Use the positive exponent in the PDF. The TeX introduces a minus sign in the numerator while retaining the positive-sign normalizing constant, which would not define the stated <probability distribution>. The <Curie–Weiss model> assigns equal weights to $s$ and $-s$, since $(\sum_js_j)^2$ is unchanged by simultaneous reversal. This <spin inversion symmetry> pairs every configuration with one having the opposite value of $S_i$, so
$$
\boxed{\mathbb ES_i=0,\qquad\mathbb P(S_i=1)=\mathbb P(S_i=-1)=\frac12.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 11F / a / Solution

Let $C_{12}=\mathbb E[S_1S_2]\geq0$, using the supplied correlation property of the <Curie–Weiss model>. Since $\mathbf1_{\{S_i=1\}}=(1+S_i)/2$ and each spin has zero <expected value>,
$$
\mathbb P(S_1=1,S_2=1)
=\frac14\mathbb E[(1+S_1)(1+S_2)]=\frac{1+C_{12}}4.
$$
Divide by $\mathbb P(S_1=1)=1/2$ to obtain
$$
\boxed{\mathbb P(S_2=1\mid S_1=1)=\frac{1+C_{12}}2
\geq\frac12=\mathbb P(S_2=1).}
$$
The expression assumes $n\geq2$, as required for $S_2$ to exist. It makes the connection between the spin correlation and the requested <conditional probability> explicit.

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 11F / b / Solution

If exactly $k$ spins equal $+1$, then $M=(2k-n)/n$. Therefore its possible values are $E_n$, with $k=n(1+m)/2$ for $m\in E_n$. Choosing the $k$ positive spins gives the <binomial coefficient>
$$
\binom nk=\frac{n!}{[n(1+m)/2]!\,[n(1-m)/2]!}.
$$
Every such configuration has the same <Curie–Weiss model> weight because $\sum_{i,j}s_is_j=(\sum_is_i)^2=n^2m^2$. The <probability mass function> of this <spin magnetization> is consequently
$$
\boxed{\mathbb P(M=m)=\frac1{Z_{n,\beta}}
\binom{n}{n(1+m)/2}e^{\beta nm^2/2},\qquad m\in E_n,}
$$
and zero elsewhere. Grouping the original <partition function> by the same count also gives
$$
Z_{n,\beta}=\sum_{k=0}^n\binom nk
\exp\!\left[\frac{\beta(2k-n)^2}{2n}\right],
$$
which verifies normalization without counting any configuration twice.

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 11F / c / Solution

{c}
{title2=$\mathbb P(s)=Z_{n,\beta}^{-1}e^{\beta(\sum_i s_i)^2/(2n)}$}

The zero-field ferromagnetic Curie–Weiss model is an <Ising model> on $n$ spins with an equal interaction between every pair. For $s_i\in\{-1,1\}$ and $\beta>0$, its normalized <probability mass function> is $Z_{n,\beta}^{-1}\exp[\beta(\sum_is_i)^2/(2n)]$. The <partition function> $Z_{n,\beta}$ sums these weights over all configurations. The square includes constant diagonal terms; these affect only normalization. <Spin inversion symmetry> makes every spin's <expected value> zero, while grouping configurations by their <spin magnetization> gives
$$
\mathbb P(M=m)=Z_{n,\beta}^{-1}\binom{n}{n(1+m)/2}e^{\beta nm^2/2}.
$$
Here $m=-1+2k/n$, $0\leq k\leq n$.

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