Question

Help please will give 5 stars and amazing feedback! TRUE OR FALSE AND WHY???

(a) Suppose that X is normally distributed with mean 0 and variance 1. Then 3. X is normally distributed with mean 0 and variance 9 True False b) The CLT states that, in large enough samples, the sample average is close to the true ex- pected value with very high probability. True False (c) The assumption that E(ui| X) 0 implies that Cov(ui, Xi)0. True False

Answer 1

(a)

If X ~ Normal ( 0 , 1 )

Then,

Mean of X = E(X) = 0

Variance of X = Var(X) = 1

Now, For any constant 'a'

We have property as :

E(aX) = a*E(X)

Var(aX) = a² * Var(X)

Mean of 3X = E(3X) = 3*E(X) = 3*0 = 0

Variance of 3X = Var(3X) = 3² * Var(X) = 9*1 = 9

So, 3X is normally distributed with mean 0 and variance 9

Hence, This statement is TRUE

(b)

The Central Limit Theorem states that the sampling distribution of the sample means approaches a normal distribution as the sample size gets larger. An essential component of the Central Limit Theorem is that the average of your sample means will be the population mean ( True expected value ).

Hence, this statement is TRUE

(c)

$\dpi{100} E(u_i | X_i) =0$

$\dpi{100} Cov(u_i,X_i) = E(u_iX_i)-E(u_i)*E(X_i)$

Using Law of Iterated expectations :

$\dpi{100} E(u_iX_i) = E(E(u_iX_i|X_i))=E(X_iE(u_i|X_i)) = E(X_i*0)=0$

Now,

$\dpi{100} Cov(u_i,X_i) = 0-E(u_i)*E(X_i) =-E(u_i)*E(X_i) \neq 0$

Hence, This statement is FALSE