I see that a lot of questions are answered here for multivariate and bivariate conditional distributions. But I did not find the proof of these equations (I need just for bivariate case).      to get this    I appreciate your help
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Suppose that $\mathbf{X} \sim N_{2}(\mathbf{\mu}, \mathbf{\Sigma})$. Then the conditional distribution of $X_1$ given that $X_2 = x_2$ is multivariate normally distributed with mean:   $$ E[P(X_1 | X_2 = x_2)] = \mu_1+\frac{\sigma_{12}}{\sigma_{22}}(x_2-\mu_2)$$  and variance: $${\rm Var}[P(X_1 | X_2 = x_2)] = \sigma_{11}-\frac{\sigma_{12}^{2}}{\sigma_{22}}$$  It makes sense that the variance would decrease since we have more information. But what is the intuition behind the mean formula? How does the covariance between $X_1$ and $X_2$ factor into the conditional mean?
84330
Suppose that I have two univariate marginal distributions, say $F$ and $G$, which I can simulate from. Now, construct their joint distribution using a , denoted $C(F,G;\Sigma)$. All the parameters are known.  Is there a non-MCMC method for simulating from this copula?
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The Multivariate Gaussian pdf is given by  $$(2\pi)^{-\frac{K}{2}} \det(\Sigma)^{-\frac{1}{2}} \exp({-\frac{1}{2}}(X-\mu)' \Sigma^{-1} (X-\mu)) $$  The wikipedia for multivariate Gaussians is    However I could not find a pdf for the multivariate lognormal distribution. Does it exist? If so, what is it?
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$X$ is multivariate normal distributed with $\mu = (2, 2)$ and $\Sigma = (1, 0 ; 0, 1)$ and I have 2 vectors $A = (1, 1)$ and $B = (1, -1)$. How can I than show that the linear combinations $AX$ and $BX$ are independent?
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It is true that $P(X&gt;t+s|X&gt;t)=P(X&gt;s)$ for certain values $t$ and $s$.  However, how can I show that this still holds if:   $T$ is a continuous random variable. That is $P(X&gt;T+s|X&gt;T)=P(X&gt;s)$ Both $T$ and $S$ are continuous RVs: $P(X&gt;T+S|X&gt;T)=P(X&gt;S)$   All the random variables are independent.
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I would like to compute the conditional choice probability which is defined as follows \begin{equation} p_k=Prob(f(m_k)+u_k&gt;f(m_i)+u_i, \forall  i\neq k|m) \end{equation} where $u_i, i=1,...,N$ are independently and identically distributed normal variables with $u_i\sim N(0,\sigma_i^2)$ and $m_i, i=1,...,N$ are independently and identically distributed standard uniform random variables, $m_i\sim i.i.d. U[0,1]$.  In words, $p_k$ is the probability that out of $N$ options, the choice $k$ has the greatest value conditional on having a realization for the vector $m$. Because $u_i-u_k\sim N(0, \sigma_i^2+\sigma_k^2)$, $p_k$ can be rewritten as \begin{equation}p_k=\prod_{i\neq k}\Phi\left(\frac{f(m_k)-f(m_i)}{\sqrt{\sigma_i+\sigma_k}}|m\right) \end{equation} where $\Phi(\cdot)$ denotes the CDF of standard normal distribution. Next, I would like to find the expected value of $p_k$ by integrating out the possible values of $m$ as follows \begin{equation} E[p_k]=\int_{(0,1)^N}\prod_{i\neq k}\Phi\left(\frac{f(m_k)-f(m_i)}{\sqrt{\sigma_i+\sigma_k}}\right) d\mathbf m \end{equation} Is it possible to use the properties of the uniform and standard normal distributions to get a more compact solution for $E[p_k]?$
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Find the distribution function of this multivariate normal variable where $X \sim N(0, 1)$ $$ \begin{pmatrix} X \\ (-1)^nX \\ \end{pmatrix} $$  I need this for a proof that marginal convergence in distribution does not imply joint convergence in distribution, but I am really rusty.   Any help is much appreciated.   Edit:  Explaining notation.  This is a sequence of two-dimensional random variable vectors, where $n$ takes values in the natural numbers. The distribution function should differ for odd and even values of $n$.
