\begin{equation*}     \begin{split}         \mathbb{E}[X_t] &amp; =\mathbb{E}[\alpha \circ X_{t-1} + \epsilon_t] \\      &amp;=\mathbb{E}[\sum_{j=1}^{X_{t-1}}{Y_j + \epsilon_t}] \\     &amp;=\sum_{n=0}^{\infty}{\mathbb{E}[\sum_{j=1}^{X_{t-1}}Y_j|X_{t-1}=n] \mathbb{P}[X_{t-1}=n]} +\mu_{\epsilon}\\     &amp;= \sum_{n=0}^{\infty}{{\sum_{j=1}^{n}{\mathbb{E}[Y_j] \mathbb{P}[X_{t-1}=n]}}} +\mu_{\epsilon}     \end{split}        \end{equation*}      \\ We can use recursion to express it in terms of $X_0$:     \begin{equation*}     \begin{split}     \mathbb{E}[X_t] &amp; =\mathbb{E}[X_{t-1}] + \mu_{\epsilon} \\     &amp;= \alpha [\alpha  \mathbb{E}[X_{t-2}+ \mu_{\epsilon}]+ \mu_{\epsilon}\\     &amp;=\alpha^2  \mathbb{E}[X_{t-2}]+ \alpha \mu_{\epsilon}+\mu_{\epsilon} \\     &amp;=....\\     &amp;=\alpha^t\mathbb{E}[X_0]+\alpha^{t-1}\mu_{\epsilon}+...+\alpha\mu_{\epsilon}+\mu_{\epsilon}\\     &amp;=\alpha^t\mathbb{E}[X_0]+\mu_{\epsilon}\sum_{j=0}^{\infty}{\alpha^j}     \end{split}     \end{equation*}          \begin{equation*}    \begin{split}     Var[X_t] &amp; = Var[\alpha \circ X_{t-1}+\epsilon_t]                                                                           \\              &amp; = Var[\sum_{j=1}^{X_{t-1}}Y_j]+Var[\epsilon_t]                                                                   \\              &amp; =\mathbb{E}[{[\sum_{j=1}^{X_{t-1}}Y_j}]^2]-{\mathbb{E}\big[\sum_{j=1}^{X_{t-1}}Y_j]}\big]^2+ \sigma_{\epsilon}^2 \\              &amp; =\mathbb{E}[{[\sum_{j=1}^{X_{t-1}}Y_j}]^2]-\alpha^2\mathbb{E}[X_{t-1}]^2+\sigma_{\epsilon}^2              \end{split}              \end{equation*}              \\                            \begin{equation*}                \begin{split}              \mathbb{E}[{[\sum_{j=1}^{X_{t-1}}Y_j}]^2] &amp; =\mathbb{E}[\sum_{j=1}^{X_{t-1}}{Y_j}^2 + \sum_{i\neq j}^{X_{t-1}}Y_iY_j]\\              &amp; =\mathbb{E}[\sum_{j=1}^{X_{t-1}}{Y_j}^2]+\mathbb{E}[\sum_{i\neq j}^{X_{t-1}}Y_iY_j]                                        \end{split}              \end{equation*}
122527
\documentclass{article} \usepackage{mathtools,amssymb}  \begin{document}  \begin{gather}     \begin{split}  (1-\underset{(0.034)}{0.060}{\textit{B}}^{12}-\underset{(0.032)}{0.123}{\textit{B}}^{2(12)}){y}_t &amp;=(1-\underset{(0.016)}{0.878}\textit{B}^{12})(1-\underset{(0.012)}{0.969}\textit{B}){e}_t\\   \end{split}  \notag\\  Box-Ljung \quad test \quad for\quad residuals\quad {\chi}^2= 15.151 \quad df= 12 \quad p-value = 0.233 \notag\\  \CENTERLINE {(Standard\quad errors\quad in\quad parentheses)}\notag  \end{gather}   \end{document}
20751
I'm not sure how to go about this. Does it involve using MVT?  I got as far as saying $e = \frac{e^x - e^y}{x-y}$.
