Assume we have two continuous Normal RV "X" and "Y". how can I show the conditional PDF f(X|Y) and f(Y|X) is Normal?
Somebody asked me this question in a job interview and I replied that their joint distribution is always Gaussian. I thought that I can always write a bivariate Gaussian with their means and variance and covariances. I am wondering if there can be a case for which the joint probability of two Gaussians is not Gaussian?
How does vendor restocking work in Fallout 4? Choosing between $z$-test and $t$-test Background: I'm giving a presentation to colleagues at work on hypothesis testing, and understand most of it fine but there's one aspect that I'm tying myself up in knots trying to understand as well as explain it to others.  This is what I think I know (please correct if wrong!)   Statistics that would be normal if variance was known, follow a $t$-distribution if the variance is unknown CLT (Central Limit Theorem): The sampling distribution of the sample mean is approximately normal for sufficiently large $n$ (could be $30$, could be up to $300$ for highly skewed distributions) The $t$-distribution can be considered Normal for degrees of freedom $&gt; 30$   You use the $z$-test if:   Population normal and variance known (for any sample size) Population normal, variance unknown and $n&gt;30$ (due to CLT) Population binomial, $np&gt;10$, $nq&gt;10$   You use the $t$-test if:   Population normal, variance unknown and $n&lt;30$ No knowledge about population or variance and $n&lt;30$, but sample data looks normal / passes tests etc so population can be assumed normal   So I'm left with:    For samples $&gt;30$ and $&lt;\approx 300$(?), no knowledge about population and variance known / unknown.   So my questions are:   At what sample size can you assume (where no knowledge about population distribution or variance) that the sampling distribution of the mean is normal (i.e. CLT has kicked in) when the sampling distribution looks non-normal? I know that some distributions need $n&gt;300$, but some resources seem to say use the $z$-test whenever $n&gt;30$... For the cases I'm unsure about, I presume I look at the data for normality. Now, if the sample data does looks normal do I use the $z$-test (since assume population normal, and since $n&gt;30$)? What about where the sample data for cases I'm uncertain about don't look normal? Are there any circumstances where you'd still use a $t$-test or $z$-test or do you always look to transform / use non-parametric tests? I know that, due to CLT, at some value of $n$ the sampling distribution of the mean will approximate to normal but the sample data won't tell me what that value of $n$ is; the sample data could be non-normal whilst the sample mean follows a normal / $t$.  Are there cases where you'd be transforming / using a non-parametric test when in fact the sampling distribution of the mean was normal / $t$ but you couldn't tell? Just need help understanding this question, as I don't really understand what a continuous random vector is. My limited understanding is that a continuous random vector must be completely continuous (so for continuous X and Y this is satisfied) and that to get the probability of the random vector occurring, we double integrate over the supports (of X and Y obviously). My guess is then, that the c.r.v (X,Y) IS a continuous random vector.  Also, what of the converse? If (X,Y) is a continuous random vector, then are X and Y continuous random variables? I also think this is correct, as the continuity of the vector MUST imply the continuity of X and Y, although I'm honestly not sure.  Thanks Wrap Caption Around TikZ Figure How to use the Bevel node in 2.79b What geometric object is given by this equation? When is it correct to use the "-wise" suffix? Show group of order $4n + 2$ has a subgroup of index 2. Let $n$ be a positive integer. Show that any group of order $4n + 2$ has a subgroup of index 2. (Hint: Use left regular representation and Cauchy's Theorem to get an odd permutation.)  I can easily observe that $\vert G \vert = 2(2n + 1)$ so $2 \mid \vert G \vert$ and 2 is prime. We have satisfied the hypothesis of Cauchy's Theorem so we can say $G$ contains an element of order 2.  This is where I am stuck  I am confused about how the left regular representation relates to the group. So my understanding at this point is that every group is isomorphic to a subgroup of some symmetric group. My question: is the left regular representation $\varphi : G \to S_G$ an isomorphism? where $G \cong S_G$ or is $S_G$ the same thing as $S_{\vert G \vert}$ and $\varphi$ is only an injection? I'm using Dummit and Foote for definitions.  I saw an argument online that said that since we have an element of order 2, there is a $\sigma \in S_G$ of order 2, but it is a product of $2n + 1$ disjoint 2-cycles. I don't understand how they could claim this and tried working it out on my own but didn't get there. -- They then went on to use the parity mapping $\varepsilon : S_G \to \{\pm1\}$ and since we have an odd permutation $\sigma$, we have $[S_G:\text{ker }\varepsilon] = 2$. I understood their computation but not how that directly shows that $G$ has a subgroup of order 2? unless $G \cong S_G$ because of how left regular representation is defined. (but again, I'm not understanding that concept very well yet.)  So, to be clear about my questions:   What is meant by left regular representation, is it an isomorphism or just an injection? and how would it be used here. If it is an isomorphism, the argument online starts to make more sense, but how can they say that since $\sigma$ is even, it is made up of $2n + 1$ disjoint transpositions? If you have a full, proof, I'd appreciate it, but good hints are just as good!   