Assume a linear relationship for a company that has several shops is not known. Let $Y_i$ be the profit the shop number $i$ makes in the coming year. Let $x_i$ be the size of the shop number $i$. We assume that for all these shops the following relationship holds. $$Y_i = \alpha + \beta x_i + \epsilon_i$$ where $\epsilon_i$ is a random term for which $E[\epsilon_i] = 0$ and such that $\epsilon_1,\epsilon_2,...$ are $i.i.d$.

$\alpha, \beta$ and $\sigma$ are unknown and must be estimated.

a.) If we want to open a shop with size 3, what is the expected profit in terms of $\alpha$ and $\beta$?

b.) Estimate $\alpha$ and $\beta$ using linear regression for the following data $x_i = 1,2,3,4,5,6,7,8,9$ corresponding to $y_i = 0,1,1,2,1,3,3,3,4$.

For part a, is the expected profit going to be $Y_i = \alpha + \beta x_i$, where $x_i = 3$, since $\epsilon_i$ has mean $0$?

For part b, we have not yet been taught linear regression but I did google it and I know that I must find $a$ and $b$ minimizing the sum of the distance square: $$\sum^n_i (Y_i -(a - bx_i))^2$$ then the estimate of $\alpha$ and $\beta$ are then $a$ and $b$ which minimize the above expression. Also, why would the given data values be necessary to find $\alpha$ and $\beta$? Isn't those values only necessary to find the standard deviation?

  • 2
    $\begingroup$ Yes, the expected profit is $\alpha+3\beta$. $\endgroup$ – Gerry Myerson Dec 3 '12 at 5:51

As Gerry said, the expected profit would be $\alpha+3\beta$ and then as for finding $\alpha$ and $\beta$ you have to do each one separatly. $\beta=\frac{Cov(x_i,y_i)}{Var(x)}$ and $\alpha=\bar y-\beta \bar x$ at least if you were going to calculate it manually that's what I was taught. If you arent sure how to find those: $$Cov(x_i,y_i)=\frac{\sum_{i=1}^n (x_i-\bar x)(y_i-\bar y)}{n-1}$$ and $$Var(x)=\frac{\sum_{i=1}^n (x_i-\bar x)^2}{n-1}$$ So really you're always going to want to find the coefficient of the $x$ variable first in order to find the intercept. You are right though about the minimizing the residuals, but I haven't seen that equation that you posted. But this formula came out of a "pre" econometrics book.

  • $\begingroup$ Also maybe consider adding the tag for "regression" and "statistics" I wouldn't put this in the same category as probability. But that's just me. $\endgroup$ – TheHopefulActuary Dec 3 '12 at 6:42
  • $\begingroup$ Thanks a lot Kyle! What does $\bar x$ stand for here? $\endgroup$ – Q.matin Dec 3 '12 at 6:45
  • $\begingroup$ $\bar x$ is just the average. Which is nothing more than what you learned in basic 3rd grade math. It's literally the average of all the $x$'s and all the $y$'s. What also helped me when I first learned of regression was to use the form $Y=\beta_0+\beta_1X_1+\varepsilon$ but by all means if using $\alpha$ and $\beta$ is your preference do so. It's helpful to me just because it reminds me of which I need to calculate first. Which would be $\beta_1$, hence the $1$. But that's just me :-) $\endgroup$ – TheHopefulActuary Dec 3 '12 at 6:50
  • $\begingroup$ Im confused, the average of what though? Because I have to estimate and it doesnt give me any avg? Is it the avg of the data values given in my question? $\endgroup$ – Q.matin Dec 3 '12 at 6:52
  • $\begingroup$ Yep it is indeed the average of the data values, which you will need to calculate yourself. If seeing it in a mathematical form helps, try using this: $$\bar x=\frac{\sum_{i=1}^n x_i}{n}$$ When you need to find the average of $y$ for covariance just swap out the $x$'s for $y$'s. When it comes to regressino analysis, you will always need to have a data set or a given value for the mean of a variable. Unless you are doing regression theory then that's an exception. $\endgroup$ – TheHopefulActuary Dec 3 '12 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.