59116
I know I can transform the above product as follows  \begin{eqnarray*}\lim_{n\rightarrow\infty}\left(\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{2n-1}{2n}\right)&amp;=&amp;\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)\\&amp;=&amp;\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1-\frac{1}{2k}\right)\\&amp;=&amp;\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right) \end{eqnarray*}  My thoughts are this is going to go to $e^{-x/2}$, but I can't figure out how to go from   $$\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right)$$  to  $$\lim_{n\rightarrow\infty}\left(1+\frac{-1/2}{n}\right)^n$$
35720
I am reading the Pattern Recognition and Machine Learning book by Bishops. I have some difficulties proving the (6.21) equation. Can someone help me?   N.B: If someone has a better title for this question, i'm opened to it as i couldn't find the good words.
34837
Being $X$ and $Y$ independent random variables distributed as $\mathrm{Exponential}(1)$ and let $T=X+Y$ Calculate $E(X|T)$.  This is my attempt at solving this:  Being $Z=\sum_{i=1}^{n}P_{i}$ where $P_{i}$~$\mathrm{Exp}(\lambda)$ then using it's MFG you have that $$M_{P_i}(t)=\frac{\lambda}{\lambda-t}$$ then $$M_{Z}(t)=M_{\sum{P_{i}}}(t)=(M_{P_{1}})...(M_{P_{n}})=(\frac{\lambda}{\lambda-t})^n$$ with this being the MFG of a $\mathrm{Gamma}$ distribution with parameters $(n,\lambda)$  With this result is easy to see that $T$~$\mathrm{Gamma}(2,1)$  Now $E[X|T]=\int xf_{X|T}(x|t)dx$  Then using Bayes' theorem we can make $$f_{X|T}(x|t)=\frac{f_{T|X}(t|x)f_{X}(x)}{f_{T}(t)}$$ then $$f_{T|X}(t|x)=Pr(T=t|X=x)$$ $$=Pr(X+Y=t|X=x)$$ $$=Pr(x+Y=t|X=x)$$ $$=Pr(Y=t-x)=f_{Y}(t-x)$$  Using this result we can now rearrenge $$f_{X|T}(x|t)=\frac{f_{Y}(t-x)f_{X}(x)}{f_{T}(t)}$$ $$=\frac{e^{-(t-x)}.e^{-x}}{\frac{te^{-t}}{\Gamma(2)}}$$ $$=\frac{1}{t}$$ then $$E[X|T]=\int (x)(\frac{1}{t})dx$$  And this where I got stuck because $1_{x\epsilon(0,\infty})$, I made this explanation step-by-step hoping that someone can se easily check where I went wrong. I would aprecciate a step-by-step answer, thanks.
92828
Suppose $\left\{a_n\right\}_{n=0}^{\infty}$ is sequence . and we have below recursion relation$$\begin{cases}a_n=\sum_{k=0}^{n-1}a_ka_{n-k-1}\\\\a_0=1\end{cases}$$  If $f(x)=\sum_{n=0}^{\infty}a_nx^n$  is generating function of $a_n$ which one is correct .  $$1)\space f^2(x)+xf(x)-1=0\\ 2)\space f^2(x)-xf(x)+1=0 \\ 3)\space xf^2(x)+f(x)-1=0 \\ 4)\space xf^2(x)-f(x)+1=0 $$ I am sorry to ask this question , But as honestly as possible it was my student's question. I forgot this type of question . Can you help me or give me a clue . Thanks in advanced.
37847
Actually I have no idea how to begin. I tried to represent it as a product of two series or as a sum of two series but have no result with this. Any idea ?
