I've been working on this problem and got stuck.  Assume that X and Y have joint normal distribution, that each $X, Y \sim N(0,1)$, and that their correlation is $\frac{1}{2}$. Calculate $Pr(X \ge 0, Y \ge 0)$.  I know that I can rewrite (X,Y) in terms of independent normal variables (W,T) as follows  $$    \begin{bmatrix}    X\\    Y\\    \end{bmatrix} =    \begin{bmatrix}    1 &amp; 0\\    \rho &amp; \sqrt{1-\rho^2}\\    \end{bmatrix}    \begin{bmatrix}    X\\    Z\\    \end{bmatrix}    $$ where $\rho$ is the correlation. This has left me with $$ X=X\\ Y=\frac{1}{2}X + Z\sqrt{\frac{3}{4}} $$ And so I have $$ Pr(X \ge 0, Y \ge 0) = Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0) $$ So I just realized that the T variable I had before was superfluous, so I removed it. I am wondering from here if it would make sense to put this: $$ Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0)=Pr(X \ge 0, Z\sqrt{\frac{3}{4}} \ge -\frac{1}{2}X) $$ Still not sure how to proceed from here. Maybe simplify to $$ Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0)=Pr(X \ge 0,  Z\sqrt{\frac{3}{4}} \ge 0) $$ Since we are looking for $Pr(X \ge 0)$ anyway?
57564
$X$ and $Y$ have a bivariate normal distribution with $\rho$ as covariance. $X$ and $Y$ are standard normal variables.  I showed that $X$ and $Z= \dfrac{Y-\rho X}{\sqrt{1-\rho^2}}$ are independent standard normal variables.  Using this I need to show that   $$P(X &gt;0,Y&gt;0) = \frac14 + \frac{1}{2\pi} \cdot\arcsin(\rho).$$
72848
My random variable $X$ is normally distributed with mean $b$ and variance $p$. I defined a new RV, $Y$, such that $Y=1/X$. Does anyone know how to find the mean of $Y$ and $Y^2$?
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):  $Z \sim \mathcal{N}(-3, 22)$
105744
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.
31422
Suppose that $X\sim N(0,1)$, $Y\sim N(0,1)$ with correlation $(X, Y) =ρ$ where $ρ ∈ (−1, 1)$.  Show the following,   Correlation $(Φ(X),Φ(Y))=\dfrac6π \arcsin\dfrac ρ2 $.  Here $Φ(X)$, and $Φ(Y)$ denote the CDF of Random variables $X$ and $Y$ respectively.  What I know so far:  $$\text{Cov}( Φ(X),Φ(Y)) =E(Φ(X)Φ(Y)) - E(Φ(X)) ×E(Φ(Y)) $$  We also know since  $X \sim N(0,1)$ and $Y\sim N(0,1)$;  $Φ(X)\sim\text{unif}(0,1)$, $Φ(Y)\sim\text{unif}(0,1)$. Hence $$E(Φ(X)) =E(Φ(Y)) =\frac12.$$  I am stuck in finding $E(Φ(X)Φ(Y))$ [I tried to find it using double expectation].  What I have written may be the wrong way of attempting the question. So I am thankful for any help.
100706
I have four random variables, A, B, C, D, with known mean and variance. As well:   Cov(A,B) is known and non-zero  Cov(C,D) is known and non-zero  A and C are independent   A and D are independent  B and C are independent   B and D are independent   I then create two new random variables:   X = A*C Y = B*D   Is there any way to determine Cov(X,Y) or Var(X+Y)?   If not exactly, is there any way to estimate it? What if I could determine the distributions of A and B and C and D?
16601
I have a question regarding correlation inference. Consider, I have two sets of variables X and Y. For an x element of X I know the correlation to an unknown variable z. I also have the covariance matrix X and Y. Moreover, I know a set of (y1,y2,y3) element of Y, which model best my variable x. So how to infer the correlation of (y1,y2,y3) and z?  I need a function: f(corr(x,z), corr(x,(y1, y2, y3)), cov(X,Y)) --> corr((y1,y2,y3),z)  This might be a very simple question, but I have no idea how to do it or for what I can search in the literature.  Thank you in advance!
43115
Suppose that  $X$ and $Y$ are two i.i.d. uniform random variables on the interval $[0,1]$  Let $Z=X/Y$, I am finding the cdf of $Z$, i.e. $ \Pr(Z\leq z) $.  Now, I came up with two ways of doing this. One produces a correct answer consistent with the pdf here: , the other does not. Why is the second method wrong?  First Method  $\newcommand{\rd}{\mathrm{d}} \Pr(Z\leq z) = \Pr(X/Y\leq z) = \Pr(X\leq zY) = \int^{1}_{0}\int^{\min(1,zy)}_{0} \rd x \rd y = \int^{1}_{0}\min(1,zy)\ \rd y$ $  = \left\{      \begin{array}{lr}        \int^{1/z}_{0}zy\ \rd y + \int^{1}_{1/z} \rd y&amp; : z &gt; 1\\        \int^{1}_{0}zy\ \rd y &amp; : z \leq 1      \end{array}    \right. $ $  = \left\{      \begin{array}{lr}        1 - \frac{1}{2z} &amp; : z &gt; 1\\        \frac{z}{2} &amp; : z \leq 1      \end{array}    \right. $  This appears correct.  Second Method  $\Pr(X/Y\leq z) = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(zY \geq 1) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(zY &lt; 1) $ by total probability  $ = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(Y \geq 1/z) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(Y &lt; 1/z)$  Taking $z&gt;1$ yields $ (1)(1-\frac{1}{z}) + (\int^{1/z}_{0}\int^{zy}_{0} \rd x \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + (\int^{1/z}_{0}zy\ \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + \frac{1}{2z^{2}}$  This is already different. Why is this wrong?  Thanks!
