{k=1}^n X_k \geq na \right)
\nonumber \\
& = P \left( e^{t \sum_{k=1}^n X_k} \geq e^{tna} \right)
\nonumber \\
& \leq e^{- tna} E (e^{t \sum_{k=1}^n X_k}) \qquad \qquad (\text{Chernoff bound})
\nonumber \\
& = e^{- tna} \prod_{k=1}^n E (e^{t X_k}) \qquad (\text{independent random variables}) \quad (*)
\end{align*}

Let $E(X_k) = 0$, $E(X_k^2) = \sigma_k^2 \leq \sigma^2$ and that $| X_k | \leq B$. We have for $i > 2$ that $E (X_k^i) = E (X_k^{i-2} X_k^2) \leq B^{i-2} \sigma_k^2$, so that

\begin{align*}
R_k & = \sum_{i=2}^\infty \frac{t^{i-2} E(X_k^i)}{i! \sigma_k^2} \leq \sum_{i=2}^\infty \frac{t^{i-2} B^{i-2} \sigma_k^2}{i! \sigma_k^2}
\nonumber \\
& = \frac{1}{(t B)^2} \sum_{i=2}^\infty \frac{t^i B^i}{i!}
\nonumber \\
& = \frac{e^{tB} - 1 - tB}{(t B)^2}
\end{align*}

Then

\begin{align*}
E (e^{t X_k}) & = E \left( 1 + t X_k + \sum_{i=2}^\infty \frac{t^i X_k^i}{i!} \right)
\nonumber \\
& = 1 + t^2 \sigma_k^2 R_k
\nonumber \\
& \leq e^{t^2 \sigma_k^2 R_k}
\nonumber \\
& = \exp \left\{ t^2 \sigma_k^2 \frac{e^{tB} - 1 - tB}{(t B)^2} \right\}
\end{align*}

Using this in $(*)$

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & \leq e^{- tna} \prod_{k=1}^n \exp \left\{ t^2 \sigma_k^2 \frac{e^{tB} - 1 - tB}{(t B)^2} \right\}
\nonumber \\
& = e^{- tna} \exp \left\{ n t^2 \tilde{\sigma}^2 \frac{e^{tB} - 1 - tB}{(tb)^2}
\right\} \quad (**)
\end{align*}

(where we have defined $\tilde{\sigma}^2 = \frac{1}{n} \sum_{k=1}^n \sigma_k^2 \leq \sigma^2$). We minimise the RHS of $(**)$ with respect to $t$:

\begin{align*}
& \frac{d}{dt} \exp \left\{ n t^2 \tilde{\sigma}^2 \frac{e^{tB} - 1 - tB}{(tB)^2} - tna \right\}
\nonumber \\
& = \exp \left\{ n t^2 \tilde{\sigma}^2 \frac{e^{tB} - 1 - tB}{(tb)^2} - tna \right\} n (\tilde{\sigma}^2 \left( \frac{B e^{tB} - B}{B^2} \right) - a) = 0
\end{align*}

We have a tighter bound when $t = \frac{1}{B} \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right)$. Using this in $(**)$:

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & \leq \exp \left\{ n \tilde{\sigma}^2 \frac{ (\frac{aB}{\tilde{\sigma}^2} + 1) - 1 - \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right)}{B^2} - \frac{na}{B} \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right) \right\}
\nonumber \\
& = \exp \left\{ \frac{n \tilde{\sigma}^2}{B^2} \left[ \frac{aB}{\tilde{\sigma}^2} - \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right) - \frac{aB}{\tilde{\sigma}^2} \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right) \right] \right\}
\nonumber \\
& = \exp \left\{ - \frac{n \tilde{\sigma}^2}{B^2} f \left( \frac{aB}{\tilde{\sigma}^2} \right) \right\}
\end{align*}

where $f(u) = (1+ u) \log (1+ u) - u \geq 0$ for $u \geq 0$ (as $f(0) = 0$ and $f'(u) = \log (1 + u) \geq 0$). We can use the inequality $u - (1+u) \log (1+ u) \leq - \dfrac{3u^2}{2(u + 3)}$ ($u > -1$), corresponding to problem #7, to obtain

