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MakcuM [25]
3 years ago
14

7. Classical linear model assumptions for time series Consider the following stochastic process {(x1, x2, x3, . . . , xk, yt): t

= 1, 2, . . . , n}{(x1, x2, x3, . . . , xk, yt): t = 1, 2, . . . , n} that follows the linear model: y=β0+β1xt1+β2xt2+β3xt3 + . . . +xtk+uy=β0+β1xt1+β2xt2+β3xt3 + . . . +xtk+u {(ut: t =1, 2, . . . , n)}{(ut: t =1, 2, . . . , n)} = sequence of error terms nn = number of observations (time periods) What are the minimum Gauss–Markov assumptions needed for the OLS estimates of βˆjβ^j , for j = 1, 2, . . . , kj = 1, 2, . . . , k , to be the best linear unbiased estimators (BLUE) conditional on the explanatory variables for all time periods ( XX )? Check all that apply.
Mathematics
1 answer:
iogann1982 [59]3 years ago
3 0

Answer:

Options are missing.

The options for the above question are:

TS.1: Linear in parameters.

TS.2:No perfect collinearity

TS.3: Zero conditional mean.

TS.4: Homoskedasticity.

TS.5: No serial correlation

TS.6: Normality.

Hence the correct answer is TS1 to TS 5

Step-by-step explanation:

Assumptions TS 1 to TS 5 are the minimum set of assumptions needed to for the OLS estimates to be the best linear unbiased estimators conditional on explanatory variables for all time periods.

The assumptions of Normality is not needed for the estimators to show the BLUE property

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the probability that a truck drives less than 159 miles in a day = 0.9374

Step-by-step explanation:

Given;

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If the mileage per day is normally distributed, we use the following conceptual method to determine the probability of less than 159 miles per day;

1 standard deviation above the mean = m + d, = 120 + 23 = 143

2 standard deviation above the mean = m + 2d, = 120 + 46 = 166

159 is below 2 standard deviation above the mean but greater than 1 standard deviation above the mean.

For normal districution, 1 standard deviation above the mean = 84 percentile

Also, 2 standard deviation above the mean = 98 percentile

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\frac{159-143}{166-143} = \frac{x-84}{98-84} \\\\\frac{16}{23} = \frac{x-84}{14} \\\\23(x-84) = 224\\\\x-84 = 9.7391\\\\x = 93.7391\ \%

Therefore, the probability that a truck drives less than 159 miles in a day = 0.9374

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I'm gonna go with rectangle and square

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