Thanks you! Is it possible to restore a MySql database only using files from Data directory? When I login, nothing happens.   I am presented with my desktop wallpaper.    No Dash, no Launcher, nothing. You can also contact me to customize your template, or create a new unique one.   Do I need to put a comma between the new and unique adjectives in this sentence?  In my native language (if I am not mistaken) we don't put a comma if the adjectives are of a different type. Not sure about English though, and if it's the case here anyway. Any help would be appreciated! Does a Vietnamese citizen need a transit visa for South Korea? I am a Vietnamese passport holder. I'm going back to Viet Nam from Montreal this December, and I will transit at Incheon Airport for a few hours. Do I need a Korean visa just for transiting, either from Montreal to Hanoi or from Hanoi back to Montreal? using gedit through ssh connection - should I have gedit installed on the server? I wanted to try out using graphical tools over ssh. Just to learn a new thing :)  so my understanding is this, I connect like this: ssh -port -x user@server  But I'm running ubuntu server so it has no gui and thus no gui tools. My understanding is I've to install gedit on server. But also xserver?  Can someone clear up this concepts to me? Ubuntu in VirtualBox wants to install on every boot I have installed Ubuntu 11.10 using Virtual Box in my Windows XP. I have use Ubuntu 11.10 iso to install it. First time when I installed every thing was fine but when I shutdown the VirtualBox and start Ubuntu again, it asks me to install Ubuntu and try Ubuntu without installing.  Do I need to do some extra settings? Build a basic Python iterator How would one create an iterative function (or iterator object) in python? Given this (very) tricky determinant, how can we calculate it easily?  $$\begin{pmatrix} \alpha + \beta &amp; \alpha \beta &amp; 0 &amp; ... &amp; ... &amp; 0 \\ 1 &amp; \alpha + \beta &amp;  \alpha \beta &amp; 0 &amp; ... &amp; 0 \\ 0 &amp; 1 &amp; \alpha + \beta &amp; \alpha \beta &amp; ... &amp; ... \\ ... &amp; ... &amp; ... &amp; ... &amp; ... &amp; 0 \\ ... &amp; ... &amp; .... &amp; ... &amp; ... &amp; \alpha \beta \\ 0 &amp; 0 &amp; 0 &amp; ... &amp; 1 &amp; \alpha + \beta   \\ \end{pmatrix} \in M_{n\times n}$$  EDIT:     I have to prove it is equal to $\frac{{\alpha}^{n+1} - {\beta}^{n+1}}{\alpha - \beta}$   Any help is appreciated, I just could not find a trick to ease it up! Book series where they collect either stones or gems I'm looking for a fantasy book I read about 10 years ago. I don't remember a lot of it, but there are a few things I still know:   The setting was in medieval times. There was no traveling dimensions. I can't recall whether there was magic or not. It wasn't on Earth, but on some invented world. I believe it was a trilogy, with a focus on the son of some nobility. The goal for the characters was to collect a certain number of stones or gems (I believe there were 5 of them), all in certain colours. Every stone has a certain power attached to it. I believe it started out with one or two characters, but the group grew and every character had a certain colour stone at the end. What is important about the gems is that they are unique. It's not part of a bigger magic system, but they are artefacts from somewhere.   There are some minor incidents I can recall vividly:   They build up a fellowship with some stereotypical fantasy builds: an old school fighter, a minstrel, a female archer and probably more. At one point, they mention that if one of the guards in a country falls asleep on duty, he will be kept awake by his garrison/brothers in arms for seven days. At some point the minstrel falls asleep, and the old fighter teaches him how to stay awake. They travel through a forest full of leeches and other insects to reach one of the stones. Also, I seem to recall there was a member of an underground evil organisation among the nobility, who tried to poison someone and were able to do that because they built up an immunity to the poison.   Does anyone has a clue what this series might be?   Suggestions by kind people, but not what I'm looking for, are:   The Seventh Tower Deltora Quest Diadem Westria Sword of Heavens series Cannot see files in my external HDD Where's the error in this $2=1$ fake proof? I'm reading Spivak's Calculus:  2 What's wrong with the following &quot;proof&quot;? Let $x=y$. Then $$x^2=xy\tag{1}$$ $$x^2-y^2=xy-y^2\tag{2}$$ $$(x+y)(x-y)=y(x-y)\tag{3}$$ $$x+y=y\tag{4}$$ $$2y=y\tag{5}$$ $$2=1\tag{6}$$  I guess the problem  is in $(3)$, it seems he tried to divide both sides by $(x-y)$. The operation would be acceptable in an example such as:  $$12x=12\tag{1}$$ $$\frac{12x}{12}=\frac{12}{12}\tag{2}$$ $$x=1\tag{3}$$  I'm lost at what should be causing this, my naive exploration in the nature of both examples came to the following: In the case of $12x=12$, we have an imbalance: We have $x$ in only one side then operations and dividing both sides by $12$ make sense. Also, In  $\color{red}{12}\color{green}{x}=12$ we have a $\color{red}{coefficient}$ and a $\color{green}{variable}$, the nature of those seems to differ from the nature of $$\color{green}{(x+y)}\color{red}{(x-y)}=y(x-y)$$ It's like: It's okay to do the thing in $12x=12$, but for doing it on $(x+y)(x-y)=y(x-y)$ we need first to simplify $(x+y)(x-y)$ to $x^2-y^2$. Possible Duplicate:          Proof to the formula  $$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$