so I need to obtain the conditional distribution of a multivariate normal. However, I can only find it for the bivariate case:  $$(x_1|x_2=a) \sim N(\bar{\mu}, \bar{\Sigma})$$ $$\bar{\mu}= \mu_1 + \Sigma_{12}\Sigma_{22}^{-1}(a-\mu_2)$$ $$\bar{\Sigma}= \Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$$   is the example given in more depth.  However, how would these formulas look like if I would have 3 or 4 variables instead of 2?  Edit: I would like the conditional for $(x_1| x_2=a,x_3=b)$ given I have mulitvariate distribution with 3 variables.
123942
gives details on the conditional distribution of the multivariate normal:     If $\mu$ and $\Sigma$ are partitioned as follows      $\boldsymbol\mu = \begin{bmatrix}  \boldsymbol\mu_1 \\  \boldsymbol\mu_2 \end{bmatrix}$      $\boldsymbol\Sigma = \begin{bmatrix}  \boldsymbol\Sigma_{11} &amp; \boldsymbol\Sigma_{12} \\  \boldsymbol\Sigma_{21} &amp; \boldsymbol\Sigma_{22} \end{bmatrix} \quad$       then, the distribution of $x_1$ conditional on $x_2= a$ is   multivariate normal $(x_1|x_2=a) \sim N(\bar{\mu}, \bar{\Sigma})$   where      $\bar{\boldsymbol\mu} = \boldsymbol\mu_1 + \boldsymbol\Sigma_{12} \boldsymbol\Sigma_{22}^{-1} \left(  \mathbf{a} - \boldsymbol\mu_2 \right) $      and covariance matrix      $\overline{\boldsymbol\Sigma} = \boldsymbol\Sigma_{11} - \boldsymbol\Sigma_{12} \boldsymbol\Sigma_{22}^{-1} \boldsymbol\Sigma_{21}.  $   How can I prove this result? Wikipedia cites Eaton, Morris L. (1983). Multivariate Statistics: a Vector Space Approach. John Wiley and Sons. pp. 116–117., but I don't have this book handy...
57420
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!
19186
Let $X_1$ and $X_2$ be two normal random variables. Write $X_1\sim N(\mu_1, \sigma^2_1)$ and $X_2\sim N(\mu_2, \sigma^2_2)$, to fix ideas.   Consider the corresponding log-normal random variables: $Z_1 = \exp(X_1)$, $Z_2 = \exp(X_2)$.  Question: what is the distribution of the product of the two random variables, i.e., the distribution of $Z_1Z_2$?  If the normal random variables $X_1, X_2$ are independent, or they have a bivariate normal distribution, the answer is simple: we have $Z_1Z_2 = \exp(X_1+X_2)$ with the sum $X_1+X_2$ normal, hence the product $Z_1Z_2$ is still lognormal.  But suppose that $X_1, X_2$ are generally $not$ independent, say with correlation $\rho$. What can we say about the distribution of $Z_1Z_2$?
46328
$X$ 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?
47819
A random vector $V \in \mathbb{R}^n$ is said to be spherically symmetric. If for every orthogonal matrix  $A$   the distribution of $AV$ is the same as the distribution of $V$.   My question: Let $W$ be  i.i.d. zero mean Gaussian random variable.     Can we show that  for a spherically symmetric $V$ the conditional expectation  \begin{align} E[V|V+W=t]=   g(\|t\|).  \end{align}  That is the conditional expectation is only a function of the magnitude.   My attempt:  \begin{align} E[V|V+W=t]= \int  V  \frac{f_W(t-v)}{f_{V+W}(t)} dP_V(v)= \frac{\int  v  f_W(t-v) dP_V(v)}{f_{V+W}(t)}= \frac{E[ V  f_W(t-V) ]}{f_{V+W}(t)} \end{align}  Therefore, we can show that the two functions in the above ratio only depend on $\|t\|$.   Observe that we can write $t= \|t\| u_t$  where $u_t$ is a unit vector. Then, for any orthogonal matrix $A$ \begin{align} f_{V+W}(t)=f_{V+W}(\|t\|, u_t)&amp;= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t-V  \|^2}{2}} ]\\ &amp;= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t-AV  \|^2}{2}} ]\\ &amp;= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|tA^{-1}-V  \|^2}{2}} ]\\ &amp;= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t A^T-V  \|^2}{2}} ]\\ &amp;= f_{V+W}(tA^T)\\ &amp;= f_{V+W}(\|t\|,  u_t A^T) \end{align}  Since, $A$ was arbitrary  we have that $f_{V+W}(t)=h(\|t\|)$(i.e., only depends on the magnitute).  Now I am not sure how to show that $E[ V  f_W(t-V) ]$ only depends on $\|t\|$?
