Dec

27

What are the chances by randomness that a market with 55% up days and 250 trading days will end at the high, and does that have any evolutionary significance e.g. the battle of males to be at a maximum relative to competitors in other fields?

Gary Phillips writes: 

The "unsinkable" S&P is reminiscent of Molly Brown. Her husband's fortune made in silver was lost, but resurrected in his discovery of gold. She even survived the sinking of the Titanic! If disaster were to befall the S&P, it would most certainly survive; and present one a buying opportunity, once again.

Ralph Vince writes: 

It will have 3 1/2 market days to recover what it will give up tomorrow in order to do that.

Alston Mabry writes: 

Just to see some stats on the SPY days YTD:

249 trading days
149 Up days, or 59.8%
mean Up day: +0.57%
mean Up day: 1.61 pts
sd: 1.32

mean Dwn day: -0.57%
mean Dwn day: -1.63 pts
sd: 1.83

Jared Albert writes:

Using  with gratitude Big Al's numbers:

up =np.random.normal(1.61, 1.32, 138)
down = np.random.normal(-1.63, 1.83, 112)
Total number of max high finishes divided by total runs:      0.1357 on 10,000 runs

from random import sample, seed
import numpy as np
seed = 10
#55% of 250 give 137.5, so to avoid half days went with 55.2% up days
'''249 trading days

149 Up days, or 59.8%
mean Up day: +0.57%
mean Up day: 1.61 pts
sd: 1.32
mean Dwn day: -0.57%
mean Dwn day: -1.63 pts

sd: 1.83'''
up =(np.random.normal(1.61, 1.32, 138))
down = np.random.normal(-1.63, 1.83, 112)
total_days = list(np.concatenate((up,down)))
win_count = 0
total_runs= 10000
for _ in range(total_runs):
    m=0
    running_total = []
    test_population = sample(total_days, len(total_days))
    #print(test_population)
    for i in test_population:
        m = m + i
        running_total.append(m)
    #print(f'running_total,{running_total}')
    if max(running_total) == running_total[-1]:
        win_count += 1
        #print(f'win_count: {win_count}')
print(f'Total number of max high finishes divided by total runs: \
     {win_count/total_runs}')


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