#### Proof

We begin by observing that the exchangeability of players implies

$$p({n}_{l}\mid L;n,N)=p({n}_{l}\mid {\text{Top}}_{n,N},L;n).$$(34)

Integrating (34) over the handedness of the top *n* players yields

$$\begin{array}{ccccc}& \underset{N\to \mathrm{\infty}}{lim}p({n}_{l}\mid L;n,N)\hfill & & & \\ & =\underset{N\to \mathrm{\infty}}{lim}\sum _{{h}_{1:n}\in {\{0,1\}}^{n}}p({n}_{l},{H}_{1:n}={h}_{1:n}\mid {\text{Top}}_{n,N},L;n)\hfill & & & \\ & =\sum _{{h}_{1:n}\in {\{0,1\}}^{n}:{\sum}_{i=1}^{n}{h}_{i}={n}_{l}}\underset{N\to \mathrm{\infty}}{lim}p({H}_{1:n}={h}_{1:n}\mid {\text{Top}}_{n,N},L;n).\hfill & & & \end{array}$$(35)

We can expand each term in the summation by conditioning on the top players’ skills,

$$\begin{array}{ccccc}& \underset{N\to \mathrm{\infty}}{lim}p({H}_{1:n}={h}_{1:n}\mid {\text{Top}}_{n,N},L;n)\hfill & & & \\ & =\underset{N\to \mathrm{\infty}}{lim}{\int}_{{\mathbb{R}}^{n}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},{\text{Top}}_{n,N},L)\hfill & & & \\ & p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & & \\ & =\underset{N\to \mathrm{\infty}}{lim}{\int}_{{\mathbb{R}}^{n}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)\hfill & & & \\ & p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}.\hfill & & & \end{array}$$(36)

Most of the proof will focus on showing that the r.h.s. of (36) equals $\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}$ where $\alpha \left(L\right):=q/\left(q+\left(1-q\right)c\left(L\right)\right)$, ${n}_{r}:=n-{n}_{l}$ is the number of top right-handers and $c\left(L\right)$ is the tail-length of the innate skill distribution as defined in the statement of the proposition. We will write ${S}_{\left[i\right]}$ for the *i*^{th} order statistic of the skills ${S}_{1:N}$ and write ${H}_{\left[i\right]}$ for the corresponding induced order statistic. Given that the innate skills have support ℝ we have $F(k\mid L)<1$ for all $k\in \mathbb{R}$, where *F* denotes the conditional CDF of a player’s total skill given *L*. Note that for any values of *k*, *L* and *ϵ* > 0, we can find ${N}_{k,L,\u03f5}\in \mathbb{N}$ such that ${N}^{n}F{(k\mid L)}^{N-n}<\u03f5$ for all $N\ge {N}_{k,L,\u03f5}$. It therefore follows that for such *k*, *L*, *ϵ* and *N* we have

$$\begin{array}{ccccc}& {\int}_{{S}_{\left[n\right]}\le k}p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & & \\ & ={\int}_{{S}_{\left[n\right]}\le k}\left(\genfrac{}{}{0pt}{}{N}{n}\right)F{({S}_{\left[n\right]}\mid L)}^{N-n}\prod _{i=1}^{n}f({S}_{i}\mid L)d{S}_{1:n}\hfill & & & \\ & \le \left(\genfrac{}{}{0pt}{}{N}{n}\right)F{(k\mid L)}^{N-n}{\int}_{{\mathbb{R}}^{n}}\prod _{i=1}^{n}f({S}_{i}\mid L)d{S}_{1:n}\hfill & & & \\ & \le {N}^{n}F{(k\mid L)}^{N-n}\hfill & & & \\ & <\u03f5\hfill & & & \end{array}$$(37)

where *f* denotes the PDF of *F*. The conditional distribution of player handedness in (36) factorizes as

$$p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)=\prod _{i=1}^{n}p({H}_{i}={h}_{i}\mid {S}_{i},L)$$(38)

and consider now the *i*^{th} term in this product. We have

$$\begin{array}{ccccc}p({H}_{i}=1\mid {S}_{i}=s,L)\hfill & =\frac{p({S}_{i}=s\mid {H}_{i}=1,L)p({H}_{i}=1\mid L)}{p({S}_{i}=s\mid L)}\hfill & & & \\ & =\frac{g\left(s-L\right)q}{qg\left(s-L\right)+\left(1-q\right)g\left(s\right)}\hfill & & & \\ & =\frac{q}{q+\left(1-q\right)\frac{g\left(s\right)}{g\left(s-L\right)}}\hfill & & & \end{array}$$(39)

Assuming $c\left(L\right):={lim}_{s\to \mathrm{\infty}}\frac{g\left(s\right)}{g\left(s-L\right)}$ exists as stated in the proposition we can take limits across (39) to obtain

$$\underset{s\to \mathrm{\infty}}{lim}p({H}_{i}=1\mid {S}_{i}=s,L)=\frac{q}{q+\left(1-q\right)c\left(L\right)}$$

which we recognize as $\alpha \left(L\right)$ which we defined above. A similar argument for the case *H*_{i} = 0 yields ${lim}_{s\to \mathrm{\infty}}p({H}_{i}=0\mid {S}_{i}=s,L)=1-\alpha \left(L\right)$. Since the limit of a finite product of functions, each having a finite limit, is equal to the product of the limits, it therefore follows that for all *ϵ* > 0 there exists a ${k}_{\u03f5}\in \mathbb{R}$ such that if ${s}_{i}>{k}_{\u03f5}$ for $i=1,\mathrm{\dots},n$ then

$$\begin{array}{ccccc}\u03f5\hfill & >|p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n}={s}_{1:n},L)\hfill & & & \\ & -\prod _{i=1}^{n}\alpha {\left(L\right)}^{{h}_{i}}{(1-\alpha \left(L\right))}^{1-{h}_{i}}|\hfill & & & \\ & =|p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n}={s}_{1:n},L)-\alpha {\left(L\right)}^{{n}_{l}}{(1-\alpha \left(L\right))}^{{n}_{r}}|.\hfill & & & \end{array}$$(40)