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I have a question. Let's suppose that the two random variables $X1$ and $X2$ follow two Uniform distributions that are independent but have different parameters:  $X1 \sim Uniform(l1, u1)$  $X2 \sim Uniform(l2,u2)$  If we define X3 as the maximum of X1, and X2, i.e., $X3 = max(X1, X2)$, what kind of distribution would it be? It is certainly not a uniform and I can calculate the cumulative probability, i.e., $p(X3 &lt;= a)$, because $p(X3 &lt;= a) = p(X1 &lt;= a) p(X2 &lt;= a)$. But, I wonder if there exists any density function that can model X3.
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When $X$, $Y$ are bivariate normal, how can we calculate the conditional expectation $E(X\mid X&gt;Y)$? A more concrete setting of this general problem can be referred .
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Let $A$ and $B$ be two constant matrices and let $x$ and $ y$ be two random vectors, what is the general formula for $Var(Ax+By)$? I know the formula for when $x$ and $y$ are scalar random variables and $A$ and $B$ are constants, but what about the matrix case?
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Show that $(X_1,X_2)$ has a bivariate normal distribution with means $\mu_1, \mu_2$, variances $\sigma _1^2 $ and $\sigma _2^2$, and correlation coefficient $\rho $ if and only if every linear combination $t_1X_1+t_2X_2$ has a univariate normal distribution with mean $t_1\mu_1+t_2\mu_2$, and variances  $t_1^2\sigma _1^2+t_2^2\sigma _2^2+2t_1t_2\rho \sigma_{12}$ where $t_1, t_2$ are non zero constants  My work:  Joint mgf of $X_1 $and $X_2=e^{[t_1\mu_1+t_2\mu_2+\frac{1}{2}(t_1^2\sigma _1^2+t_2^2\sigma _2^2+2\rho \sigma_1 \sigma_1t_1t_2 )]} $  I am confused here to find the proper grouping to get the forward result.  For the backward result, I know I have to find the m.g.f. of $t_1X + t_2Y$, for any $t_1 $and $t_2$ real,and plug in the $E(t_1X + t_2Y)$ and $Var(t_1X + t_2Y)$.  I am struggling to get the correct backward result.  Second part:   Let $Y_i=X_i/\sigma_i, i=1,2$ Show that $Var(Y_1-Y_2)=2(1-\rho)$  I have found, $Var(Y_1)=\sigma_1, Var(Y_2)=\sigma_2$ since $X_1/\sigma_1=\sigma _1$  How can I prove that $Var(Y_1-Y_2)=2(1-\rho)$? Thank you so much for your help!
73924
Suppose, $y_i|u_i\sim MN(X_i(t_i), \sigma_e^2I_{m_i})$ and $u_i\sim N(0,1)$. Now how to find the marginal distribution of $f(y_i)=\int f(y_i|u_i)f(u_i)du_i$? Since one is multivariate normal and other is univariate normal, can we actually find the product before the integration? Or it will be a no-closed form expression? Even if it's the case, how do we come up with the no-closed form expression? Here $X_i(t)=\dfrac{exp[Z_i(t)]}{\int_a^dexp[Z_i(s)]ds}$ and $Z(t)= b(t)^Tc+b(t)^TAU $ So $X_i(t)$ is determined by $u_i$ via $Z_i$
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I have 2 i.i.d Uniform random variables $\theta_{1}$ and $\theta_{2}$ on $[0, 2\pi)$. Here the problem pushes us to consider that all operations are made modulo $2\pi$. My question is how to obtain the joint PDF of $\theta_{1} - \theta_{2}$ and $\theta_{1}$ and then obtain the conditional PDF of $\theta_{1}$ given $\theta_{1} - \theta_{2}$??. For the second part It can be deduced from the first by writing it as the joint PDF divided by the PDF of $\theta_{1}$. But for the first part, A common idea that I saw is to derive the $\text{CDF}(x, y) = P(\theta_{1} - \theta_{2} \le x$, $\theta_{1} \le y$) and then perform derivations respect to $x$ and $y$. Is this a right approach? If yes, how to proceed with CDF especially with our uniform distribution restrictions?
