quester
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what you did - you created confidence intervals under assumption that chicken weights are drawn from normal disrtibution (with value range $(-\infty, \infty)$) - in fact these can be drawn from other ...

$$\sum{(x_i-\overline{x})^2} = \sum{(x_i-\overline{x})x_i} - \sum{(x_i-\overline{x})\overline{x}} = \sum{(x_i-\overline{x})x_i} - \overline{x}\sum{x_i} + n\overline{x}^2 = \sum{(x_i-\overline{x})x_i} -... View answer 3 votes [X>0 => P(X>t) = P(\phi(X) > \phi(t))] \\ \land [X<0 => P(X<t) = P(-X>-t) = P(\phi(-X) > \phi(-t)) = P(\phi(-X) > \phi(t)) = P(\phi(X) > \phi(t))] \\ => P(|X| > ... View answer 2 votes just create list of all files, then sample this list and voila!, as for calculations: probability sampling image from class A sampling from whole set is P(class\_A)=size(A)/size(whole\_set) but in ... View answer Accepted answer 2 votes since X, Y are iid then for joint distribution we have:$$\forall_{(x,y) \in \Omega} f_{XY}(x, y) = f_{XY}(y,x)$$then:$$E[\frac{Y}{X+Y}] = \iint_{\Omega} \frac{y}{x+y}f_{XY}(x,y)dxdy=\iint_{\...

In general you are writing pipelines and you relay on fact that "you run n+1 cell only if previous n completed with success" and you don't have to rely on super structurized code. On the other hand ...

why we are using softmax in first place - when we have on output [.5, -1] we would want to have positive values that sums to 1 so softmax was invented but you can develop your own function for this ...

in python hist sometimes bins tend to "cluster" and then for one value you will have observations from many values example: import numpy as np import matplotlib.pyplot as plt q = np.random.choice([...

let's presume that you have 10000 boxes, and you place blue balls in these boxes at random, and then you are placing red balls at random, let's count probability of success: $$p_b(i) = p(\text{10 blue ... View answer 1 votes one of ideas is to add \delta to one population (raising mean) and in second test substracting \delta and then computing statistic and figure out in two "one-sided tests" p-values, after ... View answer 1 votes https://besjournals.onlinelibrary.wiley.com/doi/10.1111/1365-2656.12382 please read this article, here you will see differences in different calculation methods for confidence interval, of course ... View answer 1 votes you can do it like this:$$P(Y \ge 8| succ == 3) = P(Y == 7| succ == 0) + P(Y == 7| succ == 1) + P(Y == 7| succ == 2) = \binom{7}{0}(\frac{5}{6})^7 + \binom{7}{1}\frac{1}{6}(\frac{5}{6})^6 + \binom{7}...

according to Lindeberg-Levy CLT https://en.wikipedia.org/wiki/Central_limit_theorem let $S_n = \frac{\sum_{k=1}^n X_k}{n}$ Suppose {X1, X2, …} is a sequence of i.i.d. random variables with E[Xi] ...

it can be done either way but more important is gradient and it also can be done either way, but one thing to clarify you should: start epoch, compute loss, then gradient(sgd step), then next epoch ...

$$P(N(t_2)>N_2|N(t_1)<N_1)=P(N(t_1)<N_1 \land N(t_2)>N_2)/P(N(t_1)<N_1)=\sum_{k=0}^{N_1-1} [P(N(t_1)=k)P(N(t_2) >N_2)]/P(N(t_1)<N_1)= \frac{\sum_{k=0}^{N_1-1} [P(N(t_1)=k)(1-P(N(... View answer 0 votes from markov inequality:$$P(X \ge a) \le E(X^q)/a^qP(X \ge a) = 1 - P(X \lt a) \le E(X^q)/a^q => P(X < a) \ge 1 - E(X^q)/a^q$$from https://en.wikipedia.org/wiki/... View answer 0 votes In general polynomial regression yields better results than neural networks, but they suffer from combinatorial explosion, so in order to tackle this we use neural networks with non-linear activations ... View answer 0 votes correlation will get you nowhere since correlation 0 can be yield for 2 variables that will form V-shape in scatter plot, so it's bad idea to say corr==0 <=> 2 variables are independent all ... View answer 0 votes best indicator of change would be avg vertical distance between these two lines after all you want to know what is the improvement for given traffic value, so avg vertical distance would be avg ... View answer 0 votes if you will permute input you would have to also permute coeficients in first layer of network and if these permutations "will agree" then yes, otherwise you will just generate face on random input ... View answer 0 votes since machine learning consists of stochastic methods unless you will set specific random seed on beginning, your model will be initialized with different values and this will lead to different values ... View answer 0 votes I assume you sample vertors (p1, p2, q, x), gibbs sampling says "keep n-1 coordinates from last result and sample next coordinate" here I would add "keep x=1" View answer 0 votes one of the idea is to have a lot of clusters let's say 10000, run KNN on all training observations and this will provide "tree search" in your samples then when you will have new observation take ... View answer 0 votes you can experiment with https://keras.io/callbacks/#learningratescheduler to find propper setting, general idea is that at beginning you are probably in area that cause your NN to perform badly and ... View answer 0 votes technically if we would treat sigmoid S(x) as CDF then \frac{d\,S(x)}{dx} = \frac{e^{-x}}{(1+e^{-x})^2} is PMF of Logistic distribution en.wikipedia.org/wiki/Logistic_distribution // name was ... View answer 0 votes one weak point is sorting, and second is creating new collection so you could just make another collection (list) and then keep topK items until some point this way instead of sorting everytnig and ... View answer Accepted answer 0 votes \mathbb P (X \in [Y, Z)) = \int_{-\infty}^{\infty} \mathbb P (x \in [Y, Z)) f_X(x) dx = \int_{-\infty}^{\infty} (\underset{(x, \infty) \times (-\infty, x)}{\iint} f_{YZ}(y, z) dydz) f_X(x) dx ... View answer 0 votes if these duplicates are drawn from same random variable then yes (experiment can yield same observation sometimes and it is normal) - although it would be better to give answer in terms of probability ... View answer -1 votes I would say that p_{\hat{\theta}_nX}(x)=p_{\theta X}(x(1+o_p(\frac{1}{n})/\theta)) because you are affecting characteristics of distribution by shifting mean(and scaling variance),$$p_{\hat{\theta}...