Plancherel expectations of the shifted character observables

Expectation by characters: for $n\ge r$, expanding the expectation against the Plancherel weights and inserting the definition of $p_{\rho }^{\mathrm{\#}}$ gives ${\mathbb{E}}_{P_n}[p_{\rho }^{\mathrm{\#}}]={\sum }_{\lambda ⊢n}\frac{n^{\downarrow r}{\chi }_{\rho \cup 1^{n-r}}^{\lambda }}{f^{\lambda }}\cdot \frac{(f^{\lambda }{)}^2}{n!}=\frac{n^{\downarrow r}}{n!}{\sum }_{\lambda ⊢n}{\chi }_{\rho \cup 1^{n-r}}^{\lambda }{f}^{\lambda };$ for $n<r$ both $p_{\rho }^{\mathrm{\#}}$ and $n^{\downarrow r}$ vanish, so the formula also gives $0$ there. All quantities are finite, and $n!>0$.

The class sum: by [F2] the character of $\mathbb{C}[S_n]$ is the class function ${\chi }_{\mathrm{r}\mathrm{e}\mathrm{g}}={\sum }_{\lambda ⊢n}{f}^{\lambda }{\chi }^{\lambda }$; evaluated at a permutation of cycle type $\mu ⊢n$ and compared with [F3] this gives ${\sum }_{\lambda ⊢n}{f}^{\lambda }{\chi }_{\mu }^{\lambda }={\chi }_{\mathrm{r}\mathrm{e}\mathrm{g}}(\mu )=\{\begin{array}{ll}n!,& \mu =(1^n),\\ 0,& \mu \mathrm{\ne }(1^n),\end{array}$ the class $(1^n)$ being exactly the identity class.

Case evaluation: applying step 1.2 with $\mu =\rho \cup 1^{n-r}⊢n$ in step 1.1, the sum is $n!$ precisely when $\rho \cup 1^{n-r}=(1^n)$, i.e. when $\rho =(1^r)$, and is $0$ otherwise; dividing by $n!$ and multiplying by $n^{\downarrow r}$ gives ${\mathbb{E}}_{P_n}[p_{\rho }^{\mathrm{\#}}]=n^{\downarrow r}$ for $\rho =(1^r)$ and $0$ otherwise, including $n<r$ by step 1.1.

Consequences: $n^{\downarrow r}=n(n-1)\cdots (n-r+1)$ is $0$ for $n<r$ and of modulus at most $n^r$ for $n\ge r$, so ${\mathbb{E}}_{P_n}[p_{\rho }^{\mathrm{\#}}]=O(n^{\mathrm{\mid }\rho \mathrm{\mid }})$ uniformly in $n$; and if $m_1(\rho )=0$ with $\rho \mathrm{\ne }\mathrm{\varnothing }$ then $\rho \mathrm{\ne }(1^r)$, so the expectation vanishes for every $n$, as asserted.