68378
Say $X_1, X_2, \ldots, X_n$ are independent and identically distributed uniform random variables on the interval $(0,1)$.  What is the product distribution of two of such random variables, e.g., $Z_2 = X_1 \cdot X_2$?  What if there are 3; $Z_3 = X_1 \cdot X_2 \cdot X_3$?  What if there are $n$ of such uniform variables? $Z_n = X_1 \cdot X_2 \cdot \ldots \cdot X_n$?
119958
An exercise asks me to show that if $X_1,X_2,\ldots,X_n$ is a random sample from a Poisson distribution with parameter $\theta$, the conditional distribution of $X_1,X_2,\ldots, X_{n-1}$, given $Y=\sum_{i=1}^n X_i$, is multinomial. Normally, this would have been an easy one but unfortunately the joint distribution here throws me off a bit. This is where I would appreciate some help.
5214
Suppose $x_1\sim\mathcal N(2,0.5),x_2\sim \mathcal N(2,3),$ and $x_3\sim \mathcal N(2.5,7)$ with correlations $\rho_{(1,2)}=0.3,\rho_{(1,3)}=0.1,$and $\rho_{(2,3)}=0.4.$ What is the distribution  of    $x_1+x_2+x_3?$ $x_1+ (3\times x_2)+x_3?$  $x_1+x_2+(0.5\times x_3)?$  Solutions: The distribution of $x_1+x_2+x_3 =\mathcal{N}(6.5,15.275)$. The distribution of $x_1+(3\times x_2)+x_3=\mathcal{N}(10.5,30.077)$ The distribution of $x_1+x_2+(0.5\times x_3)=\mathcal{N}(5.25,8.005)$  But I want to know what is the distribution of $(2\times x_1)-(3\times x_2)-x_3$ and $x_1+x_2-(2\times x_3)?$  The distribution of $(2\times x_1)-(3\times x_2)-x_3$ is $\mathcal{N}(-4.5,41.8407).$   The distribution of $x_1 +x_2-(2\times x_3) $ is $\mathcal{N}(-1,24.1544)$
74916
In a problem that I'm solving I find that:  "Let data (yi,xi) be sampled randomly from a two-dimensional distribution such that y|x is N(ɑ,x^2σ^2)".  Are y and x i.i.d? maybe just identically distributed but not independent?  Finally, if my model is: y = ɑ - u (where u is the error term and ɑ is a constant), and I also know that the distribution of x does not depend on either a or σ (I don't know it this assumption is needed for answer at the following question), is u independent from x?  My question is due to the fact that I thought that if the data are sampled randomly this mean that the r.v. are independent, but how is possible that the conditional distribution (his variance) depend then on x?
