Optimizing stakes in simultaneous bets

We follow the proof of [2, Lemma 1]. The following Paley-Zygmund type inequality for random variables of zero mean was proved in [13, lemma 2.2] and extended in [14]:

Applying this to $S_{\gamma}-p$ we have

The second moment of $S_{\gamma}-p$ is equal to $p(1-p)\sum c_{i}^{2}$ and the fourth moment is equal to

This can be bounded by

The Paley-Zygmund type inequality produces a lower bound on $\mathbf{P}(S_{\gamma}<p)$. Its complementary probability $\pi(p,p)$ is bounded by:

It is possible to improve on this bound for small $p$ by using Feige’s theorem. We write $S_{\gamma}=c_{1}\beta_{1}+S$. Then $\mathbf{P}(S_{\gamma}<p)\geq\mathbf{P}(\beta_{1}=0)\mathbf{P}(S<p)=(1-p)\mathbf{P}(S<p)$. Note that $\mathbf{E}[S]=p(1-c_{1})$ and we write $\mathbf{P}(S<p)=\mathbf{P}(S<\mathbf{E}[S]+pc_{1})$. We can approximate this probability by a truncated sum $\mathbf{P}(S_{n}<\mathbf{E}[S]+pc_{1})$, where $S_{n}=c_{2}\beta_{2}+\cdots+c_{n}\beta_{n}$ is a sum of independent random variables of expectation $\leq pc_{1}$. By dividing by $pc_{1}$ and applying the bound $0.1798$ of [12] we find

We have two upper bounds. The first is more restrictive for large $p$ and the second is more restrictive for small $p$. The lower bound follows from bold play. Let $k\in\mathbb{N}$ be such that $\frac{1}{k+1}<p\leq\frac{1}{k}$. If $\bar{S}_{k}$ is the average of $k\geq 1$ Bernoullis, then $\mathbf{P}(\bar{S}_{k}\geq p)=\mathbf{P}(\bar{S}_{k}\geq\frac{1}{k})=1-(1-p)^{k}>1-(1-\frac{1}{k+1})^{k}$. This is minimal and equal to $\frac{1}{2}$ if $k=1$. ∎

According to Jessen and Wintner’s law of pure type [3, Theorem 3.5], either $\mathbf{P}(S_{\gamma}=s)=0$ for each $s\in\mathbb{R}$ or there exists a countable set $\mathcal{C}$ such that $\mathbf{P}(S_{\gamma}\in\mathcal{C})=1$. In other words, the random variable $S_{\gamma}$ is either non-atomic or discrete. If $X$ and $Y$ are independent, and if $X$ is non-atomic, then the convolution formula implies that $X+Y$ is non-atomic.

Suppose that $\gamma$ is infinite. We prove that $S_{\gamma}$ is non-atomic. Let $(c_{i_{j}})$ be a subsequence such that $c_{i_{j}}>2\sum_{k=j+1}^{\infty}c_{i_{k}}$. Let $I$ be the set of all $i_{j}$ and let $J$ be its complement. Then both $S_{I}=\sum_{I}c_{i}\beta_{i}$ and $S_{J}=\sum_{J}c_{i}\beta_{i}$ are either discrete or non-atomic. By our choice of $c_{i_{j}}$, $S_{I}$ has the property that its value determines the values of all $\beta_{i}$ for $i\in I$. This implies that $S_{I}$ is non-atomic. Therefore, $S_{\gamma}=S_{I}+S_{J}$ is non-atomic. In particular $\mathbf{P}(S_{\gamma}\geq t)=\mathbf{P}(S_{\gamma}>t)$.

