Suppose $X$ and $Y$ are i.i.d. Standard Normal random variable. I need to find $\Pr(Y^2/X^2&lt;3)$. Instead of doing integral, I am thinking whether there is a simpler way. Since $Y/X$ is $\tan(\theta)$ where $\theta$ is the angle of the point $(X,Y)$ to the $X$-axis, but in order to solve this probability, we need to know the distribution of $\theta$ and I do not know how to proceed.
120914
I need to use the ratio of two variables as the dependent variable in a regression. Both variables are normally distributed but with positive values. I can either center them or use as it is.  If I center them: the ratio of two centered normals is a Cauchy distribution. Is it therefore meaningful to do a "Cauchy" regression (e.g. using JAGS or Stan?)   If I leave as they are: What would be the distribution of their ratio?
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 $Y^2$?
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?
85658
Let random variables $X$ and $Y$ be independent Normal with distributions $N(\mu_{1},\sigma_{1}^2)$ and $N(\mu_{2},\sigma_{2}^{2})$. Show that the distribution of $(X,X+Y)$ is bivariate Normal with mean vector $(\mu_{1},\mu_{1}+\mu_{2})$ and covariance matrix  $$ \left( \begin{array}{ccc} \sigma_{1}^2 &amp; \sigma_{1}^2  \\ \sigma_{1}^2 &amp;\sigma_{1}^2+\sigma_{2}^2  \\  \end{array} \right).$$  Thanks .
22712
... and why ?  Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:    $E(X_1-X_2)=\mu_1-\mu_2$ $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$   Now let's say $X_1$, $X_2$  are t-distributions with $n_1-1$, $n_2-2$ degrees of freedom. What is the distribution of $X_1-X_2$ ?   This question has been edited: The original question was "What are the degrees of freedom of the difference of two t-distributions ?". mpiktas has already pointed out that this makes no sense since $X_1-X_2$ is not t-distributed, no matter how approximately normal $X_1,X_2$ (i.e. high df) may be.
114820
So I know that if we want to find the probability distribution of a sum of independent random variables $X + Y$, we can compute it from the probability distributions of $X$ and $Y$, by saying  $$f_{X + Y}(a) = \int_{x = -\infty}^{\infty} f_{X, Y}(X = x, Y = a - x)~dx = \int_{x = -\infty}^{\infty} f_X(x) f_Y(a - x)~dx$$  Intuitively, this makes sense, because if we want to find the probability that two random variables sum to $a$, it's basically the sum of the probabilities of all the events that lead to those variables summing to $a$.  But how can I formally prove this statement?
8182
I need to find the distribution of the random variable  $$Y=\sum_{i=1}^{n}(X_i)^2$$ where $X_i\sim{\cal{N}}(\mu_i,\sigma^2_i)$ and all $X_i$s are independent. I know that it is possible to first find the product of all moment generating functions for $X_i$s, and then transform back to obtain $Y$'s distribution. However, I wonder whether there is a general form for $Y$ like the Gaussian case: we know the sum of independent Gaussian is still a Gaussian, and thus we only need to know the summed mean and summed variance.   How about all $\sigma^2_i=\sigma^2$? Will this condition make a general solution?
83455
Assume we have $n$ distinct points $x_1,x_2,...,x_n\in \Bbb{R}$, and observations which are independent random variables $y_1,y_2,...,y_n$ depending on formula: $$ y_i=\theta_0+\theta_1x_i^2+\theta_3e^{x_i}+\varepsilon_i $$ where $\varepsilon_i$ are distributed as $N(0,\sigma^2)$. Derive the maximum likelihood estimator for $\theta_{0},\theta_{1},\theta_{2}$ .  I know how to find the MLE for a known distribution like normal or Poisson, but for this question, I don't how I can find the likelihood function, should I consider it as normal distribution? And I don't know what the mean and variance would be since I can't figure out the distribution and likelihood function. Also, should I consider $\varepsilon_i$ as a useful variable, or just treat it as an error in estimation? Thanks for any help.