88502
Question  Suppose we have a multinomial distribution with $N$ possible outcomes, with probabilities $p_1,\ldots,p_N$. We sample this $n$ times, and denote the observed frequency of the $i$th outcome as $\xi_i$. In [1] the author claims that the distribution of the $\xi_i$ in the limit of large $n$ is:  $$p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right).\;\;\;\;\;(1)$$  We can see immediately that this must be an approximation, as it assigns nonzero probabilities for $\xi_1+\cdots+\xi_N&gt;1$. However we can see that these have vanishing probability in the limit $n\rightarrow\infty$. My question is how do we derive (1) from the multinomial distribution, and show that they match in the $n\rightarrow\infty$ limit?  My thoughts  My first thought would be to appeal to the central limit theorem. The multinomial distribution has mean $\mu_i=p_i$ and covariance matrix $\Sigma_{ij}=\delta_{ij}p_i-p_ip_j$, so we would expect this in the large $n$ limit to be described by a multivariate Gaussian with mean $\mu$ and covariance $\frac{1}{n}\Sigma$. However, things are complicated by the fact that the multinomial covariance is singular (since $\xi_N$ is determined by the other $\xi_i$s), and so the multivariate Gaussian is not defined.  To address this, we may try and consider only the first $\xi_1,\ldots,\xi_{N-1}$, which have a non-singular covariance matrix and hence well-defined multivariate Gaussian distribution. Let's take the Binomial distribution $N=2$. The frequency $\xi_1$, this has mean $p_1$ and variance $p_1(1-p_1)$, so this would be described the the Gaussian: $$\propto\exp\left(-\frac{n}{2}\frac{(\xi_1-p_1)^2}{p_1(1-p_1)}\right).\;\;\;\;\;(2)$$ The expression (1) gives: $$\propto\exp\left(-\frac{n}{2}\left(\frac{(\xi_1-p_1)^2}{p_1}+\frac{(\xi_2-p_2)^2}{p_2}\right)\right).\;\;\;\;\;(3)$$ If we substitute $\xi_2\rightarrow 1-\xi_1$, $p_2\rightarrow 1-p_1$ into (3), we can verify that this gives the same answer as (2). I have verified that this also works for $N=4$.   I'm sure that if I just bashed out the algebra for general $N$ we would get agreement between the central limit theorem and (1) when we restrict the latter to $\xi_1+\cdots+\xi_N=1,p_1+\cdots+p_N=1$. However, how can we start with the multinomial distribution and derive (1) as a limit which is valid everywhere? One idea would be to say that (1) goes to zero as $n\rightarrow\infty$ when you are not on that plane, however I am a bit uncomfortable with this as it goes to zero everywhere except the mean as $n\rightarrow\infty$, so I don't know if that argument is good enough.  [1] Wootters, William K. "Statistical distance and Hilbert space." Physical Review D 23.2 (1981): 357.
36651
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?
142
Let $X_1,\dots,X_{k+1}$ be mutually independent random variables, each having a gamma distribution with parameters $\alpha_i,i=1,2,\dots,k+1$ show that $Y_i=\frac{X_i}{X_1+\cdots+X_{k+1}},i=1,\dots,k$, have a joint ditribution as $\text{Dirichlet}(\alpha_1,\alpha_2,\dots,\alpha_k;\alpha_{k+1})$   Joint pdf of $(X_1,\dots,X_{k+1})=\frac{e^{-\sum_{i=1}^{k+1}x_i}x_1^{\alpha_1-1}\dots x_{k+1}^{\alpha_{k+1}-1}}{\Gamma(\alpha_1)\Gamma(\alpha_2)\dots \Gamma(\alpha_{k+1})}$.Then to find joint pdf of $(Y_1,\dots,Y_{k+1})$ I can not find jacobian i.e.$J(\frac{x_1,\dots,x_{k+1}}{y_1,\dots,y_{k+1}})$
97436
If I have the expression below, how can I use y~N($\mu$, $\Sigma$) to show it is a $\chi^2$ distribution with ($_n-1$) degrees of freedom?  ($n$-1)($s^2$)/$\sigma^2$ how is this ~$\chi^2$ $_n$-$_1$?     I recognize this contains SSE. I know that I should show the square root produces a distribution of $N$($\mu$, $I$) . I am lost at this point. I am thinking perhaps to substitute quadratic form into the original expression, then see what cancels? Or is this totally wrong?
16546
Suppose, $X\sim N(\mu,\sigma^2)$.   Can anyone help in finding the following : $\text{Var } \bigg(\Phi\big(\frac{X + c}{d}\big) \bigg)$ ? Here, c and d are positive.  Here, $\Phi(x)$ is the "Cumulative Distribution Function" of the above-mentioned normal distribution.  Thanking you in advance.
