Let {$x_t$} be the process given by  $x_t$=$A\cos(wt)+B\sin(wt)$ where   $A$ and $B$ are uncorrelated random variables with mean $0$ and   variance $1$ and $w$ is a fixed frequency in the interval $[0, \pi]$.   Find its autocovariance function at lag $k$, where $k=0,1,2,...$   I did:  $$\gamma{(k)}=E(x_tx_{t-k})=E[(A\cos(wt)+B\sin(wt))(A\cos(w(t-k))+B\sin(w(t-k)))]$$  Is there a way to make this easier to compute?
34054
Hey I need some help with this exercise:  Assume $z_{ t }$ be a sequence of independent normal random variables, each with mean 0 and time independent variance $\sigma^{ 2 }$, and let c be a constant. Is the following process for a time series stationary? If it is stationary specify the mean and the autocovariance function:  $$ x_{ t }=z_{ 1 }*cos\left( ct \right)+z_{ 2 }*sin\left( ct \right) $$  So for a weak stationarity time series the mean and covariance function should be independent of time. I'm not sure how to calculate them by hand without R.  Thanks for help.
36274
Let $(\Omega,\Sigma,P)$ be a probability space and $X: [0,\infty) \times \Omega \to \mathbb{R}$ be a Gaussian process (i.e. all finite linear combinations $\sum_i a_i X_{t_i}$ are Gaussian random variables). If the process is continuous, it seems to be clear that the process $Y_t (\omega) = \int_0^t X_s(\omega) ds$ is a Gaussian process.   Is it true that $Y_t$ is a Gaussian process even if $X$ is only assumed to be measurable? I am working through the derivation of Kalman Filter in the text by Bernt Oksendal Eq 6.2.10 (fifth edition) and this seems to be a way to show that $M_t$ (in the book) is a Gaussian process.
42977
I'm sorry if this is a duplicate, but I can't seem to find the answer to this.  If $Z_t$ is a white noise process and $X_t$ satisfies  $$ \phi(B) X_t = \theta(B) Z_t $$  (where $B$ is the lag operator), then when is $X_t$ stationary?  If $\theta(z)$ and $\phi(z)$ have no common roots and $\phi(z)$ has a root with $|z| = 1 $ then $X_t$ is necessarily non-stationary [Brockwell and Davis, Remark 3 in Chapter 3].  I also know that a moving average of a stationary process with absolutely summable coefficients is also stationary [Brockwell an Davis, Proposition 3.1.2]. So if $\phi(z)$ has no roots in with $|z| \leq 1$ then the Maclaurin series for $1/\phi(z)$ has radius of convergence greater than 1. Therefore $X_t$ is a moving average of $\theta(B) Z_t$ with absolutely summable coefficients, and hence stationary.  By stationarity, I mean weak (=second order) stationarity.  The lecturer has a habit of putting this on the final, so I'd like to know if there's a general theorem governing it.  From a past paper:
88043
We are given two independent standard normal random variables $X$ and $Y$. We need to find out the M.G.F of $XY$.  I tried as follows : \begin{align} M_{XY}(t)&amp;=E\left(e^{(XY)t}\right)\\&amp;=\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(XY)t}f_X(x)f_y(y)dxdy \\ &amp;=\dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(xy)t}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &amp;= \dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-\dfrac{(x-ty)^2}{2}}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &amp;= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dy \\ &amp;= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{-y^{2}(\frac{1}{2}+t^2)}dy \\ &amp;= \dfrac{1}{\sqrt{1+2t^2}} \end{align} Is this correct ?
107370
I would like to know how to show that the functions  $$r_n(t)=\operatorname{sgn}\big(\sin(2^n \pi t)\big)$$ (where $\operatorname{sgn}$ is the sign function) form an orthonormal system but not an orthonormal basis from $L_2([0,1])$.    Progress: I pick two different variables $i,k$ and show that $\langle r_i (t),r_k (t)\rangle =0$, so I can take the integral $\int_0^1  r_i (t) r_k (t)\,dt$, but I do not know how to show that this is 0.
