demo_tsscpp_probabilities.sagews

Demonstration of tsscpp_probabilities.sage

The following worksheet demonstrates some of the main features oftsscpp_probabilities.sage, which contains the following functions for computing probabilities associated with the TSSCPP point process:

• rho(S, K, data_type) computes $\rho_n(S)$, where K is either the matrix of interpolated entries of $K_n$ (in which case one sets data_type = 's', symbolic) or the limiting kernel $K_\infty$ (data_type = 'n', numeric)

• rho_generalized(S, T, K, data_type) computes $\rho_n(S;T)$

• first_k_vert(k, K, data_type) computes $\mathbb P(V_k^{(n)})$ or its limit $\mathbb P(V_k^{(\infty)})$, depending on whether K is gives the interpolated matrix or $K_\infty$

• first_k_hor(k, K, data_type) computes $\mathbb P(H_k^{(n)})$ or its limit $\mathbb P(H_k^{(\infty)})$

• prob_dist_path_k(k, K, data_type) computes the probability distribution of the endpoint $x$ of path $k$; i.e., $(\mathbb P(\pi_k(\mathscr X_n) = x))_{k \leq x \leq 2k}$ or its limit as $n \to \infty$

1. Loading the Interpolated $K_n$ Matrix and its Limit $K_\infty$

We first load the matrices K_interpolated and K_infinity computed in the worksheet demo_tsscpp_interpolation.sagews.

N = 100
dim = 2*floor((2*N - 9) / 5)

print('Now loading K_interpolated and K_infinity:')
load('tsscpp_data/tsscpp_K/K_interpolated_{0}_by_{0}.sage'.format(dim))
load('tsscpp_data/tsscpp_K/K_infinity_{0}_by_{0}.sage'.format(dim))

print('Done')

Now loading K_interpolated and K_infinity: Done

2. Computing Horizontal and Vertical Probabilities $\mathbb P(V_k^{(n)})$ and $\mathbb P(H_k^{(n)})$

Below, we demonstrate first_k_vert and first_k_hor for both the interpolated $K_n$ matrix and for $K_\infty$.

load('tsscpp_probabilities.sage')

for k in [1 .. 5]:
    prob = first_k_vert(k, K_interpolated, 's')
    print('Probability that the first {0} paths are vertical:'.format(k))
    show(prob)

print('\nLimits of the above probabilities as n goes to infinity:')
for k in [1 .. 5]:
    prob = first_k_vert(k, K_infinity, 'n')
    print('k={0}: {1}'.format(k, prob))

Probability that the first 1 paths are vertical:

$\displaystyle \frac{5 \, {\left(n^{2} - 1\right)}}{2 \, {\left(4 \, n^{2} - 1\right)}}$

Probability that the first 2 paths are vertical:

$\displaystyle \frac{n^{2} - 4}{4 \, n^{2} - 1}$

Probability that the first 3 paths are vertical:

$\displaystyle \frac{33 \, {\left(n^{6} - 14 \, n^{4} + 49 \, n^{2} - 36\right)}}{8 \, {\left(4 \, n^{2} - 1\right)}^{2} {\left(4 \, n^{2} - 9\right)}}$

Probability that the first 4 paths are vertical:

$\displaystyle \frac{11 \, {\left(n^{6} - 29 \, n^{4} + 244 \, n^{2} - 576\right)}}{16 \, {\left(16 \, n^{4} - 40 \, n^{2} + 9\right)} {\left(4 \, n^{2} - 1\right)}}$

Probability that the first 5 paths are vertical:

$\displaystyle \frac{2431 \, {\left(n^{12} - 64 \, n^{10} + 1518 \, n^{8} - 16852 \, n^{6} + 89881 \, n^{4} - 204084 \, n^{2} + 129600\right)}}{512 \, {\left(256 \, n^{8} - 2304 \, n^{6} + 4704 \, n^{4} - 1936 \, n^{2} + 225\right)} {\left(4 \, n^{2} - 1\right)} {\left(4 \, n^{2} - 9\right)}}$

