Let $A$ and $B$ be independent and normally distributed variables where $A \sim \mathcal{N}(\mu_1,\sigma_1^2)$ and $B \sim \mathcal{N}(\mu_2,\sigma_2^2)$.    What will be the density of $A^2 + B^2 -2AB\cos(\phi)$, where $\phi$ is a constant?  I read the answer in  , applying the same trick will result in the sum of two scaled non-central and dependent chi-squares.
120952
Suppose $X=[X_{1},X_{2}]$ and $X$~$N_2(μ,Σ)$. I wish to find the distribution of  $5X_{1}^2+2X_{1}X_{2}+X_{2}^2$. Since this is of a quadratic form I do not know a way of solving this. However I kind of feel like its chi-squared distributed with parameter 2. Any answers will be welcomed. Thanks
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Given that $a$~$\mathcal{N}(\mu, \Sigma)$ and $\mu$~$\mathcal{N}(0, \sigma^2 I)$, Is $p(a|\Sigma, \sigma)=\int p(a|\mu, \Sigma)p(\mu|\sigma)d\mu$ again a Normal distribution? And how to estimate the mean and variance respectively? Thanks a lot!
60383
If $X \sim \mathrm{Normal}(\mu,\sigma^2)$ and $Y \sim \mathrm{Normal}(\mu,\sigma^2)$ are independent random variables, how do I prove that $X+Y$ and $X-Y$ are also independent?  What happens with the independence between $X+Y$ and $X-Y$ when $X \sim \mathrm{Normal}(\mu_x,\sigma_x^2)$ and $Y \sim \mathrm{Normal}(\mu_y,\sigma_y^2) $  Thank you
8579
Let $X$ and $Y$ be independent and uniform random variables on $(0,1)$. Find the density of $X+Y$.  I know $$f_{X+Y}=\int_{-\infty}^{\infty}\,f_X(x)\,f_Y(z-x)dx$$  Now when $x\in(0,1)$, I know $f_X(x)=1$. Assuming $z\in (0,1)$, I need $z&gt;x$ in order for $f_Y(y)=1$. In that case, $$f_{X+Y}=\int_{0}^{z}\,(1)\,(1)dx=z.$$   Now when $x\in(0,1)$,  $f_Y(y)$ can also equal $1$ if $z \in (1,2)$. In that case, I need $z-x&lt;1 \implies x&gt;z-1.$ So I have: $$f_{X+Y}=\int_{z-1}^{1}\,(1)\,(1)dx=2-z.$$   Does this all look correct?
112321
Let $X_1, \ldots, X_n$ be i.i.d. exponentially distributed random variables with  density  $$\eqalign{\theta^{-1} e^{-x/\theta}, &amp;x \ge 0 \\ 0, &amp;x \lt 0} $$  and let $Y_i = X_{(i)}$ denote the order statistics such that $Y_1 \leq \cdots \leq Y_n$.  How to show that $$     2\frac{\left(\sum_{i=1}^{r}Y_i\right) + (n-r)Y_r}{\theta} $$     has a chi-square distribution with $2r$ degrees of freedom?  I wrote the joint density of $(Y_1,Y_2,...,Y_r)$ but nothing became apparent.
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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?
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Let $X_1$ and $X_2$ be two independent random variables with distribution $U(0,1)$. Let $Y=|X_1-X_2|$. Find the distribution function and density of $Y$.   Solution:  f(x1)=1 whenever x1 between 0 and 1          =0 Else where f(x2)=1 whenever x2 between 0 and 1          =0 Else where f(x1,x2)=1 whenever x1,x2 between 0 and 1           =0 Else where G(y)= P[-y &lt;= x1-x2 &lt;= y]       =\int_{-y}^y \int_{x1_y}^{y+x1} 1 dx2 dx1       =\int_{-y}^y -2y dx1       =4y^2 when y between 0 and 1       =0 when y less than 0       =1 when y greater than 1 Thus, g(y)= \diff G(y) =8y  Is it correct?
110317
I am wondering what is the probability density function for the normal cdf $\Phi (aX+b)$, where $\phi$ is the usual standard normal cumulative distribution function  I want to calculate $\mathbb{E}[\Phi(aX+b)]$ but i am stuck on how to get the distribution. thank you =]  note: X is normally distributed
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The question is-     Let $X_1,X_2,..,X_n$ be iid random variables from a continuous distribution whose density is symmetric about $0$. Suppose $\mathbb{E}(|X_1|)=2$ and define $Y=\sum_{i=1}^{n}X_i$ and $Z=\sum_{i=1}^{n}I(X_i&gt;0)$. Then calculate covariance between $Y$ and $Z$.   My attempt:  $E(X_i)=0$ for all $i=1(1)n$ because $X$ is symmetric about $0$ and $E(|X|) $ exists.   Now,  $Cov (Y,Z)=E(YZ)-E(Y)E(Z)$ $=E(YZ)-0$ $=E[(\sum_{i=1}^{n}X_i)(\sum_{i=1}^{n}I(X_i&gt;0)]$  $=(\sum_{i=1}^{n}E[(X_i.I(X_i&gt;0))]$ $+\sum\sum_{i \neq j}E(X_i)E(I(X_j&gt;0)$ as $X_i,X_j$ are independent.   $=\sum_{i=1}^{n}E[(X_i.I(X_i&gt;0)] +0 $ as $E(X_i)=0$  $ =\sum_{i=1}^{n}\{E[X_i.I(X_i&gt;0)|I(X_i&gt;0)=1]×1/2] + E[X_i.I(X_i&gt;0)|I(X_i&gt;0)=0]×1/2]\}$  $=\sum_{i=1}^{n}E[X_i.I(X_i&gt;0)|I(X_i&gt;0)=1]×1/2] +0$  $=\sum_{i=1}^{n}E[X_i|X_i&gt;0]×1/2]$  $=2n×(1/2)$ $=n$  Is my reasoning correct ? Thanks in advance!
