If $Y|X \sim N(0,1)$, does that mean $Y$ itself is also $N(0,1)$? If $Y|X \sim N(0,1)$, does that mean $Y$ itself is also $N(0,1)$? It looks like it as I can write $$ E[Y] = E[E[Y|X]] = E[0] = 0, $$ and $$ Var[Y] = Var[E[Y|X]] + E[Var[Y|X] = Var[0] + E[1] = 1.  $$ But although I get the correct expectation and variance I don't see how we can say $Y$ is normally distributed? If it is true, does this only hold because we are dealing with the normal distribution?
If $Y|X \sim \mathcal{N}(0,1)$ then is $Y^2|X \sim \chi^2(1)$? Suppose we have random variables $X$ and $Y$ such that $Y|X \sim \mathcal{N}(0,1)$. Can we then say that $Y^2|X \sim \chi^2(1)$? If we can, then what about when $Y|X \sim \mathcal{N}(0,\sigma^2/4)$, can we use the results of  to say that $Y^2|X \sim \frac{\sigma^2}{4}\chi^2(1)$? Essentially I want to know if the results of squaring a normal distribution can be applied to the case where a random variable is conditionally normally distributed?
How to revert a do-release-upgrade? I can update my Ubuntu to the new LTS version. As you can think, its not a highly important server since I run 13.10 (not a LTS version). Anyway - I actually want to upgrade my server, but would like to be able to revert this upgrade.   I would be able to reinstall Ubuntu 13.10 and overwrite it with my backup. But I want to know if there is a more comfortable way to do this. TBH I guess theres no command as undo-release-upgrade but I thought it would be worth to ask for it.  This is my current daily backup script  tar cvpzf $backup_file --exclude=/proc --exclude=/lost+found --exclude=$backup_file --exclude=/mnt --exclude=/sys --exclude=/media --exclude=/tmp / This morning I upgraded my system to Windows 10 2004, the upgrade went smooth and was through Windows Update. I am using Hyper-V as my virtualization program for virtual machines and to provide extra layer of protection to the system through core isolation. After the upgrade, I noticed that a new Virtual Switch was created with the name &quot;vEthernet (Ethernet)&quot; where &quot;Ethernet&quot; within the parentheses is the name of my primary network adapter. Also noticed this behavior has a specific pattern of creating new vSwitches correspond to the adapters you have, e.g. if you created a new Virtual Switch and restarted the system, a new additional switch will appear with the name &quot;vEthernet (your-new-switch-name)&quot;. Tried to remove Hyper-V and reinstall it but no luck. I am aware of a similar  on SuperUser but has no answers on it. Any clues for the root of the problem? The bevel depth (the radius of the tube) is set to 9cm but it actually isn't when compared to the background grid. This is is confirmed by importing it into sketchup and scaling it to the proper units it turns out the radius is only around 2.286cm. This problem only shows up on certain projects. What's wrong and how do I fix it?  (sorry that I can't upload the file, it's a long story) What size images do I need for common print sizes, such as wallet size, 8x10, and 16x20?  Is there a general formula?  How could that size change if, for example, a 16x20 was printed on canvas rather than photopaper? Badge "multiplier" a little bit misaligned on the profile page Well, you can see it for yourself. The badge multiplier is a bit misaligned. Can you fix this please? (I know it's kinda a minor issue but it's bugging me and it's somehow off)    If it helps, I'm using Chrome version 44.0.2403.125 m. (Also repro'ed on FireFox 39.0 and )    For no good reason, that's a screenie from 's profile page. Detect whether Excel workbook is already open I encountered this particular combinatorial identity :$$\begin{align}&amp;1\dbinom{n-1}{r-1}+2\dbinom{n-2}{r-1}+\cdots+(n+1-r)\dbinom{r-1}{r-1}\\=&amp;\dbinom{n}{r}+\dbinom{n-1}{r}+\cdots+\dbinom{r}{r}\\=&amp; \dbinom{n+1}{r+1}\end{align}$$How does one figure it out? Do ground seeds contain oil? Proof of "the continuous image of a connected set is connected" None of the existing questions is exactly answering my question so I'm posting a new question, but feel free to refer me to some already answered question!   In Rudin Theorem 4.22, we know that      If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, and $E$ is a connected subset of $X$, then $f(E)$ is connected.    In the proof, we started with consider $f(E) = A \cup B$, where $A$ and $B$ are nonempty separated subsets. Then put $G = E \cap f^{-1}(A)$ and $H = E \cap f^{-1}(B)$. Then Rudin is claiming that $E = G \cup H$. I'm a little suspicious about this. What if $f$ is non-surjective, then $f^{-1}(A) \cup f^{-1}(B)$  is only a proper subset of $E$? Is there a property of $f$ being continuous that forces $f$ to be 1-1? "Has been" vs "was" Would you please explain the difference between these two sentences?     Spain ceded (or gave) Puerto Rico to the United States. Since then, Puerto Rico has been a U.S. territory.      Spain ceded(or gave) Puerto Rico to the United States. Since then, Puerto Rico was a U.S. territory.   