Sample complexity of mean estimation using empirical estimator and median-of-means estimator?

Given a random variable $$X$$ with unknown mean $$\mu$$ and variance $$\sigma^2$$, we want to produce an estimate $$\hat{\mu}$$ based on $$n$$ i.i.d. samples from $$X$$ such that $$\rvert \hat{\mu} – \mu \lvert \leq \epsilon\sigma$$.

Empirical estimator: why are $$O(\epsilon^{-2}\cdot\delta^{-1})$$ samples necessary? why are $$\Omega(\epsilon^{-2}\cdot\delta^{-1})$$ samples sufficient?

Median-of-means estimator: why are $$O(\epsilon^{-2}\cdot\log\frac{1}{ \delta})$$ samples necessary?

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Big O notation – estimation of run time [migrated]

I am running very computationally intensive tasks and wish to adjust the parameters respective of how long it takes.

The function I am running is PLINK – for those who don’t know, it is used for genotype data.

The function is said to follow a O(n*m^2) w.r.t. big O.

I have the run time for two time points with different parameters for m and a constant n, they are: 3 hours and 648 hours.

From this I wish to estimate the run-time for different parameters of m, that would respect the O(n*m^2) relationship.

Can anybody provide some insight as to methods for estimating run-time with the constant n parameters however also, for running tests with different parameters as well in order to achieve an optimal run-time with respect to accuracy of results?

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Magento2 How to remove estimation call on state change action in Cart page?

I need to remove estimation call on state change action on the Cart page. actually when a user on cart change Regio estimation call execute when to add zip code again estimation call so shipping calculation takes a too time.

So How I can achieve it?

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Estimation of the number of solutions by Counting

This is a question from a quantum computation textbook.

Consider a classical algorithm for counting the number of solutions to a problem. The algorithm samples uniformly and independently $$k$$ times from the Search Space of size $$N$$ for solutions using an Oracle that outputs 1 or 0, and let $$X_1,X_2,X_3,…X_k$$ be the results of the Oracle calls. So $$X_j=1$$ if the $$jth$$ Oracle call found a solution and $$X_j=0$$ otherwise. This algorithm estimates the number of solutions $$S$$:

$$S=N * \sum_{j}\frac{X_j}{k}$$

Assuming the number of solutions is $$M$$ and this is not known in advance. The Standard Deviation of $$S$$ is stated and found to be:

$$\Delta S=\sqrt{\frac{M(N-M)}{k}}$$

The question is:
Prove that to obtain a probability at least $$\frac{3}{4}$$ of estimating $$M$$ correctly to within an accuracy $$\sqrt{M}$$ for all values of $$M$$, we must have $$k=\Omega(N)$$.

I know how to get the 2nd equation from the 1st, which is by moving $$N$$ and $$k$$ to the left, thus treating $$kS/N$$ as a Binomial Distribution $$B(k,\frac{M}{N})$$. Then finding the variance of the Binomial Distribution and some algebraic manipulation will lead to the 2nd equation. I’m clueless in proving of $$k=\Omega(N)$$. Only thing I tried writing is:

$$P\Big(\sqrt{\frac{M(N-M)}{k}}\leq \sqrt{M}\Big)\geq \frac{3}{4}$$

Can someone help me with this?

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Netherlands Taxi Fare Estimation

Is there a rule of thumb that can be applied to estimating Taxi cost (maybe linear cost per kilometer + fixed fee)? Maybe there is a good (authoritative) web application?

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Is there’s anyway to get fee estimation with vbyte instead of KB using “estimatesmartfee” RPC command?

I’m using Bitcoin-core RPC command to get current fee rate, but it returns the fees in KB, I would like to get fee rates in vbyte is that possible?

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Using O-notation for asymptotic estimation of the number of additions in recursive function

the number of additions that are executed during the calculation is a(n). How can i find an asymptotic estimation for the function mystery(n) with the help of the O-notation and master theorem. Note: here the question is not asked for the value of mystery(n), but rather for the number of additions!

def mystery(n):     if n==0:         return n * n     return 2 * mystery(n/3) + 4 * n
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Monte Carlo errors estimation routine

I would value your opinion on the following piece of code. I am rather new to both Python and Monte Carlo analysis, so I was wondering whether the routine makes sense to more experienced and knowledgeable users.

def MC_analysis_a():     x = spin_lock_durations     y_signal_a = (a_norm1, a_norm2, a_norm3, a_norm4, a_norm5, a_norm6, a_norm7, a_norm8)     x = np.array(x, dtype = float)     y_signal_a = np.array(y_signal_a, dtype = float)          def func(x, a, b):         return a * np.exp(-b * x)      initial_guess = [1.0, 1.0]     fitting_parameters, covariance_matrix = optimize.curve_fit(func, x, y_signal_a, initial_guess)     print(round(fitting_parameters, 2))      # ---> PRODUCING PARAMETERS ESTIMATES      total_iterations = 5000     MC_pars = np.array([])      for iTrial in range(total_iterations):         xTrial = x         yTrial = y_signal_a + np.random.normal(loc = y_signal_a, scale = e_signal_a, size = np.size(y_signal_a))         try:             iteration_identifiers, covariance_matrix = optimize.curve_fit(func, xTrial, yTrial, initial_guess)         except:             dumdum = 1             continue      # ---> STACKING RESULTS          if np.size(MC_pars) < 1:             MC_pars = np.copy(iteration_identifiers)         else:             MC_pars = np.vstack((MC_pars, iteration_identifiers))      # ---> SLICING THE ARRAY      print(np.shape(MC_pars))     # print(np.median(aFitpyars[:,1]))     print(np.std(MC_pars[:,1]))

The output I get is apparently satisfactory and plausible.

Many thanks in advance to any contributor!

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