49984
I'm trying to make a block of equations where the first two are standalone equations, and the second two are subequations, ie  X=XXXX (1)  X=XXXX (2)  X=XXXX (3a)  =XXXX (3b)  I'm using the amsmath package, with align and subequation, but I can't figure out how to do this without making a separate block for Eqs 1-2 and one for Eqs 3a-3b. That solution is unsatisfactory, because then there is a big vertical gap between the two blocks, and I want them to appear all as one.  Here's the code I have so far, in case it's unclear.  \begin{align} T(x) &amp;= \frac{m}{2{k_{B}}} \left\langle \mathbf{\tilde{v}}{(\mathbf{x}_{i},t)}^{2} \right\rangle\\  \mathbf{u}(x) &amp;= \left\langle \mathbf{v}_{i}(\mathbf{x}_{i},t) \right\rangle \end{align} \begin{subequations}  \begin{align} P_{xy}(x) &amp;= \frac{1}{A} \left\langle \sum\limits_{i} {m \tilde{v}_{i,x} \tilde{v}_{i,y}} + \frac{1}{2} \sum\limits_{i} {\sum\limits_{j\ne i} {r_{ij,y} F_{ij,x}}} \right\rangle \\ \begin{split}  &amp;=\frac{m}{A} \Bigg\langle \sum\limits_{i}{{{{\tilde{v}}}_{i,x}}{{{\tilde{v}}}_{i,y}}} + \frac{a^3\omega _{pd}^2}{4} \\  &amp;\quad \times \sum\limits_{i}{\sum\limits_{j\ne i}{r_{ij,y}{\left\{ \frac{e^{(-{{r}_{ij}}/\lambda )}}{{{r}_{ij}}}\left( \frac{1}{r_{ij}}+\frac{1}{\lambda} \right) \right\}_{x}}}} \Bigg\rangle \end{split} \end{align} \end{subequations}   Any suggestions would be greatly appreciated!
77357
This problem has baffled me. I have tried using $t = \tan(x/2)$ identities but end up with a 4 th order polynomial I cant solve.
63131
I am trying to find a two term expansion for the root of $$1+\sqrt{(x^2+\epsilon)}=e^x$$. Since $$\epsilon &lt;&lt; 1$$, I can tell that this equation behaves like $$1+x=e^x$$ which has a root close to zero. That gave me the idea that the root might look like $$x \approx x_0 + \epsilon^{\alpha} x_1 + \dots  $$. I used Taylor expansion for both functions $\sqrt{(x^2+\epsilon)}$ and $e^x$ and then plugged my guess into the Taylor expanded equation and balance of O(1) terms gave me $$x_0=0$$ as expected. My problem begins now. For the higher orders of $\epsilon$. Long story short this expansion ends up with some inconsistencies. So I thought maybe $x \approx x_0+\mu(\epsilon)$. But then again I get nowhere. I am getting frustrated and I would really appreciate some help.  Thank you in advance.
11378
I have managed to use the root test and get as far as getting the limit to equal to $1$. But, I still need clarification on why the answer for the limit to this questions is $\alpha &lt; 1$. \begin{align} \sum _{n=0}^{\infty }\left( \frac {\alpha n} {n+1}\right) ^{n}\end{align} \begin{align} \lim _{n\rightarrow \infty }\left( \left( \frac {\alpha n} {n+1}\right) ^{n}\right) ^{1/n} =\left| \frac {\alpha n} {n+1}\right| =\alpha \lim _{n\rightarrow \infty }\left| \frac {n} {n+1}\right|=1.\end{align}
81503
I know that  $A_{[a} B_{b]} = \frac{1}{2!}(A_{a}B_{b} - A_{b}B_{a})$  But how can write $E_{[a} F_{bc]}$ like the above?  Can you provide a reference where this notational matter is discussed?