Denote a truncated sum by $S_{\gamma,n}=\sum_{i\leq n}c_{i}\beta_{i}$. By monotonic convergence $\mathbf{P}\left(S_{\gamma}>t\right)=\lim_{n\to\infty}\mathbf{P}(S_{\gamma,n}>t)$. Therefore, for any infinite $\gamma$, $\mathbf{P}(S_{\gamma}\geq t)$ can be approximated by tail probabilities of finite convex combinations. ∎

We write $\pi(p,t^{-})=\lim_{s\uparrow t}\pi(p,s)$. Since $\pi(p,t)$ is decreasing in $t$, it suffices to show that there exists an $S_{\gamma}$ such that $\mathbf{P}(S_{\gamma}\geq t)\geq\pi(p,t^{-})$. Let $\gamma_{n}=(c_{n,i})_{i}$ be such that $\mathbf{P}(S_{\gamma_{n}}\geq t_{n})$ converges to $\pi(p,t^{-})$ for an increasing sequence $t_{n}\uparrow t$. By a standard diagonal argument we can assume that $(c_{n,i})_{n}$ is convergent for all $i$. Let $c_{i}=\lim_{n\to\infty}c_{n,i}$ and let $\gamma=(c_{i})_{i}$. Then $\gamma$ is a non-increasing sequence which adds up to $\sum c_{i}=1-c\leq 1$. Observe that $\gamma$ cannot be the all zero sequence, since this would imply that $c_{n,1}\to 0$ and $\mathrm{Var}(S_{\gamma_{n}})=p(1-p)\sum c_{n,i}^{2}\leq p(1-p)c_{n,1}\to 0$, so $S_{\gamma_{n}}$ converges to $p$ in distribution. Since we limit ourselves to $p<t$, this means that $\mathbf{P}(S_{\gamma_{n}}\geq t)\to 0$ which is nonsense. Therefore, $1-c>0$.

We first prove that $\pi(p,t^{-})\leq\mathbf{P}(S_{\gamma}\geq t-cp)$. Fix an arbitrary $\epsilon>0$. Let $i_{0}$ be such that $\sum_{j\geq i_{0}}c_{j}<\frac{\epsilon}{4}$ and $c_{i_{0}}<\epsilon^{4}$. Let $n_{0}$ be such that $\sum_{j\leq i_{0}}|c_{n,j}-c_{j}|<\frac{\epsilon}{4}$ and $c_{n,i_{0}}<\epsilon^{4}$ for all $n\geq n_{0}$. Now

so by our assumptions

where we write $T_{n}=\sum_{j\geq i_{0}}c_{n,j}\beta_{j}$. Observe that

and

By Chebyshev’s inequality, we conclude that $\mathbf{P}\left(T_{n}\geq cp+\epsilon\right)<\epsilon$ for sufficiently small $\epsilon$. It follows that

By taking limits $n\to\infty$ and $\epsilon\to 0$ we conclude that

Let $\bar{\gamma}=\frac{1}{1-c}\gamma$. Then $S_{\bar{\gamma}}=\frac{1}{1-c}S_{\gamma}$ is a convex combination such that

Therefore, $\pi(p,t)=\mathbf{P}(S_{\bar{\gamma}}\geq t)$ and these inequalities are equalities. ∎

For any $\epsilon>0$ choose a finite $\gamma$ such that $\mathbf{P}(S_{\gamma}^{p}\geq t)\geq\pi(p,t)-\epsilon$. If $p_{n}$ converges to $p$ then $\beta^{p_{n}}$ converges to $\beta^{p}$ in probability. Since $\gamma$ is finite

It follows that $\limsup_{n\to\infty}\pi(p_{n},t)\geq\pi(p,t)$ for any sequence $p_{n}\to p$. Since $\pi(p,t)$ is increasing in $p$, it follows that $\pi(p,t)$ is left-continuous in $p$.

We need to prove right continuity, i.e., $\pi(p^{+},t)=\pi(p,t)$. This is trivially true on the hypothenuse, because this is the right-hand boundary of the domain. Consider $p<t$. Let $p_{n}\downarrow p$ and $\gamma_{n}$ be such that $\lim_{n\to\infty}\mathbf{P}(S_{\gamma_{n}}^{p_{n}}\geq t)=\pi(p^{+},t)$. By a standard diagonal argument we can assume that $\gamma_{n}$ converges coordinatewise to some $\gamma$, which may not sum up to one. It cannot be the all zero sequence, i.e., the sum is not zero, by the same argument as in the proof of theorem 7. The sequence $\gamma$ therefore sums up to $1-c$ for some $0\leq c<1$. Again, we split $S^{p}_{\gamma}=H+T$ where $H=\sum_{j\leq i_{0}}c_{j}\beta_{j}^{p}$ and $T=\sum_{j>i_{0}}c_{j}\beta_{j}^{p}$. We choose $i_{0}$ such that $\mathbf{E}[T]<\frac{\epsilon}{4}$ and $c_{i_{0}}<\epsilon^{4}$. Similarly, $S_{\gamma_{n}}=H_{n}+T_{n}$ where $H_{n}$ converges to $H$ in probability, $\mathbf{E}[T_{n}]<pc+\frac{\epsilon}{2}$ and $\mathrm{Var}(T_{n})<\epsilon^{4}$ for sufficiently large $n$. As in the previous proof, Chebyshev’s inequality and convergence in probability imply

