How can I prove that $X+Y$ is a Poisson process with parameter $\lambda_X+\lambda_Y$? For 2 independent Poisson processes $X,Y$, with parameters $\lambda_X, \lambda_Y$ respectively, how can I prove that $X+Y$ is a Poisson process with parameter $\lambda_X+\lambda_Y$?  To do this, I suppose you would use the rule $$ P(X+Y = k) = \sum_{n=0}^k P(X = n) \cdot P(Y = k-n) \\ = \sum_{n=0}^k\frac{\lambda_X^n e^{\lambda_X}}{n!} \cdot \frac{\lambda_Y^n e^{\lambda_Y}}{(k-n)!} $$  However, I am unsure of how to continue from here. Can anyone prompt me in the right direction?
Poisson Distribution of sum of two random independent variables $X$, $Y$ $X \sim \mathcal{P}( \lambda) $ and $Y \sim \mathcal{P}( \mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \sim \mathcal{P}( \lambda + \mu)$ but I don't understand how to derive it.
Is it feasible to transform each variable differently while doing multiple regression I have a dataset with 10 variables ...is it feasible to transform each variable differently while doing multiple regression...  for example new_V1 = log(v1)  New_V2= V2^2  New_V3= 1/V3  Likewise differently for different variables and then applying multiple regression? What is the variance of Y = AX where A is a matrix? Here $X,Y$ are vector of length $n$ and $A$ is  a $n\times n$ matrix. Suppose the covariance matrix $D(X)$ is known? I want to express $$\sum_{n=-\infty}^\infty \dfrac{1}{(z+n)^2+a^2}$$ in closed form. What comes to mind is the formula $$\pi\cot\pi z = \dfrac{1}{z}+\sum_{n\ne 0}\left(\dfrac{1}{z-n}+\dfrac1n\right)=\dfrac{1}{z}+\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}$$  and also $$\dfrac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty \dfrac{1}{(z-n)^2}.$$ But neither of these gives the term $(z+n)^2+a^2$ that we want. Perhaps we can adjust somehow? I would like to manipulate the HTML inside an iframe using jQuery.  I thought I'd be able to do this by setting the context of the jQuery function to be the document of the iframe, something like:  $(function(){ //document ready     $('some selector', frames['nameOfMyIframe'].document).doStuff() });   However this doesn't seem to work. A bit of inspection shows me that the variables in frames['nameOfMyIframe'] are undefined unless I wait a while for the iframe to load. However, when the iframe loads the variables are not accessible (I get permission denied-type errors).  Does anyone know of a work-around to this? Remove association with blank file extension? I want to change the Message-ID header or better yet, discard it. I know everyone is going to ask "Why would you want to do that?" because I know it sounds weird to even write it.  Well I am developing a system that saves an incoming e-mail in EML format, changes it and sends it to the user, at some point the user can request to have the original EML sent to him/her. The issue is that when I re-send the EML it has the same Message-ID header which causes exchange servers to drop the re-send.  Of course I am hoping there is some option through postfix to do this, otherwise I will need to parse and change the EML but I rather not do that.  I tried to do this with "header_checks" and it works BUT I need to be able to do this conditionally, meaning only on a specific occasion, which I can identify by looking at the other headers, I want to remove the Message-ID. From :      ‘We all believe, up at the Lodge, Rachael, that he will be freed from suspicion, sooner or later.’    ‘The better I know it to be so believed there, my dear,’ said Rachael, ‘and the kinder I feel it that you come away from there, purposely to comfort me, and keep me company, and be seen wi’ me when I am not yet free from all suspicion myself, the more grieved I am that I should ever have spoken those mistrusting words to the young lady. And yet I— ’   We usually use should have to say that something was not done, despite being desirable. But in this example, the mistrusting words were indeed spoken (earlier in the book). How would one explain the use of the should have construction then?   Would the following be fully equivalent to the above quote:     ‘We all believe, up at the Lodge, Rachael, that he will be freed from suspicion, sooner or later.’    ‘The better I know it to be so believed there, my dear,’ said Rachael, ‘and the kinder I feel it that you come away from there, purposely to comfort me, and keep me company, and be seen wi’ me when I am not yet free from all suspicion myself, the more grieved I am that I have spoken those mistrusting words to the young lady. And yet I— ’ Optimization used in latex I run a large bulletin board and I get spammers every now and again. My moderation team does a good job filtering them out but every time I IP ban them they seem to come back (I'm pretty sure it's the same person on some occasions, as the post patterns are exactly the same as are the usernames) but I'm afraid to ban them by IP address every time.  If they are on a dynamic IP address, I could be banning innocent users later down the line when they try to get to my forum through SERPs, but if I ban only via static IPs I know that I'm only banning that one person.  So, is there a way to properly determine if an IP address is static or dynamic?  Thanks. What does 'per unit frequency' refer to? We were discussing Kirchoff's theorem on blackbody radiation in class today and the equation was: $$e_{f}= J(f,T)$$  $e_f$ was defined as power per unit area per unit frequency. What does per unit frequency refer to here? I know what per unit time means but don't understand 'per unit frequency'. All operating systems freeze sometimes, and Ubuntu is no exception. What should I do to regain control when...   just one program stops responding? nothing at all responds to mouse clicks or key presses? the mouse stops moving entirely? I have an    In what order should I try various solutions before deciding to pull the power plug?  