Why is the area under the ROC curve better than raw accuracy as an out-of- sample evaluation metric?
I am a little bit confused about the Area Under Curve (AUC) of ROC and the overall accuracy.  Will the AUC be proportional to the overall accuracy? In other words, when we have a larger overall accuracy will we definitely a get larger AUC? Or are they by definition positively correlated?  If they are positively correlated, why do we bother reporting both of them in some publications?  In real case, I performed some classification task and got the results as follows: classifier A got an accuracy 85% and AUC of 0.98 and classifier B got an accuracy of 93% and AUC of 0.92. Question is, which classifier is better? Is it possible to get results similar to these (or do my results indicate a bug in my implementation)?
Update to Windows 10: What is going to happen to my data and apps? I have Windows 8.1 Update, if I update to Windows 10 when it becomes available what is going to happen to my data and apps?  Of course I have to do a backup, but I wonder if the data will be overwritten or not. Game states & passing data between scenes To hyphenate or not? $A \oplus B = A \oplus C$ imply $B = C$? I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.  For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$  implies $B = C$ is true.   I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:    I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof.  On the other hand, I could be completely wrong about my reasoning.    Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well. zsh: excluding files from a pattern Say I have the following files:   |-- bar `-- foo     |-- type_A_1     |-- type_A_2     |-- type_B_1     |-- type_B_2     |-- type_B_xx     |-- type_B_xx     `-- something_else    I thought the following command  print -l foo/*~{type_B*}   would print everything under foo except things that start with type_B but it doesn't, instead it prints everything under foo:  foo/type_A_1 foo/type_A_2 foo/type_B_1 foo/type_B_2 foo/type_B_xx foo/something_else   I also tried print -l foo/*~type_B and got the same thing.  How does the exception wildcard ~ work in zsh? I have a very long equation that must be separated into two lines, and it has several pairs of nesting \left \right delimiters. Unfortunately, it seems that they need to be on the same line in order for them to work.  Also I'm using the align environment because I need the aligning functionality. Is there a way to have the size of delimiters automatically adjusted over multiple lines? Exclude menu item from the collapse of bootstrap 3 navbar When would I use a voltage regulator vs voltage divider? When would you use a voltage regulator vs a resistor voltage divider?  Are there any uses for which a resistor divider is particularly bad? Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper     Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$   and find the sum if it exists.   I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you. Richard Stallman states that Ubuntu "has installed surveillance code". See:   Is this true?   I agree with him that "Any excuse Canonical offers is inadequate....". On finding polynomials that approximate a function and its derivative (extensions of Stone-Weierstrass?) The Stone-Weierstrass Theorem tells us that we can approximate any continuous $f:\mathbb{R}^n\to\mathbb{R}$ arbitrary well on a compact subset of $\mathbb{R}^n$ by some polynomial. Suppose that $f$ is continuously differentiable, under what conditions can we guarantee we can find a polynomial $p$ such that $p$ approximates arbitrarily well $f$ and the partial derivatives approximate of $p$ arbitrarily well those of $f$ (on a compact subset of $\mathbb{R}^n$)? References are welcome. See below for my (probably naive) approach to the question.    It is straightforward to find conditions if $n=1$: Let $K:=[a,b]\subseteq\mathbb{R}$ and let $K_+$ denote some open cover of $K$. Suppose that $f:K_+\to\mathbb{R}$ is continuously differentiable and let $f'$ denote its derivative. Then, by Weierstrass's approximation Theorem we have that there exists a polynomial $q$ such that   $$||f'(x)-q(x)||\leq \varepsilon$$  for all $x\in K$. Choose some $y\in K$. For any $x\in K$ define  $$p(x):=f(y)+\int_y^xq(t)dt.\quad\quad(*)$$  By the Fundamental Theorem of calculus we have that $p'=q$. So for any $x\in K$  $$||f(x)-p(x)||=\left|\left|\int_y^x (f'(t)-q(t))dt\right|\right|\leq \int_y^x ||f'(t)-q(t)||dt\leq \varepsilon||x-y||\leq \varepsilon (a-b).