I know there are a lot of questions here about ignoring the denominator in a Bayesian approach, but I don't think mine is a duplicate of any of them.  I am reading the book "Pattern recognition and machine learning" by Cristopher Bishop.  Imagine we have a set of N observations of a (single) variable, which we collect in a vector $\mathbf{x} \in \mathcal{R}^N$. We would like to find the mean $\mu$ of the probbility density function that generated that data, using a Bayesian approach. Thus, we first need to find the posterior probability $p(\mu|\mathbf{x})$  We can write:  $p(\mu|\mathbf{x}) = p(\mathbf{x}|\mu) \cdot \dfrac{p(\mu)}{p(\mathbf{x})}$  Now as the book says, we can ignore the denominator because it is just a normalizing factor  $p(\mu|\mathbf{x}) \propto p(\mathbf{x}|\mu) \cdot p(\mu) = p(\mathbf{x}, \mu) $  where the last equation follows from the product rule, or the defition of conditional density for $p(\mathbf{x}|\mu)$ if you want.  So we are approximating a conditional distribution with a joint distribution? How is that even possible?  For one, $p(\mu|\mathbf{x})$ whould be a function of $\mathbf{x}$, while $p(\mathbf{x}, \mu) $ whould be a function of both $\mathbf{x}$ and $\mu$, right?
When I studied Bayesian statistics, a question about the notation of Bayes' Theorem came to my mind. Below is the density function version of Bayes' Theorem, where $y$ is data vector and $\theta$ is the parameter vector:  $$ p(\theta|y)=\frac{p(y|\theta)p(\theta)}{p(y)} $$ The numerator on the right handside can be written as: $$ p(y,\theta) $$ which is the joint probability distribution of $y$ and $\theta$, then Bayes' theorem could be written as: $$ p(\theta|y)=\frac{p(y,\theta)}{p(y)} $$ Furthermore, $$  p(\theta|y)\propto p(y,\theta) $$ Am I right on this? I think it does not look right. Because the posterior is the proportional to the joint density function. But where is the mistake?
limsup of intersection of events as a subset of intersection of limsups In QGIS, I want to make polygons from a set of intersecting lines. Looking at the screenshot here, I want chop the purple line (it's a closed line, not a polygon) where it crosses the green and blue lines. The result hopefully would be a polygon where I have hatched in red. I have quite a few to do.    I have tried the QGIS tool previously called Polygonizer (see ), but now called Polygonize. This is a tool that, given a layer containing a set of crossing lines arranged essentially like a # symbol, should create a polygon from the enclosed space in the middle.  (In QGIS 2.01, Polygonize is in the top menu bar: go to Processing -> Commander, then type "Polygonize" on the command line, select "Processing algorithm: Polygonize" from the list that appears, and press Enter.)   At the momemt Polygonize doesn't work properly. Is there another way to do what I want to do?   For example, wouldn't it be great if the QGIS 'Split Feature' command was able to work like MapInfo's 'Polyline Split', where you select the polygons to split, then select an already-drawn polyline to split by, and voila. But Split Feature seems not to work that way.    UPDATE: Detail of problems with running Polygonize.  I have 64-bit QGIS 2.01 installed to Windows 7 from the installer package, QGIS-OSGeo4W-2.0.1-3-Setup-x86_64.exe.  The routine processed an output from my shapefile, but did not create a polygon from the region enclosed by intersections between the blue, green and purple lines. (Note: in the actual input file, all these lines are on the same layer; the screenshot just colours them differently for illustrative purposes). What I got instead was a simple conversion of the purple outline to a polygon, as though the program completely ignored the blue and green lines.  When I node-edited the closed purple line, to open it so that I had a set of open crossing lines like in the # symbol, and then ran Polygonize again, nothing happened. No crash, but no visible polygon in the output file.    When I used Nick Hopton's own  data, I got a crash; see screenshot: Are +1/+1 counters different from +1/+1 effects? How to enable numlock at boot time for login screen? I am trying to force numlock to be on upon initial boot at the login screen on Ubuntu 12.04. The only solutions I have found so far switch numlock on only after initial login.  I'm looking to force numlock to be on when the login screen is displayed, and before the user has logged in. Can anyone assist? Typesetting 144...4 with "n times" under the 4's is easy, but what about \sqrt{144...4}? Why the shape (sB) is unknown in the following code ? How to avoid this ?  \documentclass[]{standalone} \usepackage{tikz,xparse} \usetikzlibrary{calc}  \tikzset{pics/.cd,     % Marque croix     Cross/.style args={#1 and #2}{%         code = {%         \draw[#2,rotate=45,scale=1.4] (0,#1 pt) -- (0,-#1 pt) ;         \draw[#2,rotate=-45,scale=1.4] (0,#1 pt) -- (0,-#1 pt) ;         }     },     Cross/.default={2 and black} }  \NewDocumentCommand{\Segment}{mO{}}{%     \begin{scope}[#2]     \path    (0,0) coordinate (sA) pic {Cross={1.5 and gray}}         -- (#1,0) coordinate (sB) pic {Cross={1.5 and gray}} ;     \draw (sA)--(sB) ; % Can't reuse (sB)     \end{scope} }   \begin{document}     \begin{tikzpicture}     \Segment{1.5}     \end{tikzpicture} \end{document} Confused AUC ROC score I am working on binary classification problem, I try to evaluate the performance of some classification algorithms (LR,Decission Tree , Random forest ...). I am using  a cross validation technique (to avoid over-fitting) with AUC ROC as scoring function to compare the performance of the algorithms, but I am getting a weird results with Random forest and AdbBoost, I have a perfect AUC_ROC score (i.e. =1) despite the fact that the recall(TPR) and FPR of this algorithms are different from 1 and 0 respectively .  def FPR(y_true, y_pred):     tn, fp, fn, tp = confusion_matrix(y_true, y_pred).ravel()     result = fp / (fp+tn)     return result def FNR(y_true, y_pred):     tn, fp, fn, tp = confusion_matrix(y_true, y_pred).ravel()     result = fn / (tp+fn)     return result FPR_scorer = make_scorer(FPR) FNR_scorer = make_scorer(FNR)     def get_CrossValResults2(model,cv_rst,bestIndx):     best=pd.DataFrame.from_dict(cv_rst).iloc[bestIndx]     roc=&quot;{:.12f}&quot;.format(best['mean_test_roc_auc'])     acc =&quot;{:.0%}&quot;.format(best['mean_test_accuracy'])     prec =&quot;{:.0%}&quot;.format(best['mean_test_precision'])     rec =&quot;{:.0%}&quot;.format( best['mean_test_recall'])     f1 =&quot;{:.0%}&quot;.format(best['mean_test_f1'])     r2=&quot;{:.2f}&quot;.format(best['mean_test_r2'])     g_mean=&quot;{:.2f}&quot;.format(best['mean_test_gmean'])     pr_auc=&quot;{:.8f}&quot;.format(best['mean_test_pr'])     fnr=&quot;{:.0%}&quot;.format(best['mean_test_fnr'])     fpr=&quot;{:.0%}&quot;.format(best['mean_test_fpr'])     rst = pd.DataFrame([[ model, acc,prec,rec,fpr,fnr,f1,roc,pr_auc,g_mean,r2]],columns = ['Model', 'Accuracy', 'Precision', 'Recall','FPR','FNR', 'F1-Score','ROC_auc','PR_auc','gmean','r2'])     return rst        cross_val_rst = pd.DataFrame(columns = ['Model', 'Accuracy', 'Precision', 'Recall','FPR','FNR', 'F1-Score','ROC_auc','PR_auc','gmean','r2'])              scoring = {'accuracy':'accuracy','recall':'recall','precision':'precision','fpr':FPR_scorer,'fnr':FNR_scorer,'f1':'f1' ,'roc_auc':'roc_auc','pr':'average_precision','gmean':Gmean_scorer,'r2':'r2'}     param_grid = {'n_estimators': [200],               'max_depth': [80,90],               'min_samples_leaf': [2,3, 4],             'min_samples_split': [2,5,12],             'criterion': [ 'gini'],              'class_weight' : [class_weights], 'n_jobs' : [-1]}  clf = GridSearchCV(RandomForestClassifier(class_weight=class_weights), param_grid, cv=kfold,scoring=scoring,refit=refit)#Fit the model bestmodel = clf.fit(X,Y)       cross_val_rst = cross_val_rst.append(get_CrossValResults2(model='Random Forrest',bestIndx=bestmodel.best_index_,cv_rst=bestmodel.cv_results_),ignore_index=True) How can I launch screensaver with a keyboard shortcut? Non-existence of $\lim \limits_{(x, y) \to (0,0)} \frac{x^3 + y^3}{x - y} $ How to show that $\lim \limits_{(x, y) \to (0,0)} f(x, y)$ does not   exist where,         $$f(x, y) =  \begin{cases} \dfrac{x^3 + y^3}{x - y} \; ; &amp; x \neq y \\  0 \; \;\;\;\;\;\;\;\;\;\;\; ; &amp; x = y  \end{cases}  $$   I tried bounding the value of the function as $(x, y)$ approaches $(0,0)$ but was not successful. I graphed the function and saw that it was actually approaching $0$ but Microsoft Math failed to render some points around the $x = y$ plane. So I imagine I need to find points $(x, y)$ such that $x$ is extremely close to $y$ but $(x, y)$ is not as close to $(0,0)$ to show that the function does not approach $0$ around the origin. But was unable to think of any.   Hints or suggestions would be awesome. Any help is appreciated. 32-bit vs. 64-bit systems What are the differences between 32-bit and 64-bit systems?  If you have used both of them, what kind of sharp differences have you experienced?   Would it be a problem to use 32-bit programs on 64-bit systems in some cases? $L^p$ and $L^q$ space inclusion Let $(X, \mathcal B, m)$ be a measure space.  