for sufficiently large $n$. By taking limits $n\to\infty$ and $\epsilon\to 0$ it follows that $\pi(p^{+},t)\leq\mathbf{P}(S_{\gamma}^{p}\geq t-cp)$. If we standardize $\gamma$ to a sequence $\bar{\gamma}$ so that we get a convex combination, we again find that $\pi(p^{+},t)\leq\mathbf{P}(S_{\bar{\gamma}}^{p}\geq t)$. ∎

$\bigcup\mathcal{F}$ is a union of say $c\geq 1$ separate intervals, all of lengths $\geq a$. Any interval of length $b\geq a$ contains $b-(a-1)$ intervals of length $a$. Therefore, $\bigcup\mathcal{F}$ contains $|\bigcup\mathcal{F}|-c(a-1)$ intervals of length $a$. It follows that $|\bigcup\mathcal{F}|-c(a-1)\geq k$. ∎

Let $I=[b,b+k)$ be any element in $\mathcal{F}$. An interval $[c,c+a)$ intersects $I$ if and only if $c\in[b-a+1,b+k)$, which is an interval of length $k+a-1$. Therefore, the set $\mathcal{I}$ of initial elements $c$ of intervals in $\mathcal{G}$ is contained in an intersection of $k$ intervals of length $k+a-1$. The complement of $\mathcal{I}$ thus contains a union of $k$ intervals of length $n-k-a+1$. By the previous lemma, this union has cardinality $\geq n-a$. Therefore, $\mathcal{I}$ contains at most $a$ elements. ∎

∎

Bold play gives a probability $p$ of reaching $t$. We need to prove that $\mathbf{P}(S_{\gamma}\geq t)\leq p$ for arbitrary $\gamma$. By proposition 6 we may assume that $\gamma$ is finite. It suffices to prove that $\mathbf{P}(S_{\gamma}\geq t)\leq p$ for rational $p$, since $\pi(p,t)$ is monotonic in $p$.

Let $n$ be the number of non-zero $c_{i}$ in $\gamma$ and let $p=\frac{a}{b}$. Let $X_{i}$ be a sequence of $n$ independent discrete uniform $U\{0,b-1\}$ random variables, i.e, $X_{i}=c$ for $c\in\{0,\ldots,b-1\}$ with probability $\frac{1}{b}$. Let $B_{i}^{0}=1_{[0,a)}(X_{i})$ for $1\leq i\leq n$. Then $S_{\gamma}$ and $Y^{0}=\sum c_{i}B_{i}^{0}$ are identically distributed. Think of $c_{i}B_{i}^{0}$ as an assignment of weight $c_{i}$ to a random element in $\{0,\ldots,b-1\}=\mathbb{Z}/b\mathbb{Z}$. Let $\ell(j)$ be the sum of the coefficients – the load – that is assigned to $j\in\mathbb{Z}/b\mathbb{Z}$. Then $Y^{0}=\ell(0)+\cdots+\ell(a-1)$, i.e., $Y^{0}$ is the load of $[0,a)$. Instead of $[0,a)$ we might as well select any interval $[j,j+a)\subset\mathbb{Z}/b\mathbb{Z}$. If $Y^{j}$ is the load of $[j,j+a)$, then $S_{\gamma}\sim Y^{j}$, and $\mathbf{P}(S_{\gamma}\geq t)=\frac{1}{b}\sum\mathbf{P}(Y^{j}\geq t)$. We need to prove that $\sum\mathbf{P}(Y^{j}\geq t)\leq a$.