26074
I have two independent gaussian distributions to combine and I have a doubt.  Let's say we have $X \sim N(\mu_x,\sigma^2_x)$ and $Y \sim N(\mu_y,\sigma^2_y)$. I want to mix the two variables with a 0.5 weight each, so I obtain a new variable $Z=0.5*X+0.5*Y$ with mean $\mu_z=0.5*\mu_x+0.5*\mu_y$.  Is it right to say that $\sigma^2_z=0.5^2*\sigma^2_x+0.5^2*\sigma^2_y$?  Because I clearly obtain less variance than intuitively expected when combining distributions.
51403
Let $X\sim N(0,1)$ and let $Z$ be a random variable independent of $X$ such that:  \begin{equation} \Pr(Z=z) = \begin{cases} \frac{1}{2} &amp; \mbox{if $z = -1$ or $z=1$}, \\ \\ 0 &amp; \mbox{otherwise}. \end{cases} \end{equation}   Define $Y = ZX$. Show that $Y \sim N(0,1)$. Determine $\operatorname{Cor}(X,Y)$. Are $X$ and $Y$ independent?   Progress:  I'm basically stuck at the beginning. I got as far as $P(ZX &lt; y) = P(Z &lt; y/X)$, which I'm not even sure is correct. I'm also not clear on what it means by defining $Y = ZX$ - does it mean I need to state the probability density function or something? And how do I show that $Y \sim N(0,1)$?
91112
Given that $X,Y\stackrel{\text{i.i.d.}}{\sim}\text{Exp}$ with mean $\lambda(&gt;0)$ and $Z=\mathbf{1}_{X&lt;Y}$, what is $\mathbb{E}(X\mid Z=z)$ ?   We have $Z=1$ with probability $\Pr(X&lt;Y)$ and $Z=0$ with probability $1-\Pr(X&lt;Y)$.   But $\Pr(X&lt;Y)=\frac{1}{2}$ as $X$ and $Y$ are i.i.d. random variables.   Does this mean I can simply say that $Z\sim\text{Ber}\left(\frac{1}{2}\right)$ ?   Then, $\mathbb{E}(X)=\mathbb{E}(X\mid Z=0)\times\frac{1}{2}+\mathbb{E}(X \mid Z = 1) \times \frac{1}{2}$  $\implies2\lambda=\mathbb{E}(X\mid Z=0)+\mathbb{E}(X\mid Z=1)$.   But I cannot conclude anything from here. I couldn't find the conditional distribution of $X\mid Z$ either. Maybe I have to condition on another variable $U$ so that $\mathbb{E}(X\mid Z)=\mathbb{E}\,[\mathbb{E}(X\mid Z,U)\mid Z]$.
32232
If $X_i$, $i =1,2$ are independent and have normal distribution with mean $0$ and variance $\sigma_i ^2$. Show that $X_1 + X_2$ has a normal distribution with mean $0$ and variance $\sigma_1^2 + \sigma_2^2$.  Any idea is appreciated.
110290
I have a basic question.  Say I have two random variables, $X$ and $Y$.  I can standardize them by subtracting the mean and dividing by the standard deviation, i.e., $X_{standardized} = \frac{(X - E(X))}{(SD(X))}$.  Is the correlation of $X$ and $Y$, $Cor(X, Y)$, the same as the covariance of the standardized versions of $X$ and $Y$?  That is, is $Cor(X, Y) = Cov(X_{standardized}, Y_{standardized})$?
70908
I was asked this question regarding correlation recently, and although it seems intuitive, I still haven't worked out the answer satisfactorily. I hope you can help me out with this seemingly simple question.  Suppose I have three random variables $A$, $B$, $C$. Is it possible to have these three relationships satisfied? $$ \mathrm{corr}[A,B] = 0.9 $$ $$ \mathrm{corr}[B,C] = 0.8 $$ $$ \mathrm{corr}[A,C] = 0.1 $$ My intuition is that it is not possible, although I can't see right now how I can prove this conclusively.
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 ?
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?
25610
So I'm reading notes on Gaussian Processses, and came across  syntax $p(\mathbf{y} \mid \text{stuff}) = N(\mathbf{y} \mid \mathbf{0}, \mathbf{K})$ for multivariate normal distribution, and I'm not exactly sure how to decipher it the $\mathbf{y} | $ bit. I suppose it doesn't mean "mean $\mathbf{y}$ is conditional on zero vector", because I don't think that makes sense. Assuming this is standard notation for something and I'm just unfamiliar with it, what it means? E.g. how it differs from $N(\mathbf{0}, \mathbf{K})$?