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Tutor profile: Sam K.

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Sam K.
Tax Analyst, MA
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Questions

Subject: Calculus

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Question:

If a function is differentiable, is it continuous? Is the converse true? (i.e. if a function is continuous, is it necessarily differentiable)

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Sam K.
Answer:

The answer to the first is yes: in order for a function to be differentiable, it must be continuous. For example: a function may contain cusps, a point at which it is not differentiable.

Subject: Economics

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Question:

Describe a real-life example of a market with perfect competition, and a monopoly.

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Sam K.
Answer:

An example of perfect competition could be an open market in a city like Istanbul: many vendors sell identical or near identical products, and thus don't have the ability to differentiate their prices to gain a higher profit. A monopoly could be a public company such as Ontario Hydro. This type of monopoly is granted by the government, rather than by producing a unique product.

Subject: Econometrics

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Question:

Derive the OLS estimator for $$\beta_1$$ the single linear regression. min ($$\beta_0, \beta_1) \Sigma^N_{i=1} (y_i - \hat{\beta_0} - \hat{\beta_1})^2$$

Inactive
Sam K.
Answer:

We first take the partial derivative with respect to $$\beta_1$$. This yields the equation: $$\Sigma^N_{i=1} -2 x_i (y_i - \hat{\beta_0} - \hat{\beta_1})$$ To derive the first order condition, set the above equal to 0, and substitute in the fact that $$ \hat{\beta_0} = \bar{y} - \hat{\beta_1} \bar{x}$$. We divide out the -2, and we can rearrange to get this equation: $$ (\Sigma^N_{i=1}x_i y_i - \bar{y} - \hat{\beta_1} \bar{x} - \hat{\beta_1} x^2_i) =0 $$ Distribute the sum to get: $$\Sigma^N_{i=1}x_i y_i - \bar{y} \Sigma^N_{i=1} x_i + \hat{\beta_1} \bar{x} \Sigma^N_{i=1} x_i - \hat{\beta_1} \Sigma^N_{i=1} x^2_i = 0 $$ Using the fact that $$\Sigma^N_{i=1} y_i = N \bar{y}$$ and $$\Sigma^N_{i=1}(x_i - \bar{x})(y_i - \bar{y} = \Sigma^N_{i=1} x_i* y_i -N\bar{x}\bar{y}$$ we rearrange the equation above and substitute in, once we have solved for $$\hat{\beta_1}$$ as follows: $$ \hat{\beta_1} =\frac{\Sigma^N_{i=1} x_i* y_i -N\bar{x}\bar{y}}{\Sigma^N_{i=1} (x^2_i -N\bar{x}^2)} $$. Finally, substituting in the two facts above, we arrive at our final answer: the analytical solution for $$\hat{\beta_1}$$: $$\hat{\beta_1} = \frac{\Sigma^N_{i=1}(x_i - \bar{x})(y_i - \bar{y})}{\Sigma^N_{i=1} (x_i -\bar{x})^2} $$

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