Exact sampling of skew young diagrams  (Page 5/7)

Since $\pi \left({\tau }_{\downarrow }^i\right)=0$ ,

Thus, formula [link] becomes $\frac{1}{q}>0$ .

3.) If $i$ is chosen to be a box such that ${\sigma }_{\downarrow }^i,{\sigma }_{\uparrow }^i,{\tau }_{\downarrow }^i$ , and ${\tau }_{\uparrow }^i$ are all valid YD's, then

Thus, formula [link] becomes $\frac{1}{q}=\frac{1}{q}$ .

4.) If $i$ is chosen to be a box such that both ${\tau }_{\downarrow }^i$ and ${\tau }_{\uparrow }^i$ are YD's, but ${\sigma }_{\downarrow }^i$ and ${\sigma }_{\uparrow }^i$ not valid YD's, then

Formula [link] becomes $\infty \ge \frac{1}{q}$ .

5.) If $i$ is chosen to be a box such that ${\tau }_{\uparrow }^i$ is not a valid YD, then ${\sigma }_{\downarrow }^i,{\sigma }_{\uparrow }^i$ , and ${\tau }_{\downarrow }^i$ are not valid YD's either, since $\sigma \le \tau$ . Therefore formula [link] becomes $0=0$ .

Since formula [link] holds in all cases, our probability distribution is monotonic. Therefore, we may define our vertex set $V$ to be the set of all vertices $i$ in the a $\times$ b box where each vertex $i$ is a box. Then we may apply CFTP to a MC on spin configurations with a steady state distribution equal to the probability distribution stated in formula [link] and couple the histories of the smallest and the largest configurations (the configuration in which all $i$ 's spin down, which corresponds to the empty YD, and the configuration in which all $i$ 's spin up, which corresponds to the full YD) to sample from young diagrams that fit in an a $\times$ b box according to the probability distribution stated in formula [link] .

Exact sampling of syd's in an a $\times$ B box greater than or equal to a fixed yd

Here, we attempt to formulate an algorithm for exact sampling from SYD's, ${\lambda }_s$ , that fit in an a $\times$ b box where ${\lambda }_s\ge$ a fixed YD, ${\lambda }_0$ according to this probability distribution

Clearly, there is a 1-1 correspondence between YD's $\lambda \ge {\lambda }_0$ and SYD's ${\lambda }_s\ge {\lambda }_0$ that all fit in an a $\times$ b box: each $\lambda$ corresponds to one ${\lambda }_s$ where $\lambda ={\lambda }_0+{\lambda }_s$ . Using the 1-1 correspondence, we found that the probability distributions of the two sets are equal. Thus we state this theorem and its proof.

Theorem: The probability distribution $\pi \left(\lambda \right)$ on YD's $\lambda$ such that $\lambda$ is greater than or equal to a fixed YD ${\lambda }_0$ is equal to the probability distribution ${\pi }_s\left({\lambda }_s\right)$ on the corresponding SYD's ${\lambda }_s$ such that ${\lambda }_s$ is greater than or equal to ${\lambda }_0$ .

We first define

Since all YD's we are sampling from are greater than or equal to ${\lambda }_0$ , we may factor $q^{|{\lambda }_0|}$ out of Z. Therefore Z= $q^{|{\lambda }_0|}{Z}_s$ . Also, we may say $\lambda ={\lambda }_0+{\lambda }_s$ . Therefore,

Because these two probability distributions are equivalent we may sample from SYD's ${\lambda }_s\ge {\lambda }_0$ by sampling from YD's $\lambda \ge {\lambda }_0$ . We can couple the histories of the two YD's ${\lambda }_0$ and ${\lambda }_{\text{full}}$ using the heat bath algorithm ( ${\lambda }_{\text{full}}$ denotes the YD that is the full a $\times$ b box), and then delete the unit squares contained in ${\lambda }_0$ from the outputted YD. Thus, we have sampled a SYD that is greater than or equal to a fixed YD according to the probability distribution stated in formula [link] .

Exact sampling of syd's in an a $\times$ B box greater than or equal to one of two fixed yd's

Similar to our previous problem, in order to sample SYD's, ${\lambda }_s$ , that fit in an a $\times$ b box such that ${\lambda }_s$ is greater than or equal to one of two fixed YD's ${\lambda }_0$ and ${\lambda }_1$ according to this probability distribution:

we will start by sampling from YD's, $\lambda$ , that fit in an a $\times$ b box where $\lambda \ge$ at least one of ${\lambda }_0,{\lambda }_1$ . Then we will sample from the fixed YD's ${\lambda }_0$ and ${\lambda }_1$ and the YD we select here ( ${\lambda }_0$ or ${\lambda }_1$ ) will be removed from $\lambda$ . The result will be a SYD that has been sampled exactly according to the probability distribution from formula [link] .