# Christopher Dougherty EC220 - Introduction to econometrics (review chapter) Slideshow: exercise r.10 and r.12 Original citation: Dougherty, C. (2012) EC220.

## Presentation on theme: "Christopher Dougherty EC220 - Introduction to econometrics (review chapter) Slideshow: exercise r.10 and r.12 Original citation: Dougherty, C. (2012) EC220."— Presentation transcript:

R.10Calculate the population variance and standard deviation of X as defined in Exercise R.2 using the definition given by equation (R.8). R.12Using 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(X 2 ) in Exercise R.7.) 1 EXERCISES R.10 AND R.12

R.10Calculate the population variance and standard deviation of X as defined in Exercise R.2 using the definition given by equation (R.8). R.12Using 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(X 2 ) in Exercise R.7.) 2 EXERCISES R.10 AND R.12 We will start with Exercise R.10.

Population variance of X EXERCISES R.10 AND R.12 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. 3

EXERCISES R.10 AND R.12 In Exercise R.4 we found that X had the probability distribution shown above. 4 x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

EXERCISES R.10 AND R.12 5 x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715 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. 6 x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

EXERCISES R.10 AND R.12 When X is equal to 1, the deviation is –3.4722. 7 x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

8 EXERCISES R.10 AND R.12 Similarly for the other possible values. x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

9 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. x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

10 EXERCISES R.10 AND R.12 Similarly for the other values of X. x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

11 EXERCISES R.10 AND R.12 x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 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.

12 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. x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

13 EXERCISES R.10 AND R.12 We calculate all the other weighted squared deviations. x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

The sum is the population variance of X. 14 EXERCISES R.10 AND R.12 x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715

x i p i x i –  (x i –  ) 2 (x i –  ) 2 p i 11/36–3.472212.0562 0.3349 23/36–2.47226.11180.5093 35/36–1.47222.16740.3010 47/36–0.47220.22300.0434 59/360.52780.27860.0697 611/361.52782.33420.7132 1.9715 The standard deviation of X is the square root of its population variance, in this example 1.4041. 15 EXERCISES R.10 AND R.12

16 Now for Exercise R.12. R.10Calculate the population variance and standard deviation of X as defined in Exercise R.2 using the definition given by equation (R.8). R.12Using 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(X 2 ) in Exercise R.7.)

EXERCISES R.10 AND R.12 In Exercise R.7 we showed that E(X 2 ) 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. 17

EXERCISES R.10 AND R.12 We have already seen that  = E(X) = 4.4722. 18

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). 19

Copyright Christopher Dougherty 1999–2006. This slideshow may be freely copied for personal use. 26.08.06

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