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Aug
29
comment Monty Hall (goat) problem w. elementary events
@left4bread: in summary, a major issue with your counting argument is that you seem to think that $(a,a,c,?)$ is as likely to happen as $(a,b,c,?)$. This is not true
Aug
29
comment Monty Hall (goat) problem w. elementary events
... Meanwhile the case of (a,b,?,?) also has probability $\frac{1}{9}$. The moderator can only choose $c$. So the case $(a,b,c,?)$ has probability $\frac{1}{9}$: overall there are $6$ three-state cases with probability $\frac{1}{9}$ and in these the user will win by switching. So the total probability that the user wins given that the user switches is $\dfrac{6}{9}=\dfrac23$
Aug
29
comment Monty Hall (goat) problem w. elementary events
@left4bread: The case of (a,a,?,?) has probability $\frac{1}{9}$. The moderator can choose $b$ or $c$. So the case $(a,a,b,?)$ has probability $\frac{1}{18}$ and the case $(a,a,c,?)$ also has probability $\frac{1}{18}$: overall there are $6$ three-state cases with probability $\frac{1}{18}$ and in these the user will lose by switching (the user has a strategy at the end and is not picking at random). So the total probability that the user loses given that the user switches is $\dfrac{6}{18}=\dfrac13$
Aug
29
answered Monty Hall (goat) problem w. elementary events
Aug
26
comment Why splitting the data into the training and testing set is not enough
I would reverse the wording in your second and third paragraphs: I would use the validation set to find the best model and tune its hyperparameters (doing this with several validation sets which partition the training set makes this cross validation) and, once the model has been finalised, then apply it to the test set to see an example of the model's out-of-sample performance.
Aug
26
comment Why splitting the data into the training and testing set is not enough
Personally, I would not use the phrase "external cross validation", as I would see cross validation as the repeated splitting off of different validation sets from the training set for model selection and tuning purposes. You cannot meaningfully do this repeatedly with the test set, as that is as a one-off proxy for future as-yet-unknown data used to judge the performance of the final model.
Aug
21
answered Approximate normality of distribution of counts in contigency table
Aug
5
revised Expected value as a function of quantiles?
^{ mising
Aug
4
comment Derivation of the equation of ridge regression
@bill: here you need the $I$ to get a matrix of the correct dimension so the addition works with $X^TX$: $\lambda$ is just a scalar
Jul
28
comment what does linear regression actually mean?
@AlecTeal: It is not my graph, but heights are in inches and the graph type is a sunflower plot with strokes coming out of each point (strictly speaking a range) representing the number of observations there. Look at cran.r-project.org/web/packages/HistData/HistData.pdf under "Galton" for some example code
Jul
28
answered what does linear regression actually mean?
Jul
15
comment Is there a general/golden rule for appropriate binning in a histogram?
Wikipedia list some different approaches but with over a million data points you could just choose bin widths of $0.01$ starting at $0$, so about $60$ bins and see if that causes an issue with the histogram
Jul
14
comment How to solve this problem on Curse of Dimensionality problem - Nearest Neighbours
@MarkL.Stone: Call a hypersphere an $n$-sphere if you prefer. For me volume is 3-dimensional while hypervolume is the measure at the highest dimension being considered.
Jul
13
answered How to calculate standard error between means?
Jul
13
answered Why is the degree of freedom in t.test dependent on setting of digits?
Jul
10
revised Bayesian Inference and Conditional Probabilities
added 389 characters in body
Jul
10
answered Bayesian Inference and Conditional Probabilities
Jul
3
comment Statistical Significance
If you are going to use the future perfect then perhaps you should move it out of the indicative, e.g. to "would have been rejected". You may also need to say the the "1%" represents the probability of erroneously rejecting the null hypothesis when it is in fact true, so "5%" involves that possibility and more.
Jun
19
comment What can we say about population mean from a sample size of 1?
This seems to give confidence intervals covering the mean with probability about $95\%$ when $\sigma \approx | \mu | \gt 0$ but with much higher probabilities otherwise. If $\mu = 0$ then clearly the probability is $100\%$ as the confidence intervals always contain $0$.
Jun
14
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