113561
If $U$ and $V$ are independent identically distributed standard normal, what is the distribution of their difference?  I will present my answer here.  I am hoping to know if I am right or wrong.  Using the method of moment generating functions, we have  \begin{align*} M_{U-V}(t)&amp;=E\left[e^{t(U-V)}\right]\\ &amp;=E\left[e^{tU}\right]E\left[e^{tV}\right]\\ &amp;=M_U(t)M_V(t)\\ &amp;=\left(M_U(t)\right)^2\\ &amp;=\left(e^{\mu t+\frac{1}{2}t^2\sigma ^2}\right)^2\\ &amp;=e^{2\mu t+t^2\sigma ^2}\\ \end{align*} The last expression is the moment generating function for a random variable distributed normal with mean $2\mu$ and variance $2\sigma ^2$.  Thus $U-V\sim N(2\mu,2\sigma ^2)$.  For the third line from the bottom, it follows from the fact that the moment generating functions are identical for $U$ and $V$.  Thanks for your input.  EDIT: OH I already see that I made a mistake, since the random variables are distributed STANDARD normal.  I will change my answer to say $U-V\sim N(0,2)$.
111966
Let $p_{ij}=P(X=i,Y=j)$ be the joint distribution, $P(X=i)=p_i=\sum_j p_{ij}, P(Y=j)=q_j=\sum_i p_{ij}$ be the marginal distributions, and $p_{i|j}=\frac{p_{ij}}{q_j}$ be the conditional distribution. Then the conditional entropy is defined as $$E[h(X|Y)]=-\sum_j \sum_i p_{i|j} \ln p_{i|j} q_j $$ Show $E[h(X|Y)]\leq h(X)$ where $h(X)$ is the entropy of $X$. I'm at a lost as to even know where to begin.
70859
We have a multivariate normal vector $\mathbf{X} \sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, where $\mathbf{X} =  \left[ \begin{matrix} X_1 \\ X_2 \\ \dot\\\\ \dot\\\\ X_n \end{matrix} \right]$,    $\boldsymbol{\mu} =  \left[ \begin{matrix} \mu_1 \\ \mu_2 \\ \dot\\\\ \dot\\\\ \mu_n \end{matrix} \right]$, and  $\boldsymbol{\Sigma} =  \begin{bmatrix} \sigma_1^2 &amp; \sigma_1\sigma_2 &amp; \dots &amp; \sigma_1\sigma_n   \\ \sigma_2\sigma_1 &amp; \sigma_2^2 &amp; \dots &amp; \sigma_2\sigma_n  \\ \dot\\\\ \dot\\\\ \sigma_n\sigma_1 &amp; \sigma_n\sigma_2  &amp; \dots &amp; \sigma_n^2 \end{bmatrix} $ is the covariance matrix.  Y is a random scalar of the form $Y = \beta_0 + \boldsymbol{\beta^TX} + \epsilon$, where $\epsilon \sim \mathcal{N}(0, \sigma_\epsilon^2)$  How can we go about deriving the joint probability density function(pdf) of the random vector $(Y, \boldsymbol{X^T})$?  The marginal multivariate pdf of $\boldsymbol{X}$ is given by  $ pdf(\boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} \right]$  The conditional probability density function of $Y \,\mid \, \boldsymbol{X=x}$ is given by  $pdf(y\mid \boldsymbol{X=x}) = \frac{1}{2\pi\sigma_\epsilon}.exp\left[  \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2  \right]   $  To compute the joint probability density function, we can use the fact that   $pdf(y, \boldsymbol{x})=  pdf(y\mid \boldsymbol{X=x}).pdf(\boldsymbol{x})$  I am unable to do the witty manipulation required in the algebra to arrive at the result the joint distribution will have a probability density function of a normal distribution. All questions previously asked seem to address the problem of deriving the conditional density function given the joint probability density function, for example,   Edit:  Multiplying the pdfs, we get $pdf(y, \boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}. \frac{1}{2\pi\sigma_\epsilon}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} + \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2 \right]$   Like the comment below mentions, most manipulations dealing with normal pdfs involve completing squares; but since this specific algebraic expression contains $\boldsymbol{x}$ in both terms within the exp term, I'm unsure on how to go about completing the squares.