Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with st.statistics
Search options not deleted
user 9896
Applied and theoretical statistics: e.g. statistical inference, regression, time series, multivariate analysis, data analysis, Markov chain Monte Carlo, design of experiments.
0
votes
How many samples to be confident?
If you have probability distributions $p_0, p_1$, then the proper statistic is the bayes factor
$p_0(\vec x)/p_1(\vec x)$ where $\vec x$ is the vector of observations. This yields the most powerful te …
- 3,475
1
vote
Question on Maximum Likelihood Estimation
You definitely would run into identification issues. Suppose $f(x|\mid\theta) = x \theta$. Then
multiplying $\theta$ and $\sigma$ by any common factor would result in an identical situation. In this …
- 3,475
1
vote
Accepted
Standard Deviation of Averages of Random Numbers
Let $Y$ the be mean of $t$ samples of the distribution. $Y$ has mean $1/2$ and variance $1/(12 t)$. Then $f_s(t)$ is the sample standard deviation of $s$ draws from $Y$. As the sample standard deviati …
- 3,475
1
vote
Statistics - Subpopulation Random Sampling
This is not correct. We do not know that 70% of population A is male and 70% of population B is male. It may be that the populations are correlated. For example, if A = population has beards, B = popu …
- 3,475
1
vote
1
answer
142
views
Role of statistical estimation in formal proof
Consider the following scenario: There is some mathematical constant $c$ that you want to compute. You don't have a formal proof for any particular value of $c$, but you have some sound statistical pr …
- 3,475
7
votes
Accepted
name for $\underset{x}{\operatorname{argmin}} \displaystyle\sum\limits_{i=1}^n |x_i - x|$
The value of $x$ that satisfies this is the median of $x_i$. It minimizes $L_1$ loss.
Note that the mean of $x_i$ is the value of $x$ minimizing $\sum(x_i - x)^2$, the $L_2$ loss.
- 3,475