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & \leq \exp \left\{ - \frac{a^2}{\dfrac{2 \tilde{\sigma}^2}{n} + \dfrac{2Ba}{3n}}
\right\}
\nonumber \\
& \leq \exp \left\{ - \frac{a^2}{\dfrac{2 \sigma^2}{n} + \dfrac{2Ba}{3n}} \right\} \qquad (\sigma^2 \geq \tilde{\sigma}^2)
\end{align*}

Put

\begin{align*}
a = \sqrt{ \frac{2 \sigma^2 T}{n}} + \frac{2BT}{3n}
\end{align*}

then

\begin{align*}
a^2 = \frac{2 \sigma^2 T}{n} + 2 \sqrt{ \frac{2 \sigma^2}{n}} \frac{2B}{3n} T^{3/2} + \left( \frac{2B}{3n} \right)^2 T^2
\end{align*}

and

\begin{align*}
\frac{2 \sigma^2}{n} + \frac{2B}{3n} a = \frac{2 \sigma^2}{n} + \sqrt{ \frac{2 \sigma^2}{n}} \frac{2B}{3n} \sqrt{T} + \left( \frac{2B}{3n} \right)^2 T
\end{align*}

so that

\begin{align*}
- \frac{a^2}{\frac{2 \sigma^2}{n} + \frac{2B}{3n} a} \leq - \frac{a^2}{\frac{2 \sigma^2}{n} + \sqrt{ \frac{2 \sigma^2}{n}} \frac{2B}{3n} \sqrt{T} + \frac{2B}{3n} a} = - T
\end{align*}

So finally

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq \sqrt{ \frac{2 \sigma^2 T}{n}} + \frac{2BT}{3n} \right) \leq e^{-T} .
\end{align*}

 

[fresh_42]

Problem #8:

Spoiler

We have

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & = P \left( \sum_{k=1}^n X_k \geq na \right)
\nonumber \\
& = P \left( e^{t \sum_{k=1}^n X_k} \geq e^{tna} \right)
\nonumber \\
& \leq e^{- tna} E (e^{t \sum_{k=1}^n X_k}) \qquad \qquad (\text{Chernoff bound})
\nonumber \\
& = e^{- tna} \prod_{k=1}^n E (e^{t X_k}) \qquad (\text{independent random variables}) \quad (*)
\end{align*}

Let $E(X_k) = 0$, $E(X_k^2) = \sigma_k^2 \leq \sigma^2$ and that $| X_k | \leq B$. We have for $i > 2$ that $E (X_k^i) = E (X_k^{i-2} X_k^2) \leq B^{i-2} \sigma_k^2$, so that

\begin{align*}
R_k & = \sum_{i=2}^\infty \frac{t^{i-2} E(X_k^i)}{i! \sigma_k^2} \leq \sum_{i=2}^\infty \frac{t^{i-2} B^{i-2} \sigma_k^2}{i! \sigma_k^2}
\nonumber \\
& = \frac{1}{(t B)^2} \sum_{i=2}^\infty \frac{t^i B^i}{i!}
\nonumber \\
& = \frac{e^{tB} - 1 - tB}{(t B)^2}
\end{align*}

Then

\begin{align*}
E (e^{t X_k}) & = E \left( 1 + t X_k + \sum_{i=2}^\infty \frac{t^i X_k^i}{i!} \right)
\nonumber \\
& = 1 + t^2 \sigma_k^2 R_k
\nonumber \\
& \leq e^{t^2 \sigma_k^2 R_k}
\nonumber \\
& = \exp \left\{ t^2 \sigma_k^2 \frac{e^{tB} - 1 - tB}{(t B)^2} \right\}
\end{align*}

Using this in $(*)$

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & \leq e^{- tna} \prod_{k=1}^n \exp \left\{ t^2 \sigma_k^2 \frac{e^{tB} - 1 - tB}{(t B)^2} \right\}
\nonumber \\
& = e^{- tna} \exp \left\{ n t^2 \tilde{\sigma}^2 \frac{e^{tB} - 1 - tB}{(tb)^2}
\right\} \quad (**)
\end{align*}

