Consider the savings function sav=β0+β1 inc+u, u=√(i n c) ·e where e is a random variable with E

Consider the savings function sav=β0+β1 inc+u, u=√(i n c) ·e...

Consider the savings function $$\operatorname{sav}=\beta_{0}+\beta_{1} \operatorname{inc}+u, u=\sqrt{i n c} \cdot e$$ where $e$ is a random variable with $\mathrm{E}(e)=0$ and $\operatorname{Var}(e)=\sigma_{e}^{2} .$ Assume that $e$ is independent of inc. i. Show that $\mathrm{E}(u | \text { inc })=0,$ so that the key zero conditional mean assumption (Assumption SLR.4) is satisfied. [Hint: If $e \text { is independent of inc, then } \mathrm{E}(e | i n c)=\mathrm{E}(e) .]$ ii. Show that $\operatorname{Var}(u | \text { inc })=\sigma_{c}^{2}$ inc, so that the homoskedasticity Assumption SLR.5 is violated. In particular, the variance of sav increases with inc. [Hint: Var(elinc) = Var(e) if $e$ and inc are independent.] iii. Provide a discussion that supports the assumption that the variance of savings increases with family income.

Consider the savings function
$$\operatorname{sav}=\beta_{0}+\beta_{1} \operatorname{inc}+u, u=\sqrt{i n c} \cdot e$$
where $e$ is a random variable with $\mathrm{E}(e)=0$ and $\operatorname{Var}(e)=\sigma_{e}^{2} .$ Assume that $e$ is independent of inc.
i. Show that $\mathrm{E}(u | \text { inc })=0,$ so that the key zero conditional mean assumption (Assumption SLR.4) is satisfied. [Hint: If $e \text { is independent of inc, then } \mathrm{E}(e | i n c)=\mathrm{E}(e) .]$
ii. Show that $\operatorname{Var}(u | \text { inc })=\sigma_{c}^{2}$ inc, so that the homoskedasticity Assumption SLR.5 is violated. In particular, the variance of sav increases with inc. [Hint: Var(elinc) = Var(e) if $e$ and inc are independent.]
iii. Provide a discussion that supports the assumption that the variance of savings increases with family income.

Key Concepts

Heteroskedasticity

Heteroskedasticity is the situation where the variance of the error term is not constant across different levels of the explanatory variable. It violates the homoskedasticity assumption and can lead to inefficient estimates and biased standard errors in ordinary least squares estimations, which in turn affects the validity of hypothesis tests and confidence intervals.

Homoskedasticity

Homoskedasticity refers to the condition in which the variance of the error term is constant across all levels of the explanatory variable in a regression model. This assumption is important because it supports efficient estimation and valid inference; when it holds, standard errors are correctly estimated, leading to reliable hypothesis testing.

Conditional Expectation

Conditional expectation is the expected value of a random variable given that another variable is held fixed. It is a key concept in probability and statistics that helps in understanding the relationship between variables, especially in regression analysis where we are interested in the expected behavior of the error term given specific values of the explanatory variable.

Zero Conditional Mean Assumption

The zero conditional mean assumption states that the expected value of the error term, given any value of the explanatory variable, is zero. This is a critical assumption in linear regression models because it ensures that the explanatory variable is not correlated with the error term, thus leading to unbiased and consistent estimates of the regression coefficients.

Transcript

00:01 For this question, what you consider the standard sample regression model, which is y is equal to b0 plus b1x plus u, under the garst markov assumption slr .1 through slr .5, and the usual ols estimators b -o and b -1 are unbiased for the respective population.

00:31 Parameters and b1 is the estimator of b1 obtained by assuming the intercept is 0.

00:40 Now the first question, find e of the summation of b1 in terms of x1, b0 and b1 and verify that b1 is unbiased for b1 when the population intercept b0 is 0 and the other cases is where b1 is unbiased so giving the model we're giving which is this we can be noted that um that when the sample size n increases it will result in an increase in the mean and therefore the bias so that will increase the variance then and then when the value of b is equal to zero the bias is almost zero so to prove that i'm coming the estimator for b1 would be equals to to the estimator for b1 is summation of b1 is equal to the summation of n x1 y y divided by i is equal to 1 then x i squared so let this be equation 1 and then giving y y -i is equal to b -0 plus b -1 x -i plus u -i.

02:45 There we can see that therefore the summation of b -i is equal to the summation i is equal to 1 x -i b0 plus b1 xi plus b1 xi plus ui then divided by the summation i is equal to 1 x i squared so now given this we can see that the summation b1 will be equal to b0 summation of x1 i is equal to x1 i is equal to 1 i is equal to 1 i is equal to 1 divided by the summation is equals one x squared plus b1 that is a submission x squared i is equal to 1 divided by the summation x1 squared i is equal to 1 divided by the summation x 1 squared i is equal to 1 plus the summation i is equal to 1 x i u i divided by summation i is equal to 1 x1 squared so for the for the gossan process um summation of u i is equal to 0 and this is equal to 0 so therefore summation of b i z equals to b0 summation x i i is equal to one divided by summation i is equal to one x i is equal to one x i squared plus b i and this is equation two so now from equation two when b o is equal to zero then the bias is zero and when x is equal when the mean is equal to zero the bias is zero so the regression is at the mean is equal to zero comma then y minus y equals to m open brackets x minus x then this will be equals to y minus y equals to m x so now the regression is identical to the regression with the intercept and the slope is m.

06:32 So you can see that the regression is identical to regression with intercepts and the slope is m.

07:02 So now moving on to the second question.

07:13 So this one says find the variance of b i and the variance does not depend on b 0 so the variance of b i with the sign on top is equal to is equal to the summation of n i is equal to 1 x 1 i squared raised to power minus 2 then variance and this is equals to sign, i'm sorry, variance summation, i is equal to 1, x1, then ui.

08:25 So the variance of b1 is equal to that, the open bracket, summation, i is equal to 1 x i squared, raised to bar minus 2 and this is equal to squared divided by i is equal to 1 x1 squared so therefore the variance of b i is equal to that's divided by and this is not a 6 by the way divided by the summation i is equal to 1 x1 squared so this is equation i is equal to 1 x1 squared so this is equation 3 and that's the variance.

09:29 Now moving on to the third question...

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