$$  However, the above argument doesn't work in higher dimensions because the analogue of $(*)$ does not imply that $\nabla p = q$ (since not every vector of polynomials spans a gradient field). Can anyone suggest a way around this? That is, a way to to show that for continuously differentiable $f:\mathbb{R}^n\to\mathbb{R}$ and for any $\varepsilon&gt;0$, we can always find a polynomial $p$ such that   $$||f(x)-p(x)||+||\nabla f(x)-\nabla p(x)||\leq \varepsilon$$  for all $x$ in some compact set $K$? Feel free to add extra conditions on $K$, if it makes life easier. I am using postgreSQL, postGIS and pgrouting. Now I am really confused on how am I going to find the shortest path between multiple points.   I know how to find it between two points through dijkstra but I am confused how to accomplish the same result with more than two points.   I have done some research and found this:    But still I did not understand if its the right way and how to install it. What is the difference between equation and formula? Sometimes equation and formula are used interchangeably, but I was wondering if there is a difference.  For example, suppose we can calculate a car's fuel efficiency as:  mpg = distance traveled in miles / the fuel used in a gallon   Is that an equation or formula? How can I get apt to use a mirror close to me, or choose a faster mirror? Downloading from the main Ubuntu archive is slow even when it's not release day, how can I get apt-get to automatically use a mirror that is close to me? What do you call a person who started something that is later followed by everyone? In one of the concluding scenes of the movie Thor, it's shown that Loki deliberately let go of the Scepter, and falls deep into the vast intergalactic sea sort of thing. How does he survive and return in The Avengers? Turn off buffering in pipe I have a script which calls two commands:  long_running_command | print_progress   The long_running_command prints progress but I'm unhappy with it. I'm using print_progress to make it nicer (namely, I print the progress in a single line).  The problem: Connection a pipe to stdout also activates a 4K buffer, so the nice print program gets nothing ... nothing ... nothing ... a whole lot ... :)  How can I disable the 4K buffer for the long_running_command (no, I do not have the source)? I understand the idea of a focal length for a single lens, i e the distance from the lens to the point at which parallel rays of light converge. However, in the case of a photographic lens with multiple lens elements, where exactly is the focal length of the lens as a whole measured from? I was recently granted German refugee status with a travel document and, in my interview, I told them I would never go back to my home country, Iraq. Now, someone I loved died in my country.  Is it possible for me to leave and return to the Schengen area, using my Iraqi passport? Would German Immigration notice and would it revoke my refugee status?  Would I be able to use both my Iraqi passport and my German travel documents? Would the German authorities know if I did this? Why does Ridders' method work as well as it does? I've just read section 9.2.1 in Numerical Recipes Ed. 3 (Press et al. 2007), which describes Ridders' method of root finding. I understand that allowing for some curvature of the function by linearizing three points with an exponential will probably get you a better approximation and more rapid convergence, but I have no idea how they get that it converges quadratically, and the motivation behind choosing an exponential (rather than e.g. a polynomial) is not clear to me.  So, specifically, I'm wondering (1) why linearize with an exponential, and (2) why the method converges quadratically.  I tried looking for Ridders' original paper "A new algorithm for computing a single root of a real continuous function", but the only copy I could find was in a journal that my school's library does not give me free access to.  UPDATE:  I've realized that the fact that an exponential is strictly positive means that weighting the three points with an exponential won't change their signs. Hence, the root of the linearization will have to land inside the original bracket.  Polynomials, rationals, sinusoids, and logarithms all have positive and negative bits, so using them probably runs the risk of getting bad interpolated roots. One of these squared might be usable, but maybe non-monoticity gives some sort of undesirable effect. Since this is a complete list of all simple functions that I am aware of, exponentials seem to be the only easily available weighting function with this property.  A constant function is also technically single-signed and monotone, but it can't be used to linearize stuff.