Limits of the above probabilities as n goes to infinity: k=1: 5/8 k=2: 1/4 k=3: 33/512 k=4: 11/1024 k=5: 2431/2097152

for k in [1 .. 4]:
    prob = first_k_hor(k, K_interpolated, 's')
    print('Probability that the first {0} paths are horizontal:'.format(k))
    show(prob)

print('\nLimits of the above probabilities as n goes to infinity:')
for k in [1 .. 4]:
    prob = first_k_hor(k, K_infinity, 'n')
    print('k={0}: {1}'.format(k, prob))

Probability that the first 1 paths are horizontal:

$\displaystyle \frac{3 \, {\left(n^{2} + 1\right)}}{2 \, {\left(4 \, n^{2} - 1\right)}}$

Probability that the first 2 paths are horizontal:

$\displaystyle \frac{59 \, n^{6} + 299 \, n^{4} + 866 \, n^{2} + 576}{16 \, {\left(64 \, n^{6} - 176 \, n^{4} + 76 \, n^{2} - 9\right)}}$

Probability that the first 3 paths are horizontal:

$\displaystyle \frac{3 \, {\left(2579 \, n^{12} + 39364 \, n^{10} + 374412 \, n^{8} + 2174092 \, n^{6} + 6601109 \, n^{4} + 11674044 \, n^{2} + 6350400\right)}}{512 \, {\left(64 \, n^{6} - 176 \, n^{4} + 76 \, n^{2} - 9\right)} {\left(16 \, n^{4} - 136 \, n^{2} + 225\right)} {\left(4 \, n^{2} - 1\right)}}$

Probability that the first 4 paths are horizontal:

$\displaystyle \frac{45 \, {\left(151181 \, n^{20} + 5389132 \, n^{18} + 117984048 \, n^{16} + 1797839324 \, n^{14} + 19407568705 \, n^{12} + 147255615696 \, n^{10} + 810109953722 \, n^{8} + 3037474082608 \, n^{6} + 7089046189344 \, n^{4} + 9780996810240 \, n^{2} + 4682022912000\right)}}{65536 \, {\left(256 \, n^{8} - 6912 \, n^{6} + 63456 \, n^{4} - 233200 \, n^{2} + 275625\right)} {\left(64 \, n^{6} - 176 \, n^{4} + 76 \, n^{2} - 9\right)}^{2}}$

Limits of the above probabilities as n goes to infinity: k=1: 3/8 k=2: 59/1024 k=3: 7737/2097152 k=4: 6803145/68719476736

4. Computing the Limiting Distribution of Path $k$

Below, we use prob_dist_path_k to compute the distribution of the endpoint of path $k$ (which has support on $k \leq x \leq 2k$) in the limiting TSSCPP point process given by kernel $K_\infty$. One may compute individual path probabilities $\mathbb P(\pi_k(\mathscr X_n) = x)$ using the function prob_path_k_hits_x(k, x, K, data_type).

from sage.plot.bar_chart import BarChart

for k in [1 .. 8]:
    dist = prob_dist_path_k(k, K_infinity, 'n')
    print('Distribution for the endpoint of path k={0}:\n{1}'.format(k, dist))

    plot = bar_chart(dist, color='slategrey', axes=False, width=0.5)
    plot.show(ymin=0, axes=False, figsize=[4,3])

Distribution for the endpoint of path k=1: [5/8, 3/8]

Distribution for the endpoint of path k=2: [1/4, 325/512, 59/512]

Distribution for the endpoint of path k=3: [33/512, 545/1024, 791665/2097152, 54159/2097152]

Distribution for the endpoint of path k=4: [11/1024, 35679/131072, 589403/1048576, 5181180291/34359738368, 142866045/34359738368]

Distribution for the endpoint of path k=5: [2431/2097152, 188881/2097152, 16939639467/34359738368, 25652736981/68719476736, 1512910002731985/36028797018963968, 17250322859055/36028797018963968]

Distribution for the endpoint of path k=6: [85/1048576, 674088573/34359738368, 19122742785/68719476736, 2366395891945965/4503599627370496, 1515761510156805/9007199254740992, 624086381545377842663/75557863725914323419136, 2973802586179287577/75557863725914323419136]