96005
Let $X_1,\ldots,X_n$ be i.i.d. $B(1,\theta)$ random variables, $0&lt;\theta&lt;1$. Then, as an estimator $\theta$, check if $T(X_1,\ldots,X_n)= \dfrac{\sum_{i=1}^n X_i +\sqrt{n}/2}{n+\sqrt{n}}$ is consistent and/or unbiased.  $$T=\frac{\frac{1}{n}\sum_{i=1}^n X_i +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}$$  $$T=\frac{\bar{X} +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}$$  $$E(T)=\frac{E(\bar{X} +\frac{1}{2\sqrt{n}})}{1+\frac{1}{\sqrt{n}}}=\dfrac{\theta +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}.$$  So, $T$ is biased.  Consistency: $\lim\limits_{n \to\infty}E(T)=\lim\limits_{n \to\infty}\dfrac{E(\theta +\frac{1}{2\sqrt{n}})}{1+\frac{1}{\sqrt{n}}}$  So $T$ is consistent.   But The given answer is "neither unbiased nor consistent"  Where did I go wrong ? Please advise.
111865
Problem: Let $(x_i)$ be a sequence in (0,1) such that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n x_i^k$ exists for every $k=0,1,2,...$ and $f\in C[0,1]$. Show that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i)$ exists.  I've used the fact that $B_n(f)$ converges uniformly to $f$ to get   $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i) = \lim_{n \to \infty} (1/n)\sum_{i=1}^n \sum_{k=0}^n f(\frac{k}{n}) {n \choose_k }x_i^k (1-x_i)^{n-k}$  And then you can arrange to get   $\lim_{n \to \infty} (1/n) \sum_{k=0}^n \left[ f(\frac{k}{n}) {n \choose_k }\sum_{i=1}^nx_i^k (1-x_i)^{n-k}\right]$  I want to use the fact that $(1-x_i) &lt; 1$ to consolidate some of the equation but I get nowhere.
111071
I am having trouble with (a) and (b):  I evaluated (a) and I got that sun of the first three terms does not equal $\ln(5/8)$. I got that it is equal to $\ln(3/4) + \ln(2/3) + \ln(15/16)$ which is not equal to $\ln(5/8)$.  I am also having trouble with part (b). I cannot figure out how to calculate the math to cancel out the $(k+1)^2$ and turn it into $2(n  +1)$.  Any help would be amazing.
40353
I'm trying to split up the following equation so that it fits the margins of my document, so far I have:  \documentclass{article}  \usepackage{amsmath}  \begin{document}   \begin{equation} 0 = n_1^3 n_2 n_3^2(\lambda+\mu)^2 +n_1 n_2^3 n_3^2(\lambda + \mu)^2  + (\lambda + \mu)(\mu - \rho c^2)n_1^3 n_2 + (\lambda + \mu)(\mu - \rho c^2) n_1 n_2^3 + n_1 n_2 n_3^2 (\lambda + \mu)(\mu - \rho c^2) + n_1 n_2 (\mu- \rho c^2)^2 - n_1^3 n_2 n_3^2 (\lambda + \mu)^2 - n_1 n_2^3 n_3^2 (\lambda + \mu)^2  \end{equation}  \end{document}    How do I make the equation split so that it fits?
82722
$a_{n,3}$     $$1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,\cdots \cdots $$   $a_{n,5}$     $$1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,\cdots \cdots $$   $a_{n,7}$     $$1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,\cdots \cdots $$   $$\vdots $$  $a_{n,2k+1}$     $$1,\cdots \cdots\cdots \cdots,1,\cdots \cdots\cdots \cdots,\overbrace{1,\cdots \cdots\cdots \cdots}^{2k+1},1,\cdots \cdots\cdots \cdots,1,\cdots \cdots$$   I found  $$a_{n,k}= \left \lfloor \left | \cos\frac{\pi (n-1)}{2k+1} \right | \right \rfloor$$ but this term include abs and floorfucntion.  I mean i want the General Term without abs and floorfucntion.(plus sum)  Any hints will be appreciated, thank you.