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#1 Help Me ! » Strongly stationary process » 2007-09-22 14:46:19

bluesilver
Replies: 0

Let Z_t, t = ..., -2, -1, 0, 1, 2, ..., be a strictly stationary white noise process and let a, b, c be constants. Which of the following processes are stationary? For each stationary process specify whether it is strongly or weakly stationary:
a). X_t = a + b Z_t + c Z_(t-1)
b). X_t = a + b Z_0
c). X_t = cos(ct) Z_1 + sin(ct) Z_2
d). X_t = cos(ct) Z_0
e). X_t = cos(ct) Z_t + sin(ct) Z_(t-1)
f). X_t = Z_t Z_(t-1)

I understand that in order to show that a process is stationary (and hence weakly stationary) I need to show that the mean is a constant and that the autocovariance function does not depend on t. However, how can I show that a stationary process is strongly stationary? I think that it is necessary to show that it is Gaussian, but I do not know how it is possible to show that a stationary process is Gaussian in a simple yet acceptable way. Please pick any process from the list above and show me how it is possible to prove that it is strongly stationary. I would appreciate your help and hints!

#2 Re: Help Me ! » Two designs to obtain estimates » 2006-12-06 01:42:14

John,
Thank you for making a guess, but I doubt that it makes sense in this question... 

Does anybody know how to solve this problem? rolleyes

#3 Help Me ! » Two designs to obtain estimates » 2006-12-05 06:10:49

bluesilver
Replies: 2

Question:

The weights of three individual objects are to be estimated. Two different designs to obtain these estimates are being considered:

D1:
The three objects are weighed individually, seven times each.

D2:
i. Each object is weighed separately once.
ii. Then the objects are weighed two at a time in the three possible, different combinations.
iii. Finally, the three objects are weighed all together, once.
iv. This factorial design (steps i, ii, and iii) is repeated three times.

The two designs both use all of the n = 21 measurements for which funds are available.

Which design do you EXPECT to yield the more precise estimates of the individual weights of the three objects? Why? Show the details of your work.

What assumptions does your answer require?

Note: To answer this question you do no need to know, or use, the "y" values; nor do you need to estimate the residual mean squares.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

My solution:
Let w_1, w_2, w_3 be the true weights and e be the measurement error. Let's try to estimate w_1 by adding w_1 + e, w_1 + w_2 + e, w_1 + w_3 + e, w_1 + w_2 + w_3 +e (3 tries each), then subtracting the sum of 3 samples of w_2 + w_3 + e multiplied by two, and finally dividing by 12. This random variable has mean w_1 and variance (12 σ^2 + 12 σ^2) / 12^2 = σ^2 / 6. This is worse than D1.

However, I also noticed that it is possible to estimate w_1 by adding w_1 + e, w_1 + w_2 + e, w_1 + w_3 + e, w_1 + w_2 + w_3 +e (3 tries each), then subtracting the sum of 3 samples of w_2 + w_3 + e, w_2 + e, and w_3 + e, and then finally dividing by 12. The variance becomes 21 σ^2 / 12^2 instead of σ^2 / 6. This is suspicious (even though the variance is still greater than in the first design).

It turns out that it is possible to estimate w_1 in different ways? Which way is then the correct one? What do you think?

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