What should I do when starting up Ubuntu fails?  Is there a diagnostic procedure I can follow? folder content doesn't get refreshed automatically in ubuntu 14.04 I am given this problem: let $a\ge0$,$b\ge0$, and the sequences $a_n$ and $b_n$ are defined in this way: $a_0:=a$, $b_0:=b$ and $a_{n+1}:= \sqrt{a_nb_n}$ and $b_{n+1}:=\frac{1}{2}(a_n+b_n)$ for all $n\in\Bbb{N}$    To prove is that both sequences converge and that they have the same limit. I don't know how to show this. I have spent 2 hours on this, no sign of success strictly positive elements in $C^*$-algebra Let $A$ be a $C^*\text{-algebra}$ and $A_+$ denote the positive elements. An element $a\in A_+$ is called strictly positive if $\overline{aAa}=A$. Want to find the following:a)What are the strictly positive elements of $C_0(\Omega)$ where $\Omega$ is locally compact Hausdorff space, and $C_0(\Omega)$ is the space of continuous complex valued functions that vanishes at $\infty$.b)If $A$ is unital, then $a\in A_+$ is strictly positive iff $a$ is invertible.c)if $(e_n)$ is an approximate identity of $A$, then $a:=\sum_{n=}^{\infty}\frac{1}{2^n}e_n$is strictly positive.What I tried and know:a) I know that positive elements in this space are functions with positive images, and thus, strictly positive elements are functions with positive images and now real root. is this correct? and how can I show that the positive elements are these kind of functions.b) I figures out that if $a$ in invertible then $a$ is strictly positive, but no idea how to do the other direction.c)I did this but using another characterization of strictly positive elements but want to do it with this one, and I don't know how.Thank you for your help. I currently trying to join two rigged (and animated) models, which both has separate meshes, separate bones and separate animations. The first mesh is the horse while the second mesh is human:   So I am interested to join human model onto horse model to make a rider:   I think the best way would be to make horse as a parent and human as a child during the join process. In that case I would like that all horse animations (there are 3 of them - idle, walk and run) would remain as they are, while I can lose human animations, as I will need different ones for it. I was also thinking to make human root bone as a child of one horse back bones, that when horse moves up and down it would move entire human mesh together. It would be also good to keep the same weights that human bones would not start deforming horse mesh and vice versa.  I tried to use Ctrl+J by selecting various combinations between bones and meshes in object mode, however, I noticed that it messes up scaling and rotation of the child object. So I was wondering how it would be possible to properly join these models to create a rider? Winsock2's select() on fd 0 (stdin) fails Where's the energy in a boosted capacitor? Suppose I look at a parallel plate capacitor in its rest frame and calculate the electrostatic energy, $E$.  Next, I look at the same capacitor in a primed frame boosted in the direction perpendicular to the plane of the plates. In this frame, the $E$-field is the same strength, there is no magnetic field, and the volume over which the $E$-field extends is less by a factor $1/\gamma$. This suggests $E' = \frac{1}{\gamma} E$, but relativity states that energy transforms as $E' = \gamma E$.   Where is the missing energy? Let $R$ be a ring such that $x^4=x$ for every $x\in R$. Is this ring commutative? How to send array with CURL? Should I `urlencode` it? I have this table definition as part of a invoice  \begin{tabularx}{\textwidth}[ht]{lXrr}%   It worked fine for years until I have a very short word in the X column for the first time. Now my table is very small in width and centered horizontally on the page.  In my expectation the X column should fill up all space until the width of the table reaches the width of the page except the margins.  I have read about workarounds specifying something like 1.2\textwidth, but I don't like that, because the table is a template included via \include, so it would affect all my invoices.  I also tried \linewidth, \hsize and \columnwidth, but everything results in the same layout.  Is there a way to make the only X column expand so that it uses all available space?    Complete included table definition (Rechnungstabelle.tex):  \newcommand{\Rechnungstabelle}[4]{% vielen Dank für Ihren Auftrag. \noindent \begin{tabularx}{\textwidth}[ht]{lXrr}% \toprule% \bf{Anzahl} &amp; \bf{Bezeichnung} &amp; \bf{Einzelpreis} &amp; \bf{Gesamtpreis} \\% \midrule% #1% \bottomrule% &amp; \bf{Rechnungsbetrag}  &amp; &amp; \bf{\EUR{#2}} \\% \ifthenelse{\equal{#3}{}}{}{&amp; Enthaltene MwSt. 19\% &amp; &amp; \EUR{#3} \\}% \ifthenelse{\equal{#4}{}}{}{&amp; * Enthaltene MwSt. 7\% &amp; &amp; \EUR{#4} \\}% \end{tabularx}% }   Changing the table header row to  \bf{Anzahl} &amp; \bf{Bezeichnung} \hspace{\textwidth} &amp; \bf{Einzelpreis} &amp; \bf{Gesamtpreis} \\%   is a workaround, but I don't think that would be good TeX style.  Minimum example for reproducing:  \documentclass[ fontsize=12pt,                  paper=a4,                 foldmarks=TBMPL,                 fromalign=center,                 ngerman                 ]{scrlttr2} \usepackage{babel} \usepackage[utf8]{inputenx} \usepackage[right]{eurosym} \usepackage{tabularx} \usepackage{ltablex} \usepackage{booktabs} \usepackage{ifthen}  \input{Rechnungstabelle} \begin{document} \begin{letter}  \Rechnungstabelle{     1,0 h &amp; Do something &amp; \EUR{99,00} &amp; \EUR{99,00} \\      }     {99,00}     {}{} \end{letter} \end{document}