(where we have defined $\tilde{\sigma}^2 = \frac{1}{n} \sum_{k=1}^n \sigma_k^2 \leq \sigma^2$). We minimise the RHS of $(**)$ with respect to $t$:

\begin{align*}
& \frac{d}{dt} \exp \left\{ n t^2 \tilde{\sigma}^2 \frac{e^{tB} - 1 - tB}{(tB)^2} - tna \right\}
\nonumber \\
& = \exp \left\{ n t^2 \tilde{\sigma}^2 \frac{e^{tB} - 1 - tB}{(tb)^2} - tna \right\} n (\tilde{\sigma}^2 \left( \frac{B e^{tB} - B}{B^2} \right) - a) = 0
\end{align*}

We have a tighter bound when $t = \frac{1}{B} \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right)$. Using this in $(**)$:

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & \leq \exp \left\{ n \tilde{\sigma}^2 \frac{ (\frac{aB}{\tilde{\sigma}^2} + 1) - 1 - \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right)}{B^2} - \frac{na}{B} \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right) \right\}
\nonumber \\
& = \exp \left\{ \frac{n \tilde{\sigma}^2}{B^2} \left[ \frac{aB}{\tilde{\sigma}^2} - \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right) - \frac{aB}{\tilde{\sigma}^2} \log \left( \frac{aB}{\tilde{\sigma}^2} + 1 \right) \right] \right\}
\nonumber \\
& = \exp \left\{ - \frac{n \tilde{\sigma}^2}{B^2} f \left( \frac{aB}{\tilde{\sigma}^2} \right) \right\}
\end{align*}

where $f(u) = (1+ u) \log (1+ u) - u \geq 0$ for $u \geq 0$ (as $f(0) = 0$ and $f'(u) = \log (1 + u) \geq 0$). We can use the inequality $u - (1+u) \log (1+ u) \leq - \dfrac{3u^2}{2(u + 3)}$ ($u > -1$), corresponding to problem #7, to obtain

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq a \right) & \leq \exp \left\{ - \frac{a^2}{\dfrac{2 \tilde{\sigma}^2}{n} + \dfrac{2Ba}{3n}}
\right\}
\nonumber \\
& \leq \exp \left\{ - \frac{a^2}{\dfrac{2 \sigma^2}{n} + \dfrac{2Ba}{3n}} \right\} \qquad (\sigma^2 \geq \tilde{\sigma}^2)
\end{align*}

Put

\begin{align*}
a = \sqrt{ \frac{2 \sigma^2 T}{n}} + \frac{2BT}{3n}
\end{align*}

then

\begin{align*}
a^2 = \frac{2 \sigma^2 T}{n} + 2 \sqrt{ \frac{2 \sigma^2}{n}} \frac{2B}{3n} T^{3/2} + \left( \frac{2B}{3n} \right)^2 T^2
\end{align*}

and

\begin{align*}
\frac{2 \sigma^2}{n} + \frac{2B}{3n} a = \frac{2 \sigma^2}{n} + \sqrt{ \frac{2 \sigma^2}{n}} \frac{2B}{3n} \sqrt{T} + \left( \frac{2B}{3n} \right)^2 T
\end{align*}

so that

\begin{align*}
- \frac{a^2}{\frac{2 \sigma^2}{n} + \frac{2B}{3n} a} \leq - \frac{a^2}{\frac{2 \sigma^2}{n} + \sqrt{ \frac{2 \sigma^2}{n}} \frac{2B}{3n} \sqrt{T} + \frac{2B}{3n} a} = - T
\end{align*}

So finally

\begin{align*}
P \left( \frac{1}{n} \sum_{k=1}^n X_k \geq \sqrt{ \frac{2 \sigma^2 T}{n}} + \frac{2BT}{3n} \right) \leq e^{-T} .
\end{align*}

Very good!

I knew the Chernoff bound as Markov inequality ...

The Markov inequality says that for a monotone increasing function $f: \mathbb{R}\longrightarrow [0,\infty )$ and a constant $a\in \mathbb{R}$
$$
f(a)\cdot P(X\geq a) \leq E(f(X))
$$
and in particular for $f(a)>0$
$$
P(X\geq a) \leq \dfrac{E(f(X))}{f(a)}.
$$

... but I also thought nobody would solve this one. The name of the inequality in the statement (#8) is Bernstein inequality.

 

[julian]

It helped that I solved problem #7 first. I thought problem #7 was a bit easy, but now I realize it was setting you up to solve problem #8.