Let $\Omega$ be the sample space of the $X_{i}$. For $\omega\in\Omega$, let $J(\omega)$ be the cardinality of $\mathcal{J}(\omega)=\{j\colon Y^{j}(\omega)\geq t\}\subset\mathbb{Z}/b\mathbb{Z}$. In particular, $\mathbf{P}(S_{\gamma}\geq t)=\frac{1}{b}\mathbf{E}[J]$. It suffices to prove that $J\leq a$. Assume that $J(\omega)\geq a$ for some $\omega\in\Omega$. Apply lemma 11 to the counting measure to find

Note that $i\in\mathcal{J}(\omega)-j$ if and only if $[i+j,i+j+a)$ has load $\geq t$. Therefore, there are at least $(k+1)a-kb$ elements $i$ such that the intervals $[i,i+a),[i+a,i+2a),\ldots,[i+ka,i+(k+1)a)$ all have load $\geq t$. The intersection of these $k+1$ intervals is equal to

It has load $\geq(k+1)t-k$ by lemma 11. Its complement $I_{i}^{c}$ has load $\leq k+1-(k+1)t<t$. If $j\in\mathcal{J}(\omega)$ then $[j,j+a)$ has load $\geq t$ and therefore it intersects $I_{i}$ for all $i\in\bigcap_{l=0}^{k}(J(\omega)-la)$. There are $\geq(k+1)a-kb$ such intervals $I_{i}$, and we denote this family by $\mathcal{F}$. Let $\mathcal{G}$ be the family of $[j,j+a)$ with $j\in J(\omega)$. Lemma 10 applies since the length of $I_{i}$ is $(k+1)a-kb$ and since $a\leq b-\left((k+1)a-kb\right)$. We conclude that $J(\omega)\leq a$. ∎

By proposition 6 it suffices to prove that $\mathbf{P}(S_{\gamma}\geq t)\leq p$ for finite $\gamma$. We adopt the notation of the previous theorem. Let $n$ be the number of non-zero coefficients in $\gamma$, and let $X_{i}$ be $n$ random selections of $\{0,\ldots,k+1\}$. We assign the coefficients according to these selections and let $Y^{j}=1-\ell(j)$ be the load of the set $\{0,\ldots,k+1\}\setminus\{j\}$. Each $Y^{j}$ is identically distributed to $S_{\gamma}$. For $\omega\in\Omega$ let $J(\omega)$ be the number of $Y^{j}(\omega)$ that reach the threshold, or equivalently, the number of loads $\ell(j)\leq\frac{1}{k+2}$. We have $\frac{1}{k+2}\mathbf{E}[J]=\mathbf{P}(S_{\gamma}\geq t)$. In the proof above, we showed that $J\leq k+1$ if $t>p=\frac{k+1}{k+2}$. This is no longer true now that we have $t=p$. It may happen that $J(\omega)=k+2$ in which case all $Y^{j}(\omega)$ are equal to $\frac{k+1}{k+2}$ and all loads $\ell(j)$ are equal to $\frac{1}{k+2}$. Note that this can only happen if all $c_{i}$ are bounded by $\frac{1}{k+2}$, so $n\geq k+2$.

We think of the coefficients as being assigned one by one in increasing order. In particular, $c_{n-1}$ and $c_{n}$ are placed last. If $J=k+2$, then either $k$ or $k+1$ of the loads are equal to $\frac{1}{k+2}$ before $c_{n-1}$ and $c_{n}$ are placed. In the first case, there are two remaining loads $<\frac{1}{k+2}$ and the probability that $c_{n-1}$ are $c_{n}$ are placed here is $\frac{2}{(k+2)^{2}}$. In the second case, there is only one remaining load $<\frac{1}{k+2}$ and the probability that $c_{n-1}$ and $c_{n}$ are placed here is $\frac{1}{(k+2)^{2}}$. We conclude that $\mathbf{P}(J=k+2)\leq\frac{2}{(k+2)^{2}}$ and therefore

is bounded by $k+1+\frac{2}{(k+2)^{2}}$. Thus we obtain $\mathbf{P}(S_{\gamma}\geq t)\leq\frac{k+1}{k+2}+\frac{2}{(k+2)^{3}}$. This bound is reached if $k=1$ and $c_{1}=c_{2}=c_{3}=\frac{1}{3}$.