Distribution for the endpoint of path k=7: [126293/34359738368, 194001963/68719476736, 1885647099121635/18014398509481984, 8461089993529877/18014398509481984, 27847669196528437539075/75557863725914323419136, 1002981740399396437407/18889465931478580854784, 11696714564822243668097989109/10141204801825835211973625643008, 23399944624587413326693899/10141204801825835211973625643008]

Distribution for the endpoint of path k=8: [7429/68719476736, 1207343934995/4503599627370496, 29899438259811/1125899906842624, 10592810737354813451289/37778931862957161709568, 37935826865957722542213/75557863725914323419136, 56637620518355032803977444751/316912650057057350374175801344, 60230842225375535507557060843/5070602400912917605986812821504, 1247494052952693497593902443281976891/10889035741470030830827987437816582766592, 1049985344725805761378702198632901/10889035741470030830827987437816582766592]

5. Computing (Symbolically) the Distributions of the First Few Paths

Now we use prob_dist_path_k to compute the distribution of the endpoint of path $k$ in the order-$n$ TSSCPP point process, using the interpolated kernel K_interpolated.

for k in [1 .. 3]:
    dist = prob_dist_path_k(k, K_interpolated, 's')
    print('Distribution for the endpoint of path k={0}:'.format(k))
    show(dist)

Distribution for the endpoint of path k=1:

[$\displaystyle \frac{5 \, {\left(n^{2} - 1\right)}}{2 \, {\left(4 \, n^{2} - 1\right)}}$, $\displaystyle \frac{3 \, {\left(n^{2} + 1\right)}}{2 \, {\left(4 \, n^{2} - 1\right)}}$]

Distribution for the endpoint of path k=2:

[$\displaystyle \frac{n^{2} - 4}{4 \, n^{2} - 1}$, $\displaystyle \frac{5 \, {\left(65 \, n^{6} - 175 \, n^{4} - 322 \, n^{2} - 72\right)}}{8 \, {\left(16 \, n^{4} - 40 \, n^{2} + 9\right)} {\left(4 \, n^{2} - 1\right)}}$, $\displaystyle \frac{59 \, n^{6} + 299 \, n^{4} + 866 \, n^{2} + 576}{8 \, {\left(16 \, n^{4} - 40 \, n^{2} + 9\right)} {\left(4 \, n^{2} - 1\right)}}$]

Distribution for the endpoint of path k=3:

[$\displaystyle \frac{33 \, {\left(n^{6} - 14 \, n^{4} + 49 \, n^{2} - 36\right)}}{8 \, {\left(16 \, n^{4} - 40 \, n^{2} + 9\right)} {\left(4 \, n^{2} - 1\right)}}$, $\displaystyle \frac{5 \, {\left(109 \, n^{6} - 797 \, n^{4} - 1616 \, n^{2} - 360\right)}}{16 \, {\left(16 \, n^{4} - 40 \, n^{2} + 9\right)} {\left(4 \, n^{2} - 1\right)}}$, $\displaystyle \frac{35 \, {\left(22619 \, n^{12} - 112604 \, n^{10} - 550356 \, n^{8} - 2274740 \, n^{6} - 713683 \, n^{4} - 4432356 \, n^{2} - 4639680\right)}}{512 \, {\left(256 \, n^{8} - 2304 \, n^{6} + 4704 \, n^{4} - 1936 \, n^{2} + 225\right)} {\left(16 \, n^{4} - 40 \, n^{2} + 9\right)}}$, $\displaystyle \frac{21 \, {\left(2579 \, n^{12} + 39364 \, n^{10} + 374412 \, n^{8} + 2174092 \, n^{6} + 6601109 \, n^{4} + 11674044 \, n^{2} + 6350400\right)}}{512 \, {\left(16 \, n^{4} - 40 \, n^{2} + 9\right)}^{2} {\left(16 \, n^{4} - 104 \, n^{2} + 25\right)}}$]