65089
Hi I'm solving some exercise problems in my text : "A Course in Mathematical Statistics".  I'm in the chapter "Point estimation" now, and I want to find a UMVUE of $\theta$ where $X_1 ,...,X_n$ are i.i.d random variables with the p.d.f $f(x; \theta)=\theta e^{-\theta x}, x\gt0$.   I know that $E(X_i)=1/\theta,$ for each $i$, and also have that $\bar{X}$ (or equivalently $\sum_1^n X_i$) is a complete sufficient statistic for $\theta$. But I cannot go any further here. Somebody can help me?
83429
If $$Y=\sum_{i=1}^{N}X_i^2$$ where $X_i \sim \mathcal{N}(0,\sigma^2)$, i.e. all $X_i$ are i.i.d. normal random variables of zero mean with same variances, then $$Y \sim \Gamma\left(\frac{N}{2},2\sigma^2\right).$$  I know the chi-squared distribution is a special case of the gamma distribution, but could not derive the chi-squared distribution for random variable $Y$. Any help, please?
96676
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.   So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$.   We shall find the CDF of $Z$ and then differentiate it.  $F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \  dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.  At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.  How do I do this? Or otherwise is there different more elegant way to get the required pdf?
21152
I am trying to find distribution of random variable:  $$\frac{X_1+X_2 X_3}{\sqrt{1+X_3^2}}$$  where $X_i \sim \mathcal{N}(0,1)$ and independent.  My thinking so far is to use random variable algebra and to find distributions step by step. However, I don't know what to do with the variable: $\sqrt{1+X_3^2}$ (it seems like some variant of $\chi$-distribution).   Moreover, I'm worried that my thinking is to complicated and that the solution should be more elegant.
34581
I came across this question as an exercise, had a brief idea, but didn't know how to proceed.     Let $X \sim N(0, 1)$. What is the p.d.f of $|X|$ ?   I know the final p.d.f looks just like the right half of the original pdf, but extended vertically for a factor of 2.  Could someone please show mathematically what the p.d.f of variable |X| is going to be, and explain briefly? Thank you.
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 $\mathbb{E}(X\mid Z)=\mathbb{E}\,[\mathbb{E}(X\mid Z,U)\mid Z]$.
22586
$X_1$, $X_2$, $X_3$ are independent random variables, each with an exponential distribution, but with means of $2.0, 5.0, 10.0$ respectively. Let $Y$= the smallest or minimum value of these three random variables. Derive and identify the distribution of $Y$. (The distribution function may be useful).    How do I solve this question? Do I plug in each mean to the exponential distribution? I would appreciate it if someone could explain this to me, thanks.
123303
Given two independent random variables $X \sim \mathcal{U}[-1,5]$ and $Y \sim \mathcal{U}[-5,5]$, what is $$\mathbb{E}\{Y - X | X \le 1, Y &gt; X, Y \in [-1,1] \}\,?$$  I managed to compute the respective probability   $$ \mathbb{P}\{ X \le 1, Y &gt; X, Y \in [-1,1] \} = \mathbb{P}(X \le 1) \mathbb{P}(Y &gt; X) \mathbb{P}(Y \in [-1,1]) = 1/3 * 1/2 * 1/5 = 1/30$$  How do I continue from there?
86454
I am working on a paper, and have a problem regarding ellipsoid and Trivariate Normal Distribution. Suprisingly I can't find much in literature but I found your in one of your answers:     Because this construction has nothing to do with "confidence" per se,   the objective is to establish some convention for describing the shape   and relative size of the points. Using 1.96 sort of works (for three   variables): it contains about 72% of the probability of the trivariate   normal distribution. But as the number of variables increases this   method produces ellipses that are far too small. For instance, with 10   variables it will contain only 4.6% of the probability; using 4.28   instead of 1.96 in this case will contain 95% of the probability.   How did you get this number 72%? Or do you have some literature to recommend to me in which I can find this. I would appreciate it very much!