I am stuck between 'has been' and 'was' in the above sentences. Topology induced by seminorms and initial topology Let's say we have a family of seminorms $(\rho_\alpha)_{\alpha \in A}$ on a vector space $V$.  There are two ways to topologize $V$ using those seminorms:   We define topology $\mathcal S$ by a subbasis consisting of sets of the form $$ B_{x, \alpha, \epsilon} =\{v \in V : \rho_\alpha(v - x) &lt; \epsilon \} $$ for some $\epsilon &gt; 0, x \in V, \alpha \in A$. We define topology $\mathcal I$ as the smallest topology in which the seminorms are continuous.   I've showed that the seminorms are continuous in $\mathcal S$, and that  implies $\mathcal I \subseteq \mathcal S$. I'm having trouble with the other direction. It is clear to me that $B_{0, \alpha, \epsilon} = \rho_\alpha^{-1}  ((-\epsilon, \epsilon ))$, so it's enough to show that vector addition is continuous in respect to $\mathcal I$, but I don't know how to do that. Black flakes from new cast iron pan I bought a brand new Lodge 12" cast iron pan, washed it in soap and warm water and dried it with a paper towel thoroughly. Then I applied a thin layer of canola oil with a paper towel (paper towel was clean - not black after this application) and then heated the empty pan on my electric stove for 15-20 minutes. The oil on the pan smoked a bit, some parts of the pan turned slightly ashy (grey). Mostly everything looked OK.   After I turned it off and cooled the pan, I now take a clean paper towel and rub the surface and I see black flakes! What is this? Is this the oil broken down? Can I cook in the pan or wash it with soap to eliminate the black flakes? Once washed and dried should I rub canola oil again? This is my first experience with cast iron. Any pointers will help. Given the functional dependencies of a relation is there more then one way to normalize it? What is the difference between the words which and that?  For example I have the following sentences:   “I have a car which is blue.” “I have a car that is blue.”   Are there any rules specifying usage of which and that? Is there any way to use a nether portal to travel to y = 250? Show that $$\sum_{k=0}^n(-1)^k\binom nk=0$$ So for odd $n$ we have an even number of terms. So $\binom nk=\binom n{n-k}$ which have opposite signs. Thus the sum is 0.  For even $n$ we have that $$\sum_{k=0}^n(-1)^k\binom nk= \binom n0+\sum_{k=1}^{n-1}(-1)^k\binom nk+\binom nn$$ Now $$\sum_{k=1}^{n-1}(-1)^k\binom nk= \sum_{k=1}^{n-1}(-1)^k\left[\binom{n-1}k+\binom{n-1}{k-1}\right]$$ What would that sum be in the square brackets? I was thinking of the noted 1980 paper by Sidney Coleman and Frank de Luccia--"Gravitational effects of and on vacuum decay"-- about metastable vacuum states that could tunnel to a lower energy "true vacuum" with catastrophic results. I suspect the answer is that if our vacuum is a false one, we wouldn't "see" the true one til it hit us. I have from Exercise 1 pg 33 - "Algebra - T. W. Hungerford" following:      Question to proof: Show by example that the first conclusione may be false if $G,\, H$ are monoids that are not group.   I thinked:     Proof: let be for example two monoids $(G:=\{\emptyset\}, \bot)$ and $(H:=\{\emptyset, \{\emptyset\}\}, \top)$ with: $$\begin{array}{c|c} \bot &amp; \emptyset \\ \hline \emptyset &amp; \emptyset \end{array} \, \, \, \,\, \,\, \,\, \,\, \,\, \,\, \,\, \,\, \, \begin{array}{c|cc} \top &amp; \emptyset &amp; \{\emptyset\}\\ \hline \emptyset &amp; \emptyset &amp; \emptyset \\ \{\emptyset\} &amp; \emptyset &amp; \{\emptyset\} \end{array}$$ $G$ with $e_G:=\emptyset$ like neutral element and $H$ with $e_H:=\{\emptyset\}$ like neutral element. Let be $f$ a function from $G$ to $H$ with $\emptyset \mapsto \emptyset$ (namely $x \mapsto x$), the function is Homomorphismus in fact $$ \begin{align} {\color{Red}f}{\color{Red}(}{\color{Red}x}{ \color{Red}\bot }{\color{Red}y}{\color{Red})}=&amp;f( \emptyset \bot \emptyset)= {\color{Green}f}{\color{Green}(}{\color{Green}\emptyset}{\color{Green})}= \emptyset \\ {\color{Red}f}{\color{Red}(}{\color{Red}x}{\color{Red})} {\color{Red}\top} {\color{Red}f}{\color{Red}(}{\color{Red}y}{\color{Red})}=&amp;f(\emptyset) \top f(\emptyset)= \emptyset \top \emptyset= \emptyset \\ &amp;\text{with }x,y \in G \end{align}$$ but it means that $$ \begin{align} f(e_G)&amp;\neq e_H \\f(e_G)={\color{Green}f}{\color{Green}(}{\color{Green}\emptyset}{\color{Green})}=\emptyset &amp;\neq \{\emptyset\}= e_H \end{align}$$ Therefore $f$ ist Semigroup-Hom but not Monoid-Hom...   Is it correct? Pugh, "Mathematical Analysis", exercise 2.17:     Assume $f : M \to N$ is a map from one metric space to another which satisfies the following condition: for every convergent sequence $(a_n) \subset M$, the sequence $(f(a_n)) \subset N$ converges. Prove that $f$ is continuous.   (I know that if $f$ commutes with limits then the answer is affirmative, but I can't prove it with that hypothesis). Word for in-house jargon