For $1 \leq p &lt; q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample in the case the condition is not satisfied? Sample query/DDL for producing page locks on a clustered index in SQL Server (without hints) We currently have some indexes where ALLOW_PAGELOCKS is set to off. This presumably was done in order to reduce deadlocks. However I doubt that it would really had an effect back then.   Now I am trying to understand WHEN SQL Server actually chooses to start locking pages rather than keys in a clustered index. I asked Jonathan Keyhaisas recently and he told me that this could happen if I am touching rows on several subsequent pages. However I didn't manage to get any exclusive page locks by updating rows in a clustered index with a sample query.   Could you help me understand page locks better with a sample query and table? I am running SQL Server 2008 SP4.  Thanks in advance  Martin Does a Warlock receive the benefit of their familiar's Magic Resistance trait? The Warlock has access to special forms for their find familiar spell through the Pact of the Chain feature. I'm wondering if these special forms, the Quasit, Imp, and Pseudodragon in particular, allow the player character to share the familiar's Magic Resistance feature. In the Monster Manual each of these creatures has a sidebar that states that the familiar shares its Magic Resistance feature with the companion they are bonded to, but the PHB doesn't mention this in any of the creatures' stat blocks.   For example, in the Variant: Pseudodragon Familiar sidebar on page 254 of the MM it says:     "While the pseudodragon is within 10 feet of its companion,        the companion shares the pseudodragon's Magic Resistance trait."   This feature seems clear in the MM but I'm led to believe that it wasn't intended for Player Characters since mention of it is absent from the stat blocks in Appendix D of the PHB (pages 307-309).  Would a 3rd level warlock who chose one of these familiar forms through the Pact of the Chain feature benefit from advantage from saving throws from spells and other magical effects due to these special forms, or was this feature only meant for powerful NPCs and enemy spellcasters who had made this link with these familiars themselves? Why can I put an electrolytic capacitor on AC? I have seen here and there schematics with electrolytic capacitors put on AC. And this sounds weird to me.  Electrolytic capacitors have a polarity, right? If we invert the polarities on DC, bad things happen. As far as I understand, AC inverts polarity every now and then (commonly 50Hz). Why can we put such capacitors on AC without damaging them?  Example:    from demonstration here: Possible A* Optimizations How to show that $\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0$ My question isn't how they receive the energy to jump, but why.  When someone views an element's emission spectrum, we see a line spectrum which proves that they don't exist outside of their orbitals (else we would see a continuous spectrum).  Electrons can be released in the form of beta decay, thus proving that they are capable of traveling outside of orbitals contrary to the statement my teacher said that they stay within orbitals. Then, to add to the confusion, the older model of rings floating around a nucleus has, from what I can tell, been outdated, which would support this model. My teacher's explanation was that the electrons made a quantum jump of some kind. How do electrons move between orbitals or do we know how they jump, excluding the reason that energy causes them to jump, and why are positrons formed sometimes instead of electrons in Beta decay?  When I'm asking "how do electrons jump" I would like to know how an electron can jump between each orbital such as how it moves and how it knows where to jump since it appears to be a jump where the electron doesn't slow into a orbital position. Specifically how they jump what is this Atomic electron transition, I understand that they jump and that they do this through absorbing and releasing energy but what is this Atomic electron transition other than what is already on the wikipedia article . Calculation of subgroups of $(Z_{12}, +)$ How calculate all subgroups of $(Z_{12}, +)$?  I know that the order of subgroups divide the order of the group, but there is such a smart way to calculate the subgroups of order 6? Analysing the "kind(s) of problem(s) are [ones]/[to be]" construction A. These kinds of problem are to be avoided. B. These kind of problems are to be avoided.  Are both A and B grammatical? If not, why not? If so, after having noticed the parallelism between A and C, and between B and D, could we say that C and D are &quot;standard&quot; English, too?  C. These kinds of problem are ones to avoid. D. These kind of problems are ones to avoid. I have two Ubuntu Software Centers, one can be opened, another one can't