Part I Flashcards
A 5-year old BMW is being sold for 85. My regression equation is telling me that at this age, the BMW should be selling for 100. The mean price of all BMWs in my area is 90. Find the sum explained by the regressand and the error that the regressand fails to explain.
The portion explained by the regressand is the amount that should be added/subtracted from the mean. The mean is 90, so the regressand explains that we should add 10 to this mean to obtain the predicted price of 100. This is a component of SSR. The regressand fails to explain why the BMW is selling for so cheap at 85, when according to my regression calculations, it should be selling for 100. Thus, the regressand does not explain why the BMW is selling for 15 less than it should be. This is a component of SSE.
Say Yi … Yn are randomly drawn from the same population. For Yi … Yn to be identically distributed (a) the joint probability distribution of Yi must be the same (b) Yi must have the same marginal distribution for i=1,…,n (c) the conditional distribution of Y2 given Y1 must be the same as the marginal distribution of Y2 (c) Y1 must be distributed independently of Y2, …, Yn.
Answ: (b) Yi must have the same marginal distribution for i=1,…,n; Not (a) because the joint probability distribution differs from the marginal distribution; (c) and (d) only apply to independently and identially distributions or i.i.d, moreover, (c) and (d) say the same exact thing … if the cont’l prob of Y2 given Y1 is equal to the marginal distribution, then Y1 must be independent of Y2. Identically distributed, in and of itself, just means that each RV has the same probability distribution as the others. RVs can have the same mean, but may not necessarily be independent. The unbiasedness of X-bar does not rely on independence, and it’s expected value can still equal the pop. mean (mew) … X-bar is consistent if its variance tends to zero. P46
Supose we draw Y1 from the same distribution. Yet Y1 tells us nothing about the value of Y2, to prove this … we take the conditional distribution of Y2 given Y1 and find that it is the same as the marginal distribution of Y2. In this case (a) the distribution is identically distributed (b) Y1 is not a consistent estimator of Y2 (c) the RVs are i.i.d or independently and identially distributed (d) the RVs are distributed independently but not identically.
Answ: (c) i.i.d.; Note that the sample RV is drawn from the SAME population, so the RVs must all be identically distributed. Now, testing their conditional probabilities, we find that P(Y2 | Y1)= P(Y2) … so now they are also INDEPENDENT. Thus the RVs are i.i.d.
VAR(X+Y) is (a) the variance of the sum of any two variables represented by VAR(X) +2COV(X,Y) + VAR(Y) (b) the variance of the sum of two independent random variables plus twice their COV which is a constant >0 (c) equal to the sum of each of X and Y’s variances, if X and Y are independent RVs and COV is equal to zero (d) the variance of the sum of two independent RVs where the conditional probability of each X given Y must not equal the marginal probability
Answ: (c). The VAR(X+Y) assumes two independent RVs. Now, if they are independent, then by default for any ind. X and Y … COV(X,Y) always =0. Not (d) because it is pretty much stating that the variables are dependent. Not (a) because it states ‘any’ two variables, and the formula works for only independent RVs. Not (b) because if the RVs are independent, their COV=0, and is, obviously, not a ‘constant greater than zero.’
The Y (with a line on top) is (a) a random variable (b) a non-random constant representing the average of the sample from a collection of random variables (c) a variable representing the average of a set of dependent random variables
Answ: (a) an RV; Because Y1, …, Yn are random, their average is a random variable. Y-bar also has a probability distribution called a sampling distribution.
If the conditional mean of Y given X is zero, then (a) the probability weighed-average of this conditional mean is greater than the sample average (b) the mean of X is zero (c) the probability weighed-average of X is zero (d) the mean of Y must be zero.
Answ: (d) The mean of Y must be zero. This is the law of iterated expectations P33.
We have a data set of Y1,…,Yn observations that we took a sample of. Dividing the sum of all variables by the number of observations we obtain (a) a non-random average (b) the square root of the variance (c) a random variable (d) the population average.
(b) a random variable. The act of drawing a random sample has the effect of making the sample average a random variable. P46
The sample average of n observations (Y-bar) (a) is a constant and has no probability distribution (b) is a constant, but since its elements are random, it must have a normal distribution (c) is a random variable, and thus has a constant probability distribution (d) is a random variable, and thus has a sampling probability distribution
(d) it has a sampling distribution because Y-bar is the probability distribution associated with the possible values of Y-bar that could be computed for different possible samples Y1,…,Yn. P47
The law of large numbers states that (a) that the sample distribution approaches the population distribution as n increases with reasonable probability (b) the asymptopic normal distribution will be normal only if it is derrived from a normally distributed population (c) that the sample distribution approaches the population distribution as n increases with high probability (d) that when n is large, the sample distribution of the standardized sample average is aproximately normal (e) a and d
Only (c) … think of this as a limit. As n approaches infinity, Y-bar (samp. avg) approaches the pop average. Obviously, if n is 32 billion … then there is an extremely high probability that we have something that resembles that population mean.
The central limit theorem states that (a) that the sample distribution approaches the population distribution as n increases with reasonable probability (b) that when n is large, the sample distribution of the standardized sample average is aproximately normal (c) that the sample distribution approaches the population distribution as n increases with high probability (d) the asymptopic normal distribution will be normal only if it is derrived from a normally distributed population (e) all of the above
Only (b) … if we standardize all RVs of the sample average, we obtain an aproximately normal dist EVEN IF if original population is not normally distributed. So Y can have a non-normal dist., but the sample asymptopic dist will be normal.
