生成 for 循环以计算具有不同样本量的 montecarlo 积分的误差

Question

Basically, I'm trying to create a for loop which computes the estimates and errors for the integral below for different sample sizes.:

This is the code for a single calculation:

import numpy as np
truetheta = 0.2

m = 10000

x = np.random.uniform(0, 1, m)

y = (x)**4
naive = (np.sum(y)/m)
error = abs(naive - truetheta)

However, I want to produce one error for M = 2^i, with i = 1, 2, ... 10

My attempt looks as follows:

N = 10
sample_size = np.zeros(N)

# for loop creates sample size 2, 4, 8, 16 ....1024
for n in range(N):
    sample_size[n] = 2**(n+1)



# store output
naive = []
errornaive = []

# for-loop to compute error for different sample sizes
for i in (int(sample_size)):
    m = 2**i
    x = np.random.uniform(0, 1, m)
    y = x**4
    naive[i] = (np.sum(y)/m)
    errornaive[i] = abs(naive - truetheta)

When i run the code i get the error:

TypeError: only size-1 arrays can be converted to Python scalars

I'm not really sure how to work around this issue, or whether the remainder of the loop is sound. So i'm asking for a bit of assistance here, could someone offer a solution to my error message? and 2, does anyone have a better suggestion than mine to perform the calculation?

Best regards and thanks in advance.

Answer 1

With the help of some very helpful comments I managed to create the following code:

N = 20
sample_size = np.zeros(N, dtype=int)

# for loop creates sample size 2, 4, 8, 16 ....1024
for n in range(N):
    sample_size[n] = 2**(n+1)



# store our output
naive = []

# for loop to compute error for different sampel sizes
for i in (sample_size):
    m = i
    x = np.random.uniform(0, 1, m)
    y = x**4
    naive.append(np.sum(y)/m)
 
# convert to array and get errors
naivearray = np.array(naive)
naiveerror = abs(naivearray - truetheta)

This runs without errors, but it's not pretty.