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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?

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

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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. –  TheHopefulActuary Dec 3 '12 at 6:42
    
Thanks a lot Kyle! What does $\bar x$ stand for here? –  Q.matin Dec 3 '12 at 6:45
    
$\bar x$ 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 :-) –  TheHopefulActuary Dec 3 '12 at 6:50
    
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? –  Q.matin Dec 3 '12 at 6:52
    
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. –  TheHopefulActuary Dec 3 '12 at 6:57
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