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I want to prove very simple equations, but am having some trouble because I have no clue.  Definition of Conditional Probability is given as follows:  Definition $E(Y|X)$ is a conditonal expectation of $Y$ given $X$ if it is a $\sigma (X)$-measurable random variable and for any Borel set $S \subseteq R$, we have $E(E(Y|X)1_{X \in S})=E(Y1_{X \in S})$  I want to solve those two problems based on above definition.   Suppose $X$ and $Y$ are independent.   (a) Prove that $E(Y|X)=E(Y)$ with probability 1.  (b) Prove that $Var(Y|X)=Var(Y)$ with probability 1.  (c) Explicitly verify the following theorem.(with $\mathcal{G}=\sigma (X)$ in this case)  Theorem: Let $Y$ be a random variable, and $\mathcal{G}$ a sub-$\sigma$-algebra. If $Var(Y)&lt;\infty$, then $Var(Y)=E(Var(Y|\mathcal{G}))+Var(E(Y|\mathcal{G}))$   Let X and Y be jointly defined random variables.   (a) Suppose $E(Y|X)=E(Y)$ with probability 1. Prove that $E(XY)=E(X)E(Y)$  (b) Give an example where $E(XY)=E(X)E(Y)$, but it is NOT the case that $E(Y|X)=E(Y)$ with probability 1.  1.  For this part, I think I could deal with this by myself if I understand how to prove (a). So please help me with just (a).  In order to prove that $E(Y|X)=E(Y)$ with probability 1, I need to show that  \begin{equation} E(E(Y|X)1_{X})=E(E(Y)1_{X}). \end{equation}  if $X$ and $Y$ are independent.  From the above definition, conditional expectation $E(Y|X)$ is defined such that $E(E(Y|X)1_{X \in S})=E(Y1_{X \in S})$.  So maybe it suffices to show that $E(Y1_{X \in S})=E(E(Y)1_{X \in S})$. But it looks too obvious. My guess is that there is more educated way of showing why $E(Y1_{X \in S})=E(E(Y)1_{X \in S})$ holds if $X$ and $Y$ are independent.  2.  I can do (b) on my own. So for (a), "$E(Y|X)=E(Y)$ with probability 1" means that $E(E(Y|X)1_{X})=E(E(Y)1_{X})$. But, how can I prove $E(XY)=E(X)E(Y)$ with that information? I know how to prove this equation with integral sign or summation sign. But how does information about a.s. convergence give any implication about that equation?
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Let $Z$ be a random uniformly distributed variable on $[0,1]$. Show that the random variables $X = \sin 2\pi Z$ and $Y = \cos 2\pi Z$ are not independent, but nevertheless Cov$[X,Y]=0$.  This is a homework assignment, but I'm a bit stuck.  My thoughts  We can see that $X$ and $Y$ are not independent, since both depend on $Z$. If we want to show this explicitly, then we need to show that $$f_{X,Y}(a,b) \neq f_X(a)\;f_Y(b),$$ where $f_{X,Y}(a,b)$ is the joint probability distribution function. But how can I find the (joint) probability distribution function(s) $f_X, f_Y$ and $f_{X,Y}$?   If I can find these functions, I can also solve the covariance problem. Is this the right way? Or is there a 'better' way to solve this problem?
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Problem: Show that if $X$ and $Y$ are independent $N(0,1)$-distributed random variables, then $X/Y ∈ C(0,1)$.  Question: I don't know how to proceed below. I want to prove that the PDF of $X/Y$ is Cauchy. PS. I looked on wiki and they defined $C$ as $X/Y$  Attempt:   The expression looked hairy and  could not integrate it.   Note to self: problem 8.
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Question Suppose that a random variable $X$ has the property that, for any $a &gt; 1$, the conditional distribution of $\frac 1 a X$ given that $X &gt; a$ is the same as the distribution of $X$. Specify all possible distributions of $X$. My thoughts I do know that $f_{X\mid Y}(x\mid y) = \frac {f_{X, Y}(x,\ y)} {f_Y(y)}$ and I suppose I have to make use of this relationship, but I have no idea how to even begin - When I find conditional distributions, I am usually given a specific value, not something like $X &gt; a$. Any proposed solutions with intuitive explanations will be greatly appreciated :)