From a very general point of view, when you have a dataset $X$ and want to predict a label $y$, what is the purpose of beginning with a PCA (principal component analysis) first, and then doing the prediction itself (with logistic regression, or random forest or whatever) from both intuitive and theoretical reason? In which case can this improve the quality of prediction?
Suppose I am running a regression $Y \sim X$. Why by selecting top $k$ principle components of $X$, does the model retain its predictive power on $Y$?   I understand that from dimensionality-reduction/feature-selection point of view, if $v_1, v_2, ... v_k$ are the eigenvectors of covariance matrix of $X$ with top $k$ eigenvalues, then $Xv_1, Xv_2 ... Xv_k$ are top $k$ principal components with maximum variances. We can thereby reduce the number of features to $k$ and retain most of the predictive power, as I understand it.   But why do top $k$ components retain the predictive power on $Y$?   If we talk about a general OLS $Y \sim Z$, there is no reason to suggest that if feature $Z_i$ has maximum variance, then $Z_i$ has the most predictive power on $Y$.   Update after seeing comments: I guess I have seen tons of examples of using PCA for dimensionality reduction. I have been assuming that means the dimensions we are left with have the most predictive power. Otherwise what's the point of dimensionality reduction?
Forgive me for the unclear title. Some time ago, my HTC M8, which is rooted and runs on a  showed something weird. Firstly, in the Google Maps App a mysterious blue dot appeared, that captures the screen completely and makes it impossible to quit or use the application. It looks like this (I could make a screenshot)     When this happens in Maps, the only thing I can do is a hard restart with Power+Volume Up.  Since I have seen this occasionally in the camera app as well, I suspect it is not a bug of Maps itself. I know that the Venom ROM is not standard but maybe this is connected to some setting and anyone has seen this. Btw, the Venom ran smoothly for almost 3 years now.  Question: Has someone a clue what this could be and how I can get rid of it? The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \&amp; \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$   Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false. The implication [False $\iff$ False] is vacuously true.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.  $\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$  To prove:  $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$   $\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.  Hence assume that $k + 1 ≥ 3.$ We now consider three cases:  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k    \iff F_k$ odd &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$  Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.  $\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$  $\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?   $\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here. So shouldn't $3$ base cases be checked?   $\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?   $\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?  I referenced . Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al    Supplement to peterwhy's Answer:  $\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$  for both directions of the proof? How to install Wi-Fi driver for Realtek RTL8821CE on Ubuntu 18.04? I'm trying to install drivers for my Wifi on my HP All-in-one desktop. I've been told to show the output of the command "sudo lshw -C network" so here it is:  *-network description: Ethernet interface product: RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller vendor: Realtek Semiconductor Co., Ltd physical id: 0 bus info: pci@0000:01:00.0 logical name: enp1s0 version: 15 serial: 48:ba:4e:5c:06:5f size: 100Mbit/s capacity: 1Gbit/s width: 64 bits clock: 33MHz capabilities: pm msi pciexpress msix bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation configuration: autonegotiation=on broadcast=yes driver=r8169 driverversion=2.3LK-NAPI duplex=full firmware=rtl8168h-2_0.0.2 02/26/15 ip=192.168.1.21 latency=0 link=yes multicast=yes port=MII speed=100Mbit/s resources: ire:31 import:e000(size=256) memory: fea04000-fea04fff memory:fea00000-fea03fff  *-network UNCLAIMED description: Network controller product: Realtek Semiconductor Co., Ltd. vendor: Realtek Semiconductor Co., Ltd. physical id: 0 bus info: pci@0000:02:00.0 version: 00 width: 64 bits clock: 33MHz capabilities: pm msi pciexpress cap_list configuration: latency=0 resources: import:d000(size=256) memory:fe900000-fe90ffff   Output from "lspci -nnk | grep -A2 0280":  02:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8821CE 802.11ac PCIe Wireless Network Adapter [10ec:c821] Subsystem: Hewlett-Packard Company RTL8821CE 802.11ac PCIe Wireless Network Adapter [103c:831a] 03:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. RTS5229 PCI Express Card Reader [10ec:5229] (rev 01) Why are the dineutron and diproton unbound? My router can do port-forwarding based on MAC addresses. That is, a specific MAC will get a specific IP, for which I can configure a set of ports to be forwarded. In order to easily change that set of ports, I'd like to have different connections in the Network manager.  How do I change the MAC address for a network connection? This week I saw the definition of a measurable set for an outer measure.     Let $\mu^*$ be an outer measure on a set $X$. We call $A \subseteq X$ measurable if $$\mu^*(E) = \mu^*(A\cap E) + \mu^*(A^c\cap E)$$ for every $E \subseteq X$.   This is not the first time I've seen this definition. Unlike most other things in mathematics, over time I have gained absolutely no intuition as to why this is the definition.   