Let $k>1$ and let $J(\omega)=k+2$. We first consider the case that $c_{n-1}\not=c_{n}$. suppose there are two remaining loads $<\frac{1}{k+2}$ before $c_{n-1}$ and $c_{n-2}$ are placed, then each $c_{n-1}$ and $c_{n}$ can only be assigned to a unique place to complete all loads to $\frac{1}{k+2}$. Since $k>1$, there are at least two loads $\ell(i_{1})=\ell(i_{2})=\frac{1}{k+2}$ before the final two coefficients are placed. Let $\bar{\omega}$ assign $c_{j}$ for $j<n-1$ in the same way as $\omega$, but it reassigns $c_{n-1}$ to $i_{1}$ and $c_{n}$ to $i_{2}$. Then $J(\bar{\omega})=k$, because the loads at $i_{1}$ and $i_{2}$ exceed the threshold. We can reconstruct $\omega$ from $\bar{\omega}$ because the loads at $i_{1}$ and $i_{2}$ are the only ones that exceed the threshold for $\bar{\omega}$, and their values are different because $c_{n-1}\not=c_{n}$. We have a $1-1$ correspondence between $\omega\in\{J=k+2\}$ and $\bar{\omega}\in\{J=k\}$. Let $\mathcal{E}=\{J=k+2\}$ and let $\mathcal{F}=\{\bar{\omega}\colon\omega\in\mathcal{E}\}$. Then $\mathbf{P}(\mathcal{F})=\mathbf{P}(\mathcal{E})$ and $\mathcal{E}\cap\mathcal{F}=\emptyset$. This implies that

In particular $\mathbf{P}(S_{\gamma}\geq t)\leq\frac{k+1}{k+2}$ and bold play is optimal.

Finally, consider the remaining case $k>1$ and $J(\omega)=k+2$ and $c_{n-1}=c_{n}$. In this case, we may switch the assignments of $c_{n-1}$ and $c_{n}$ to complete the loads. Let $\omega^{\prime}\in\Omega$ represent this switch (it may be equal to $\omega$ if the assignments are the same). Again, let $i_{1}$ and $i_{2}$ be two locations for which the loads have already been completed before $c_{n-1}$ and $c_{n}$ are placed. Let $\{\bar{\omega},\bar{\omega}^{\prime}\}$ be the elements which assign the first $n-2$ coefficients in the same way, but assigns the final two elements to $i_{1}$ and $i_{2}$. In particular, $J(\bar{\omega})=J(\bar{\omega}^{\prime})=k$. We can reconstruct $\{\omega,\omega^{\prime}\}$ from $\{\bar{\omega},\bar{\omega}^{\prime}\}$, the correspondence is injective, so again $\mathbf{P}(\mathcal{E})=\mathbf{P}(\mathcal{F})$ and we conclude in the same way that bold play is optimal. ∎

We randomly distribute the coefficients of a finite $\gamma$ over $2k+3$ locations. Let $Y^{j}=\ell(j)+\ldots+\ell(2k+j)$ be the load of the discrete interval $[j,2k+j+1)$, where as before we reduce modulo $2k+3$. Then $S_{\gamma}\sim Y^{j}$ and the sum of all $Y^{j}$ is equal to $2k+1$. Let $J$ be the number of $Y^{j}$ that reach the threshold. Not all $Y^{j}$ can reach the threshold and therefore $J\leq 2k+2$. Then

We need to prove that $\mathbf{P}(S_{\gamma}\geq t)\leq\frac{2k+1}{2k+3}$. If $\mathbf{P}(J=2k+2)=0$ then we are done. Therefore, we may assume that $\mathbf{P}(J=2k+2)>0$. Only one of the $Y^{j}$ does not meet the threshold and without loss of generality we may assume it is $Y^{2}$, which has load $1-\ell(0)-\ell(1)$. The other $Y^{j}$ reach the threshold, and since the sum of all $Y^{j}$ is equal to $2k+1$, we find that