24330
Itō isometry from :     Let $W : [0, T] \times \Omega \to \mathbb{R}$ denote the canonical   real-valued Wiener process defined up to time $T &gt; 0$, and let $X :  [0, T] \times \Omega \to \mathbb{R}$ be a stochastic process that is   adapted to the natural filtration $\mathcal{F}_{*}^{W}$ of the Wiener   process. Then $$      \mathbb{E} \left[ \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t} \right)^{2} \right] = \mathbb{E} \left[ \int_{0}^{T} X_{t}^{2} \,  \mathrm{d} t \right], $$ where $\mathbb{E}$ denotes expectation with   respect to  $\gamma$. In other words, the Itō   stochastic integral, as a function, is an isometry of normed vector   spaces with respect to the norms induced by the inner products $$      ( X, Y )_{L^{2} (W)} := \mathbb{E} \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t} \int_{0}^{T} Y_{t} \, \mathrm{d} W_{t} \right) =  \int_{\Omega} \left( \int_{0}^{T} X_{t} \, \mathrm{d} W_{t}  \int_{0}^{T} Y_{t} \, \mathrm{d} W_{t} \right) \, \mathrm{d} \gamma  (\omega) $$ and $$      ( A, B )_{L^{2} (\Omega)} := \mathbb{E} ( A B ) = \int_{\Omega} A(\omega) B(\omega) \, \mathrm{d} \gamma (\omega). $$   I was wondering what the two normed spaces are and what their norms are, so that the two normed spaces are isometric wrt their norms?  Thanks and regards!
93264
I would like to find the characteristic function of the product of two independent brownian motions. This boils down to the characteristic function of the product of two normal random variables. This is known to follow the product-normal distribution:    but I haven't found an easy proof of this fact. This is proven making use of the Mellin transform, but this questions was in an assignment of a considerably lower level and no mention of the Mellin transform in the lectures either. I would be really happy if someone could help me.
8353
There are three random variables, $x,y,z$. The three correlations between the three variables are the same. That is,  $$\rho=\textrm{cor}(x,y)=\textrm{cor}(x,z)=\textrm{cor}(y,z)$$  What is the tightest bound you can give for $\rho$?
90516
I'm trying to understand the basic concept of random processes.I already understood that a continuous-time random process is defined by X(¥,t) where ¥ is each element in the Sample Space and t is the time index.     By this definition, we can see the random process as an ensemble of sample realizations ( time-functions defined for each outcome in the Sample Space) or as a set of random variables ( each one defined for a specific time t ). In my mind, the value of the time-function ( sample realizations ) X(¥1,t) reflects the changing value of the probability of outcome ¥1 being choosen in the sample space, is that correct ?    For example, we have a boy B in the sample space representing outcome ¥1.Having this outcome ¥1 at time t=0 equals 30%.As the time passes, the sample spaces gets more crowded and at time t=5, the probabily of choosing boy B equals 20%.   Can i think that the time-function X(¥1,t) will equal 0,3 for t=0 and 0,2 for t=5 ? Does X(¥1,0)=0,3      and     X(¥1,5)=0,2         ?  Thanks in advance
114802
I'm trying to wrap my head around linear transformations to random variables (with coefficients > 1).  Consider the two random and independent variables $X$ and $Y$ where:  $$X \sim \mathcal{N}(0,1)\qquad Y \sim \mathcal{N}(-1, 2)$$  I'm trying to determine the distribution of $Z = 2X + 3Y$.    The attempt so far:  I've defined the MGF of both $X$ and $Y$ as:  \begin{align*}M_X(s) &amp;= \exp(s \mu + \sigma^2 s^2 / 2) = \exp(s^2/2)\\ M_Y(s) &amp;= \exp(-s + s^2)\\ \end{align*}  Then the plan was to calculate $U$ as:  $Z = 2X + 3Y = 2 \exp(s^2/2) \cdot 3 \exp(-s + s^2) = 6 \exp(3s^2/2 - s)$  This lead me to a dead end and I'm not sure if this is the right approach, the "new" MGF doesn't exactly look like anything I know.  I've considered using a convolution to try and solve this but I am not 100% about that approach.   There was a mention on  about how to do it for the case of the coefficients = 1 and that there was a special case for coeffs > 1 (however it failed to expand on it).    I've done some simulations in MATLAB with random sampling from both of these Gaussians  and then applying the transformation. The results indicate the solution should be (assuming the code is correct):  $Z \sim \mathcal{N}(-3, 22)$
1225
I know that the variance of the difference of two independent variables is the sum of variances, and I can prove it. I want to know where the covariance goes in the other case.
46358
Show that $\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y}) = \sum_{i=1}^{n} (x_i - \bar{x})(y_i)$. I fell into this hole where I keep finding that $$\sum_{i=1}^n (x_i - \bar{x}) = \sum_{i=1}^n x_i - \sum_{i=1}^n \bar{x} = n\bar{x} - n\bar{x} = 0$$ This seems true, but I know that I am missing something, since this implies that covariance is always zero, which is obviously not true.
