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What I was hoping is that if I reorganize the first regression $$x = \frac{var(x)}{cov(x,y)}y - \frac{var(x)}{cov(x,y)}\epsilon_x$$ the variance of the new error term would match the variance of the … second regression. …

If we regress $y$ on $x$, the regression coefficient is $\beta_1=\frac{cov(x,y)}{var(x)}$ and if we regression $x$ on $y$, the coefficient is $\beta_2=\frac{cov(x,y)}{var(y)}$. … If we want to draw these two fitted lines on the same $(y-x)$ coordinates we know that the fitted line of regression $x$ on $y$ will be steeper. …

Simple Linear Regression. You regress $Y \sim X$, and You know $\beta_{y,x}$. Now, if you regress $X \sim Y$, what can you say about $\beta_{x,y}$, what about its range? … $\beta_{x,y} = \beta_{y,x}\frac{s_{x,x}}{s_{y,y}}$. How can I infer the range? …

If we have X and Y that are mathematically dependent: X = Y + Z, is it 'forbidden' to use Y as a predictor to X in linear regression? … But, if we can only measure Y, does the regression give us a tool to predict X? Isn't it better to find the regression between Y and Z? I'm confused. …

Consider a linear regression model y on x and x on y. We have $Y = a'X + a$ where $a' = \frac{cov(X,Y)}{Var(X)}$. Equivalently, we have $X = b'Y+b$ where $b' = \frac{cov(X,Y)}{Var(Y)}$. … It follows that: \begin{aligned} a' &= \frac{1}{b'} \\ \frac{cov(X,Y)}{Var(X)} &= \frac{Var(Y)}{cov(X,Y)} \\ \frac{1}{Var(X)} &= Var(Y) \end{aligned} Ok. …

When we consider the flipped regression of $x =b_1'y + b_0'$ can we establish any upper or lower bound on $b_1'$? Other than it being greater than 0?? … This is the only thing I can think of because we know if $b_1 = 2, cov(x,y) > 0$ …

I'm trying to think through whether the regression $y = a + b (x/y)$ Is problematic at all. It's not colinearity, because there's no linear relation? … But, as $y$ increases, then $bx$ would tend to decrease, so I'm suspicious. Is there a problem here? It's just a regression of $y^2$ on $x$, right? …

Given that the plan is to use the inverse formula $x = x(y)$, the regression is also run by considering $y$ as the independent variable and $x$ as the dependent variable. … So I thought: what if I determined $A, B$ by running the regression on $y = y(x)$, i.e. considering $x$ as the independent variable and $y$ as the dependent variable? …

Let $X$ and $Y$ be mean 0 and variance 1 random variables such that we choose $\alpha$ and $\beta$ to minimise $$\mathbb{E}(X-\beta Y)^2$$ and $$\mathbb{E}(Y-\alpha X)^2$$ after not so difficult derivation … This seems very strange, because if $y=mx$ is regression line, then surely $x = \frac{1}{m }y$. …

I read that the conditional probability density function $f_{Y|X}(y|x)$ of $Y$ can be written as $$ f_{Y|X}(y|x) = f_\varepsilon(y - g(x)), $$ where $f_\varepsilon$ is the density of $\varepsilon$. … Here is my attempt: $$ \begin{align} f_{Y|X}(y|x) = f_{Y|X}(Y=y|X=x) &= f_{Y|X}(g(X)+\varepsilon=y|X=x) \\ &= f_{g(X)+\varepsilon|X}(g(x)+\varepsilon=y|X=x) \\ &= f_{g(X)+\varepsilon|X}(\varepsilon = y …

In a multivariate context, that is with at least X or Y being a random vector, are there formulae or theorems that link (even remotely) the forward and inverse regression, $\text{E}(X|Y)$ and $\text{E} … (Y|X)$ ? …

x on y. … The equation of the line $L_{1}$ can be written in the form $x = ay + b$. (a) Find the value of $a$ and the value of $b$. Let $L_{2}$ be the regression line of $y$ on $x$. …

Suppose I have data $x$ and $y$, where $x$ is a count and $y$ is continuous. I would like to predict $x$ from $y$. … Specifically, I would use the proposed regression method to answer a question such as "What is the value of $x$ for a corresponding $y$-value of $y$ = 5?" …

I've been reading linear regression and least square estimator. … However I wonder what would be the effect if we do regression of x onto y. Would we then be able to use the least square estimate for the regression of x onto y to estimate of $\frac{1}{\beta}$? …

Given a linear regression model with all the assumptions checked and validated, I would like to obtain the probability that $Y>y|X=x$. … For example for the iris dataset, I would do the following to obtain the probability of $Y>5|X=1,2,3...7$: plot(Sepal.Length~Petal.Length, data=iris) lm1<-lm(Sepal.Length~Petal.Length, data=iris) summary …