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Time series. Mathematics & Economics. Math Problem

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time series problems
3. Review of conditional expectation using joint normal distributions
Suppose X and Y are standard normal random variables (i.e. with mean 0 and variance 1), with correlation ρ with each other. 
(a) What is the mean of Y conditional on X = x, denoted E[Y|X = x]? Hint: Write Y = ρX +p1−ρ2Z, where Z is a standard normal random variable independent of X. The answer is a function of x.
(b) What is E[Y|X]? (Just replace x with X in your answer to part (a)). Note that E[Y|X = x] is a deterministic function of x, whereas E[Y|X] is a random variable.
(c) What is Var(Y|X = x)? What is Var(Y|X)?
5. Can we do linear regression without assuming i.i.d. samples? Yes, sometimes.
Now suppose we have random variables X1,X2,...,XN and Y1,Y2,...,YN and 1,2,...N, which are related by Yi = αXi + i but the Xi’s and the Yi’s and the i’s are no longer necessarily i.i.d. Let’s discuss what assumptions we’ll need to do linear regression. 
(a) Let’s assume that for each i, that its conditional expectation dependent on all the Xi’s is zero, i.e. E[i|X1 = x1,X2 = x2,...XN = xN] = 0 for i = 1,...,N. This condition is called strict exogeneity. In other words, while we’ve relaxed the i.i.d. restrictions, we have to keep in place strict restrictions on the conditional expectations of the noise terms. Show that least square estimator ˆ α is unbiased, i.e. E[ˆ α|X1 = x1,...,XN = xN] = α.
(b) Suppose that for times t = 1,2,...,T we have a time series Xt which need not be i.i.d. Suppose t is i.i.d. with mean 0 (and each t is independent of all the Xt’s). Finally suppose Yt = αXt + t. Is the least squares estimator ˆ α unbiased? Why or why not?
(c) Suppose that for times t = 1,2,...,T we have a time series Yt which given by Yt = αYt−1 + t where Y0 = y0 is the initial value of the time series, and where the t’s are i.i.d. with mean 0. Is the least squares estimator ˆ α unbiased? Why or why not?
6. Biased or unbiased?
Let’s do a specific example of the autoregressive process discussed in the last part of the previous problem. Suppose that Yt = αYt−1 + t for t = 1,2 and where Y0 = y0. Suppose that the t are i.i.d. with t = σ or −σ with 50%–50% probability, so that E[t] = 0 and Var(t) = σ2. Then our least squares estimator for α is
ˆ α =y0Y1 + Y1Y2 y2 0 + Y 2 1
.Find an expression for E[ˆ α] of the form α− numerator denominator, where the denominator is a product of two quadratic polynomials. Assuming α > 0 and y0 is nonzero, is E[ˆ α] greater than or less than or equal to α?
TIME SERIES 9
7. Why do we use “N −1” when calculating sample variance? Suppose Xi are i.i.d. random variables with mean µ and variance σ2, for i = 1,...,N. Let
¯ X =
1 N
N X i=1
Xi
be the sample mean estimator. Let
ˆ σ2 = 1 N −1
N X i=1
(Xi − ¯ X)2
be the sample variance estimator. Show that E[ˆ σ2] = σ2.
8. Standard error for sample mean for an autocorrelated time series
Suppose that Xt is a stationary time series with (unconditional) variance σ2. Consider the sample mean estimator
¯ X =
1 N
N X t=1
Xt.
Recall that if the Xt were i.i.d. then the variance of ¯ X would be σ2/N. However, suppose that instead the stationary time series Xt is autocorrelated, with autocorrelation function ρ(k). Show that
Var( ¯ X) =
σ2 N"1 + 2 N X r=11− r N
ρ(r)#.
9. Question regarding the sum of two strictly stationary time series
Let Xt and Yt be two strictly stationary time series. Must their sum Xt + Yt be strictly stationary? Why or why not?

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Question#3
(a).
Let’s suppose that we have been given two discrete random variables called X and Y. With x ∈ Range(X), the condition expectation of Y is X = x:
First of all, we will determine the sum of the value X = 1. This has been mentioned in the table above.
0.03 + 0.15 + 0.15 + 0.16 = 0.49.
Now we will divide every value in the X = 1 column by the total of the first step.
0.03 / 0.49 = 0.061
0.15 / 0.49 = 0.306
0.15 / 0.49 = 0.306
0.16 / 0.49 = 0.327
Next, we will multiply every answer by the corresponding Y value. This is present in the left-hand column.0.0612244898 * -1 = -0.0610.306122449 * 2.75 = 0.8420.306122449 * 3 = 0.9180.3265306122 * 4.55 = 1.486
Let’s now sum the values in the second step.
E(Y|X = 1) = -0.061 + 0.842 + 0.918 + 1.486 = 3.19
E(Y|X = 1) = 3.19
3.19 is the mean of Y conditional on X = x.
(b).
Here, E(X) can be regarded as the expectation value of the continuous random variable X
P(x) is the probability density function
and x is the value of the continuous random variable X
E(aX) = aE(X)
E(X+Y) = E(X) + E(Y)
Here c is constant:
E(c) = c
X and Y are independent random variables.
E(X ⋅Y) = E(X) ⋅ E(Y)
This is conditional expectation.
(c).
It is Var(X)+Var(Y)−2Cov(X,Y)Var(X)+Var(Y)−2Cov(X,Y), as Var(−Y)=Var(Y)Var(−Y)=Var(Y).
For a discrete random variable called X, the variance can be written as Var(X) or as Var(Y |X). Var(X) = E[ (X – m)2 ]; here m is the expected value E(X)
Question#5
(a).
Cov(X1; X2) = E(X1X2)E(X1)E(X2)
f1(x1) = Z 1 0 4x1x2dx2 = 2x1
E(X1) = Z 1 0 x1f1(x1)dx1 = 2/3 = E(X2).
E(X1X2) = Z 1 0 Z 1 0 x1x24x1x2dx1dx2 = 4/9.
x1 = √ r1r2 and x2 = p r2/r1.
Jac(r1, r2) = det ∂x1/∂r1 ∂x2/∂r1 ∂x1/∂r2 ∂x2/∂r2 = 1 4 det p r2/r1 − p r2/r3 p 1 r1/r2 p 1/r1r2 = 1 2r1 .
fY1,Y2 (r1, r2) = fX1,X2 ( √ r1r2, p r2/r1)Jac(r1, r2) = 4r2/2...
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