Christopher Dougherty EC220 - Introduction to econometrics (review chapter) Slideshow: exercise r.10...
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Christopher Dougherty EC220 - Introduction to econometrics (review chapter) Slideshow: exercise r.10 and r.12 Original citation: Dougherty, C. (2012) EC220 - Introduction to econometrics (review chapter). [Teaching Resource] © 2012 The Author This version available at: http:// learningresources.lse.ac.uk/141/ Available in LSE Learning Resources Online: May 2012 This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. This license allows the user to remix, tweak, and build upon the work even for commercial purposes, as long as the user credits the author and licenses their new creations under the identical terms. http://creativecommons.org/licenses/by-sa/3.0/ http://learningresources.lse.ac.uk/
Transcript of Christopher Dougherty EC220 - Introduction to econometrics (review chapter) Slideshow: exercise r.10...
Christopher Dougherty
EC220 - Introduction to econometrics (review chapter)Slideshow: exercise r.10 and r.12
Original citation:
Dougherty, C. (2012) EC220 - Introduction to econometrics (review chapter). [Teaching Resource]
© 2012 The Author
This version available at: http://learningresources.lse.ac.uk/141/
Available in LSE Learning Resources Online: May 2012
This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. This license allows the user to remix, tweak, and build upon the work even for commercial purposes, as long as the user credits the author and licenses their new creations under the identical terms. http://creativecommons.org/licenses/by-sa/3.0/
http://learningresources.lse.ac.uk/
R.10 Calculate the population variance and standard deviation of X as defined in Exercise R.2 using the definition given by equation (R.8).
R.12 Using equation (R.9), find the variance of the random variable X defined in Exercise R.2 and show that the answer is the same as that obtained in Exercise R.10. (Note: You have already calculated m in Exercise R.4 and E(X2) in Exercise R.7.)
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EXERCISES R.10 AND R.12
222 )( XEX
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iinnX pxpxpxXE
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R.10 Calculate the population variance and standard deviation of X as defined in Exercise R.2 using the definition given by equation (R.8).
R.12 Using equation (R.9), find the variance of the random variable X defined in Exercise R.2 and show that the answer is the same as that obtained in Exercise R.10. (Note: You have already calculated m in Exercise R.4 and E(X2) in Exercise R.7.)
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EXERCISES R.10 AND R.12
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iinnX pxpxpxXE
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22 )()(...)()(
222 )( XEX
We will start with Exercise R.10.
Population variance of X
EXERCISES R.10 AND R.12
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The expected value of the squared deviation is known as the population variance of X. It is a measure of the dispersion of the distribution of X about its population mean.
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EXERCISES R.10 AND R.12
In Exercise R.4 we found that X had the probability distribution shown above.
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xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
EXERCISES R.10 AND R.12
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xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
4722.4)( XEX
To calculate the population variance, we first calculate the deviations of the possible values of X about its population mean.
EXERCISES R.10 AND R.12
In Exercise R.4 we saw that the population mean of X was 161/36, that is, 4.47 to two decimal places. To minimize rounding error in our working, we will take it to four decimal places.
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xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
4722.4)( XEX
EXERCISES R.10 AND R.12
When X is equal to 1, the deviation is –3.4722.
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xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
4722.4)( XEX
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EXERCISES R.10 AND R.12
Similarly for the other possible values.
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
4722.4)( XEX
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EXERCISES R.10 AND R.12
Next we need a column giving the squared deviations. When X is equal to 1, the squared deviation is 12.0562.
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
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EXERCISES R.10 AND R.12
Similarly for the other values of X.
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
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EXERCISES R.10 AND R.12
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
Now we start weighting the squared deviations by the corresponding probabilities. What do you think the weighted average will be? Have a guess.
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EXERCISES R.10 AND R.12
A reason for making an initial guess is that it may help you to identify an arithmetical error, if you make one. If the initial guess and the outcome are very different, that is a warning.
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
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EXERCISES R.10 AND R.12
We calculate all the other weighted squared deviations.
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
The sum is the population variance of X.
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EXERCISES R.10 AND R.12
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
xi pi xi – (xi – )2 (xi – )2 pi
1 1/36 –3.4722 12.0562 0.3349
2 3/36 –2.4722 6.1118 0.5093
3 5/36 –1.4722 2.1674 0.3010
4 7/36 –0.4722 0.2230 0.0434
5 9/36 0.5278 0.2786 0.0697
6 11/36 1.5278 2.3342 0.7132
1.9715
The standard deviation of X is the square root of its population variance, in this example 1.4041.
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EXERCISES R.10 AND R.12
EXERCISES R.10 AND R.12
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Now for Exercise R.12.
R.10 Calculate the population variance and standard deviation of X as defined in Exercise R.2 using the definition given by equation (R.8).
R.12 Using equation (R.9), find the variance of the random variable X defined in Exercise R.2 and show that the answer is the same as that obtained in Exercise R.10. (Note: You have already calculated m in Exercise R.4 and E(X2) in Exercise R.7.)
222 )( XEX
i
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iinnX pxpxpxXE
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221
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22 )()(...)()(
EXERCISES R.10 AND R.12
9716.1
0006.209722.21
)4722.4(9722.21
)(2
222
XEX
In Exercise R.7 we showed that E(X2) was 791/36, which is 21.97 to two decimal places. To minimize rounding error, we will take it to four decimal places, 21.9722, in our working.
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9716.1
0006.209722.21
)4722.4(9722.21
)(2
222
XEX
EXERCISES R.10 AND R.12
We have already seen that = E(X) = 4.4722.
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EXERCISES R.10 AND R.12
Hence, using the alternative expression, we find that the population variance is 1.9716. Apart from rounding error affecting the last digit, this is the same as in Exercise R.10 (1.9715).
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9716.1
0006.209722.21
)4722.4(9722.21
)(2
222
XEX
Copyright Christopher Dougherty 1999–2006. This slideshow may be freely copied for personal use.
26.08.06
