Cfrgtkky

So we have $100$ observations for $(x, y)$.
The mean of $x$ is $1.06$, and for $y$ it is $3$.
The standard deviation is $0.52$ for $x$ and for $y$ it is $1.13$.
the correlation between $x$ and $y$ is $0.89$.

In the question we are told to:

• Estimate the linear regression line of the regression of $Y$ on $X$ and the standard deviation of the errors.

• estimate the regression line when we regress $X$ as dependent variable on $Y$ and obtain an estimate of the standard deviation of the errors.

• Are the two regression lines the same? If not, then explain why not.

• For the regression of $Y$ on $X$, suppose that we wish to predict the
dependent variable $y$ at $x = x^* = 0.7$. Obtain the prediction, as
well as the standard error of the prediction.

• Obtain the standard deviation of the prediction error and hence
obtain a $95%$ prediction interval for $y$ for the the given $x = x*.$

Now I thought we were supposed to generate $100$ points of data assuming $x$ and $y$ had a normal distribution with the given means and standard deviations, and then use stata to regress and find the prediction interval, etc

But I was told this was not the case by the lecturer, and was wondering if there was a way to solve this another way? I'm thinking some kind of derivation/calculations using the above info, but I have no idea where to start.

I've found estimates for $B_1$ and $B_0$ from modifying the formula used for their estimation; the numerator can be turned into $n times cov(x,y)$; and we can find $cov(x,y)$ given $corr(x,y)$ and std of x and y.

Problem now is how to find the standard deviation of the errors and the prediction errors.

Using matrix notation, you get

$ hat{beta} = (X'X)^{-1}X'Y \
Var(hat{beta}) =(X'X)^{-1}X'sigma_{y}^{2}X(X'X)^{-1} = (X'X)^{-1}sigma_{y}^{2}$

So for the simple linear regression, this will be

begin{align}
X &= (j, x) quad X'=(j,x)'\
X'X &=
begin{bmatrix}
n & sum_{k=1}^{n} x_{k} \
sum_{k=1}^{n} x_{k} & sum_{k=1}^{n} x_{k}^{2} \
end{bmatrix} \
(X'X)^{-1}sigma_{y}^{2} &=
begin{bmatrix}
sum_{k=1}^{n} x_{k}^{2} & -sum_{k=1}^{n} x_{k} \
-sum_{k=1}^{n} x_{k} & n \
end{bmatrix}frac{sigma_{y}^{2}}{nsum_{k=1}^{n} x_{k}^{2} -(sum_{k=1}^{n} x_{k})^{2}} \&= begin{bmatrix}
sum_{k=1}^{n} x_{k}^{2} & -sum_{k=1}^{n} x_{k} \
-sum_{k=1}^{n} x_{k} & n \
end{bmatrix}frac{sigma_{y}^{2}}{nsum_{k=1}^{n} x_{k}^{2} -(nbar{X})^{2}} \ quad \
&= begin{bmatrix}
frac{sigma_{y}^{2}sum_{k=1}^{n} x_{k}^{2}}{S^{2}_{X}} & -frac{sum_{k=1}^{n} x_{k}}{nS^{2}_{X}} \
-frac{sum_{k=1}^{n} x_{k}}{nS^{2}_{X}} & frac{sigma_{y}^{2}}{S^{2}_{X}} \
end{bmatrix}\
&=begin{bmatrix}
frac{sigma_{y}^{2}(S^{2}_{X}+nbar{X}^{2})}{nS^{2}_{X}} & -frac{sum_{k=1}^{n} x_{k}}{nS^{2}_{X}} \
-frac{sum_{k=1}^{n} x_{k}}{nS^{2}_{X}} & frac{sigma_{y}^{2}}{S^{2}_{X}} \
end{bmatrix}\
end{align}

So if you have the s.d. for X, you just need to find $sum_{k=1}^{n} x_{k}^{2} $ and you will have everything needed to calculate the estimate variance.

Just manipulate the variance formula a little to get it: $sum_{k=1}^{n} x_{k}^{2} = sigma_{X}^{2} +nbar{X}^{2}$

If you are predicting in-sample, you get $frac{sum_{k=1}^{n} x_{k}^{2}}{S^{2}_{X}} + frac{sigma^{2}_{Y}}{n} $

If you are predicting beyond sample range, you get $frac{sum_{k=1}^{n} x_{k}^{2}}{S^{2}_{X}} + frac{sigma^{2}_{Y}}{n} + sigma^{2}_{Y} $