We are now in a position to prove that

$$\begin{array}{ccccc}& \underset{N\to \mathrm{\infty}}{lim}{\int}_{{\mathbb{R}}^{n}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & & \\ & =\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}.\hfill & & & \end{array}$$(41)

For any *ϵ* > 0, for all $N>{N}_{{k}_{\u03f5/3},L,\u03f5/3}$ we have

$$\begin{array}{ccccc}& |\alpha {\left(L\right)}^{{n}_{l}}{(1-\alpha \left(L\right))}^{{n}_{r}}\hfill & & & \\ & -{\int}_{{\mathbb{R}}^{n}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}|\hfill & & & \\ & \le |\alpha {\left(L\right)}^{{n}_{l}}{(1-\alpha \left(L\right))}^{{n}_{r}}-{\int}_{{S}_{\left[n\right]}\ge {k}_{\u03f5/3}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)\hfill & & & \\ & p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}|\hfill & & & \\ & +{\int}_{{S}_{\left[n\right]}\le {k}_{\u03f5/3}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)\hfill & & & \\ & p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}.\hfill & & & \end{array}$$(42)

Observe that

$$\begin{array}{ccccc}& {\int}_{{S}_{\left[n\right]}\ge {k}_{\u03f5/3}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & & \\ & \le {\int}_{{S}_{\left[n\right]}\ge {k}_{\u03f5/3}}\left(\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}+\frac{\u03f5}{3}\right)\hfill & & & \\ & p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & & \\ & \le \alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}+\frac{\u03f5}{3}\hfill & & & \end{array}$$(43)

where the first inequality follows from (40). Similarly, its minimum value is bounded by

$$\begin{array}{ccccc}& & {\int}_{{S}_{\left[n\right]}\ge {k}_{\u03f5/3}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & \\ & & \ge {\int}_{{S}_{\left[n\right]}\ge {k}_{\u03f5/3}}\left(\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}-\frac{\u03f5}{3}\right)\hfill & & \\ & & p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & \\ & & =\left(\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}-\frac{\u03f5}{3}\right)\hfill & & \\ & & (1-{\int}_{{S}_{\left[n\right]}\le {k}_{\u03f5/3}}p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n})\hfill & & \\ & & \ge \left(\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}-\frac{\u03f5}{3}\right)\left(1-\frac{\u03f5}{3}\right)\hfill & & \\ & & \ge \left(\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}-\frac{\u03f5}{3}\right)-\frac{\u03f5}{3}\left(1-\frac{\u03f5}{3}\right)\hfill & & \\ & & \ge \alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}-\frac{2}{3}\u03f5\hfill & & \end{array}$$(44)

where the first inequality follows from (40) and the second inequality follows from (37). Combining the upper and lower bounds of (43) and (44) implies that the first term on the r.h.s. of (42) is bounded above by $2\u03f5/3$. The second term on the r.h.s. of (42) satisfies

$$\begin{array}{ccccc}& {\int}_{{S}_{\left[n\right]}\le {k}_{\u03f5/3}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\hfill & & & \\ & \le {\int}_{{S}_{\left[n\right]}\le {k}_{\u03f5/3}}p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}\le \u03f5/3.\hfill & & & \end{array}$$

We can therefore rewrite the bound in (42) as

$$\begin{array}{ccccc}& |\alpha {\left(L\right)}^{{n}_{l}}{(1-\alpha \left(L\right))}^{{n}_{r}}-{\int}_{{\mathbb{R}}^{n}}p({H}_{1:n}={h}_{1:n}\mid {S}_{1:n},L)\hfill & & & \\ & p({S}_{1:n}\mid {\text{Top}}_{n,N},L)d{S}_{1:n}|\hfill & & & \\ & \le \frac{2}{3}\u03f5+\frac{1}{3}\u03f5=\u03f5\hfill & & & \end{array}$$

completing the proof of (41). Substituting (41) into (36) and (36) into (35) yields

$$\begin{array}{ccccc}& \underset{N\to \mathrm{\infty}}{lim}p({n}_{l}\mid L;n,N)\hfill & & & \\ & =\sum _{{h}_{1:n}\in {\{0,1\}}^{n}:{\sum}_{i=1}^{n}{h}_{i}={n}_{l}}\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}\hfill & & & \\ & =\left(\genfrac{}{}{0pt}{}{n}{{n}_{l}}\right)\alpha {\left(L\right)}^{{n}_{l}}{\left(1-\alpha \left(L\right)\right)}^{{n}_{r}}\hfill & & & \end{array}$$

so the limiting distribution is $\text{Binomial}({n}_{l};\alpha \left(L\right),n)$ as desired. □

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