109889
The continuous random variables $X$ and $Y$ have joint density function $f_{xy} = 1$ for $0 &lt; x &lt; 1$ and $2x &lt; y &lt; 2$, and zero otherwise.  I am stuck on the above question, with parts a and b below:  a) Find the $cov(x,y)$.  Work:   $cov(x,y) = E(XY) - E(X)E(Y)$   $= \int_0^1\int_{2x}^2 xydydx - \int_0^1\int_{2x}^2 xdydx\int_0^1\int_{2x}^2 ydydx$ = $\frac{-1}{18}$  However I am given the answer $\frac{1}{18}$.  Did I make a mistake in my computation or is the answer incorrect?  b) Find $E(X|Y = y)$.  Work:   $E(X|Y=y) = \int_{-\infty}^{\infty}xf_{x|y}(x|y)dx$  This is where I am puzzled.  I know that $f_{x|y}(x|y) = \frac{f_{xy}(xy)}{f_yy}$, and $f_yy= \int_{-\infty}^{\infty}f_{x,y}(x,y)dx$.  That means $f_yy = 1$ and therefore $f_{x|y}(x|y) = 1$.  However I am given the answer $E(X|Y=y) = \frac{y}4$. Again, did I make a computational mistake or am I not getting something?  Thanks.
47003
I'm trying to find the minimal sufficient statistics for a Cauchy distributed random sample $X_1,...,X_n$, here \begin{equation} f(x|\theta) = \frac{1}{\pi[1+(x-\theta)^2]} \end{equation} I begin by guessing that the order statistics are the minimal sufficient statistics (first of all, they are sufficient). Then I try to prove that \begin{equation} \frac{f(X|\theta)}{f(Y|\theta)} = \prod_{i=1}^n\frac{1+(y_i-\theta)^2}{1+(x_i-\theta)^2}=C \end{equation} Where $C$ is a constant of $\theta$, I want to prove that the above equation holds iff $T(X)=T(Y)$, where $T(X)$ is the order statistics of $X$.  I am really stuck and don't know how to show why that if it holds then $T(X)=T(Y)$, can anyone help me on this proof? Thanks in advance!  p.s., a similar thread  discusses the problem but offers no proof for the minimal sufficiency.
38831
Find the correlation coefficient $\rho_g$ of $(G(x), H(y))$ where $G(x),H(y)$ are marginal  C.D.Fs of $X,Y$ and $X,Y$ follow  bivariate normal distribution $\text{B.N.}(\mu_1,\mu_2,\sigma_1,\sigma_2,\rho)$. I am finding it difficult to calculate $E(G(x)H(y))$, otherwise I can find all the other terms. Please help! The answer given is $\frac{6}{\pi} \arcsin{\frac{\rho}{2}}$
107393
Provided two real number sequences: $a_1,a_2,...,a_n$;$b_1,b_2,...,b_n$, define their means respectively: $$\bar a=\frac{1}{n}\sum_{i=1}^n a_i,\bar b=\frac{1}{n}\sum_{i=1}^n b_i$$ and define their variances and covariance respectively: $$var(a)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)^2,var(b)=\frac{1}{n}\sum_{i=1}^n (b_i-\bar b)^2,cov(a,b)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)(b_i-\bar b)$$ naturally leads to the definition of normalized cross correlation: $$NCC=\frac{cov(a,b)}{\sqrt{var(a)var(b)}}=\frac{\sum_{i=1}^n(a_i-\bar a)(b_i-\bar b)}{\sqrt{\sum_{i=1}^n (b_i-\bar b)^2 \sum_{i=1}^n (a_i-\bar a)^2}}$$ Now how to show that $NCC$ lies in $[-1,1]$?
8095
I was looking at the following question from "One Thousand Exercises in Probability" by Grimmett, page 25, question 16 (not homework just self-study):     Let $X$ and $Y$ be independent Bernoulli random variables with $p = 1/2$.   Show that $X + Y$ and $|X − Y|$ are dependent though uncorrelated.   Now, the solution given on page 176 goes as follows:  $\mathrm{cov}(X+Y, |X-Y|) = E[(X+Y)\cdot(|X-Y|)] - E[X+Y]\cdot E[|X-Y|]$  I can work out that   $E[X+Y] = (0)(1/4) + (1)(1/2) + (2)(1/4) = 1$  and  $E[|X-Y|] = (0)(1/2) + (1)(1/2) = 1/2$  But I can't work out how to do  $E[(X+Y)\cdot(|X-Y|)]= ?$  The solution in the book for the covariance is:  $1/4 + 1/4 - 1(1/2) = 0$  To show dependence, the book provides the following solution:   $P(X+Y=0, |X-Y|=0) = 1/4$ is not the same as $P(X+Y=0)\cdot P(|X-Y|=0) = 1/8$