The only explanation I've ever seen is that a set is measurable if it 'breaks up' other sets in the way you'd want. I don't really see why this is the motivation though. One reason I am not comfortable with it is that you require a measurable set to break up sets which, according to this definition, are non-measurable; why would you require that? Of course, you can't say what a non-measurable set is without first defining what it means to be measurable so I suppose no matter what your condition is, it will have to apply to all subsets of $X$.     Is there an intuitive way to think about the definition of measurable sets? Is there a good reason why we should use this definition, aside from "it works"? how to scale math font only There has been a trend now two persons downvote it then it goes on like that. So let there be down voting but do not display the downvotes till the user votes on it. Else the user reads it with an impression that the question is not good.  Am I missing something? I'm trying to solve this problem from Hatcher 3.3.24.  Let $M$ be a closed connected 3-manifold, and write $H_1(M, \mathbb{Z})$ as $\mathbb{Z^r} \oplus T$ where $T$ is torsion. Show that $H_2(M, \mathbb{Z})$ is $\mathbb{Z^r}$ if $M$ is orientable or $\mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$ if $M$ is not orientable. In particular $r \geq 1$ if $M$ is not orientable. Using exercise 6 construct examples showing there are no other restrictions on the homology groups of closed 3-manifolds. (In the nonorientable case consider the manifold $N$ obtained from $S^2 \times I$ by identifying $S^2 \times \{0\}$ with $S^2 \times \{1\}$ via a reflection of $S^2$.)  I think I've solved the first part, here's my proof:   First note the following theorem: If $M$ is a closed connected and orientable n-manifold then $H_n(M, \mathbb{Z}) = \mathbb{Z}$ and $Tors(H_{n-1}(M, \mathbb{Z})) = 0$. If nonorientable, then $H_n(M, \mathbb{Z}) = 0$, $Tors(H_{n-1}(M, \mathbb{Z})) = \mathbb{Z_2}$.  Now note by Poincare duality and UCT we have that $H_2(M) \cong H^1(M) \cong \mathbb{Z^r} \oplus T_0 \cong \mathbb{Z^r}$.   In the nonorientable case consider the exact sequence:  $0 \longrightarrow H_2(M) \longrightarrow H_1(M) \longrightarrow H_0(M) \longrightarrow 0$  But we know $H_1(M) = \mathbb{Z^r} \oplus Z_2$ from the earlier theorem and $H_0(M) = \mathbb{Z}$.   Therefore $H_2(M) \cong ker(\mathbb{Z^r} \oplus Z_2 \rightarrow \mathbb{Z}) \cong \mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$.   For the rest of the question I'm quite lost. Is there a single word for "moving unhinderedly"? For a Google account, you are allowed to sync Contacts and People details. What is the difference between these two? What prevents photons from getting mass from higher order Feynman diagrams The Higgs boson and gluons have no electric charge and photons couple to charge, so there is no tree level interaction between them and photons.  But what prevents higher order diagrams from contributing a non-zero mass term to the photon, for instance    where a photon couples to some fermion (say an electron, or a top quark) which can interact with the Higgs field.   Or consider that same diagram but with quarks and a gluon interacting between them? Or any higher diagram with even more loops?  I have heard charge conservation depends on gauge invariance, which in turn depends on photons being massless.  So it appears the photon has no mass, and these diagrams must all cancel somehow.  So I'm hoping there is a very nice symmetry explanation for why they all disappear, but if I say "because of gauge invariance" that would be circular logic, so there must be another symmetry at stake here?     What prevents photons from obtaining a mass from high-order self-energy loop diagrams? Do I need a Turkey transit visa for a layover? I have an Indian passport and am planning to visit Germany with a valid visa. To book a flight for Germany (Mumbai – Dusseldorf), I came across some connected flights that stop at Istanbul.  Do I need to have visa for Turkey?  This is the first time I am travelling outside my country, and any information on this will be helpful. Here are two examples of what I mean:   I am sure you all too have experienced this frustration before. A dialog box opens that displays only part of a long string (e.g. a file path), but won't allow you to select the text so you can read the full string. Neither text container will resize upon window resizing. (By the way, you can force windows —specially dialog boxes— to resize with tools such as , but helps little in this case, since controls are fixed-width.) The text is hidden, period. Copying a very long drop-down list control contents to a text file.   Do you know any way to solve these problems? Page layout assignments with XML metadata API How do I define page layout assignments for different record types via metadata XML files? How to use the Bradhley Hand font? I've checked the  that can be used through beamer class. I've successfully used most of the fonts, but when I tried to include the Bradhley Hand font, it did not work (it shows ordinary characters)?! How can I use it?  For example, when I use the following :  \documentclass[serif]{beamer}  \usepackage{mathpazo}  \renewcommand{\rmdefault}{ibh}  \begin{document} \begin{frame} $\frac{ \sqrt{ \mu(i)^{ \frac{3}{2}} (i^{2} -1) } } { \sqrt[3]{\rho(i) - 2} + \sqrt[3]{\rho(i) - 1} } $ \end{frame} \end{document}   it gives: I am a photography newbie and have only a 18 - 135mm canon kit lens at my disposal. I wanted to take macro shots but can't really afford a macro lens right now. One very optimistic goal of mine is to capture a single snow flake this winter.   Is there any cheap(er) way to take macro shots? Also, any specific advice from anyone one how to go about photographing a single snow flake. Solve a congruence $1978^{20}\equiv x\pmod{125}$ Laws of addition of Vectors limit of Holder norms: $\sup\limits_{x\in [a,b]} f(x) = \lim\limits_{n\rightarrow\infty} \left(\int_a^b (f(x))^n \;dx\right)^{\frac{1}{n}}$