87354
Let  $X\sim \mathcal{N}(6,1)$ and $Y\sim\mathcal{N}(7,1)$ be two independent normal variables. Find $Pr(X&gt;Y)$. the answer is $0.2389$ but I do not know how to do it.
102954
Let $X$ ~ $U(0,2)$ and $Y$ ~ $U(-10,10)$ be two independent random variables with the given distributions. What is the distribution of $V=XY$?   I have tried convolution, knowing that   $$h(v) = \int_{y=-\infty}^{y=+\infty}\frac{1}{y}f_Y(y) f_X\left (\frac{v}{y} \right ) dy$$  We also know that $f_Y(y) = \frac{1}{20}$,    $$h(v)= \frac{1}{20} \int_{y=-10}^{y=10} \frac{1}{y}\cdot \frac{1}{2}dy$$ $$h(v)=\frac{1}{40}\int_{y=-10}^{y=10} \frac{1}{y}dy$$  Something tells me, there is something weird here since it is discontinuous at 0. Please help.
3712
If $Y \sim N(\mu,\sigma^2)$ is normally distributed, then $X=\mathrm{e}^Y$ is lognormally distributed. To get the log-$\mu$ and log-$\sigma$ of this lognormal distribution you calculate $$\sigma^2 = \ln\left( \frac{\mathit{Var}}{E^2} + 1 \right)$$ and $$\mu = \ln(E)-\frac{\sigma^2}{2}$$.  But how is this in the case of a bivariate Normaldistribution, $Y \sim N((\mu_1, \mu_2),\Sigma)$, $\Sigma=((\sigma_1,\rho),(\rho,\sigma_2))$ How do I get $\mu, \sigma, \rho$? And more general in the case of a multivariate distribution?
36151
It's a standard result that given $X_1,\cdots ,X_n $ random sample from $N(\mu,\sigma^2)$, the random variable $$\frac{(n-1)S^2}{\sigma^2}$$ has a chi-square distribution with $(n-1)$ degrees of freedom, where $$S^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2.$$ I would like help in proving the above result. Thanks.
55129
Suppose $X_1, X_2,...,X_n$ be i.i.d from $N(\mu,\sigma^2)$ with unknown $\mu \in \mathcal R$ and $\sigma^2&gt;0$  Let $Z=\frac{X_1-\bar{X}}{S},$ S is the standard deviation here.  It is can be shown that  $Z$ has the Lebesgue p.d.f.  $$f(z)=\frac{\sqrt{n} \Gamma\left(\frac{n-1}{2}\right)}{\sqrt{\pi}(n-1)\Gamma\left(\frac{n-2}{2}\right)}\left[1-\frac{nz^2}{(n-1)^2}\right]^{n/2-2}I_{(0,(n-1)/\sqrt{n})}(|Z|)$$  My question is then how to get this pdf?  The question is from  in example 3.3.4  to find the UMVUE of $P(X_1 \le c)$. I can understand the logic and procedures to find the UMVUE but don't know how to get the pdf.  I think this question also relate to this    Thank you very much for help or point to any related references will be also appropriated.
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Let $(X, Y)$ be distributed as a multivariate normal with parameters  $$ \mu = \begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix} \qquad \Sigma = \begin{bmatrix} \sigma_X^2 &amp; \sigma_{XY} \\ \sigma_{XY} &amp; \sigma_Y^2 \end{bmatrix}. $$   I would like to calculate $E(X | y_1 &lt; Y &lt; y_2)$, where $y_1$ and $y_2$ are constants.  From , I have managed to work out that $$ E(X | Y &gt; y_1) = \mu_X + \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}{1 - \Phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}\right]\\ E(X | Y &lt; y_2) = \mu_X - \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}{\Phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}\right] ,$$ where $\phi(\cdot)$ and $\Phi(\cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 &lt; Y &lt; y_2)$, though.  I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them:
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$X\sim U[0,1]$ and $Y\sim U[-1,1]$ are two uniform-distributed R.V.'s. Are $X^2$ and $Y^2$ still uniform? Do they have explicit probability density funtion?
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Let $X, Y$ be independent standard normal random variables.  We already know that $X+Y$ is normal with mean 0 and variance 2. However, I am trying to prove this result in a slightly different way than usual, i.e., using convolutions.  We can address the problem in the following way: the sum of the two variables has a density $h(x)$ given by the convolution product of the densities of the two variables, say $f$ and $g$, so that   $$ h(x) = \int_{-\infty}^{\infty} f(x-y)g(y)dy. $$  However,  $$ f(y) = g(y) = \frac{1}{\sqrt {2\pi}}\exp(-y^2/2). $$  Hence now the integral becomes:  $$ h(x) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-(x-y)^2/2)\exp(-y^2/2)dy. $$  However, I am puzzled here. How should we go on from here? I cannot see how to transform this integral into the integral of a normal r.v of type $N(0,2)$. Thanks in advance for your help.