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Statistics for PP&A Spring 2004 Problem Set #1 Suggested Answers Stock and Watson Exercises 2.2, 2.3, 2.5.d, 2.6.b, 2.8 #" $% ) *', + &(' ) Total 2.2 Using the random variables and from Table 2.2, and . Compute (a) and corr . Rain ( ) No Rain ( Long Commute 0.15 Short Commute 0.15 Total 0.30 ! consider two new random variables ; (b) and ; and (c) and 0.07 0.63 0.70 0.22 0.78 1.00 '- !.0/1', a It might be helpful to give and interpretations. For example, you could think of as the if the commute cost of a cab ride contingent on the length of the commute: is is short and if the commute is long. The answer uses the mean of a linear function of a random variable (end of section 2.2). % 2.3/-4 5 6 7 . ## equation 2.12 67 9 8: * bybecause #;<98: =8> ? 0 @0 @ equation 2.12 0@ 9+8: A* bybecause B98: A + 'DCE8>F C b The answer uses the variance of a linear function of a random variable (end of section 2.2) and the variance of a Bernoulli random variable (section 2.2). 9 G 8>4by 9equation 2.13 H 8 * =8>F by equation 2.7, variance for a Bernoulli variable 2.13 9I 8>byA* equation 9 > 8 4 A98JC* 4A C by equation 2.7, variance for a Bernoulli variable c The answer uses the definition of covariance (equation 2.22) and the probability distribution given in the example. 1 ! L KM X N Y + ON T QP Y Y S T S Y R [Z N )\ ON E]B^ Z " \ UWV UWV < ) No Rain ( %_ ) Total Rain ( ) Long Commute ( 0.15 0.07 0.22 `'D ) Short Commute ( 0.15 0.63 0.78 Total 0.70 1.00 N =8> and N ('aCE8:FbC .0.30 Recall that Y T Y Y X T Y S S ! UWR V UWV )Z N )\ N E]B^ Z " \ 0.=8>* 3c'DC98:F C* =8M',F because 0.15 of the time and 0.=8: * '-3c'DCE8>F C1 98d',F4 because 0.15 of the time % and `'- _0.=8: * 3c'DCE8>F C1 98H*4 because 0.07 of the time %_ and _0.=8: * '-3c'DCE8>F C1 98H4* because 0.63 of the time %_ and e98>F With covariance in hand, correlation is easy to compute. #" B ! e98>F*f hg i8:F g A98JC*4A C*;(98JC4C and are negatively correlated? and are positively Is there any intuition for why j increases with , correlated: short commutes are associated with no rain. Because klm decreases with , it’s not surprising that and are negatively correlated. and is up and is down. Is this consistent When and are both up, which happens frequently, and ? with your interpretation of corr 2.3 I covered this in some detail in class. n F="o]B4^ B] ^ qp pmA4 . + r V ptsu]Bp ^ r p whenp Z is a standard normal variable. Can we ]B^ v p p(A* +8Jw CEV ' w type problem into this problem? The 4 ' I . g , find 2.5.d If Y is distributed We know how to compute convert , an example of a standard deviation of , is , or about 2 r V + w0 V .OF4N yfI xf I z 9 8>=' g r +wA0 .OF4N yfI xf I z = 8d', g ]B^ =8:='qp{s|p{=8d',4 and we can turn to the standard So we have converted the problem into normal tables. ]B^ 98>='2p<sp<=8d',4B];^ sp=8d',4}@]B^ spm=8:='-B98>4='-A F="4 will fall between 6 and 8. Around 22 percent of the time, (~nO ]B^ p=8: A* . 2.6.b If Y is distributed , find ) This problem is a direct look-up from the table and takes advantage of the fact that 7.78 happens to be a critical value. Ten percent of the time, ~u will exceed 7.78; so 90 percent p<=8>A . % 2.8 In any year, the weather can inflict storm damage to a home. From year to year, the damage is random. Let denote the dollar value of damage in any given year. Suppose that in 95 percent of the years, , but in 5 percent of the years . 49"44 + + ? 9QVH8 _*1F V * 7 =8>*F 4="o44* ='4"o 4 ) y P The variance of the damage is Kd+ KM) + x P 9V8H_* iF V 3@)= '4x"o 4* . 9 8>4 F 4 ) 9"ox4 4 6.='4"o4* '-_9"o 49"o4 a What is the mean and standard deviation of the damage in any year? The mean of the damage is . The standard deviation of the damage is the square root of the variance. I var + ; '-_9"o49"o44 z CE"o*F A98>_ 3 b A key feature of the insurance pool is that the damage to each house is uncorrelated with damage to any other house. Consider: homeowners on the Florida coast do not want to form an insurance pool together. In this case, the average damage can be thought of as adding up the total damage ( ) and dividing the cost over the 100 participating households, or /-/-/- V)h i V ' ST T ( UWV T ' S T ? UWV T ' S T ) UWV T T ' < T ) because all of the + are equal. ) T ='4"44 (we computed ) B='4"44 in part a) ]B^ 44* . We need to know the probability distribution We would like to compute ? First we can compute the expected value of the average damage: ii for the average damage. Is it possible that this is a normal distribution problem? is a mean of random variables, and the Central Limit Theorem states that means of random variables are normally distributed. So is normally distributed with mean and variance . Before we worry about these values, let’s convert the normal problem into a standard normal problem. I ]B^6 44 with `~n+ " I is the same as ];^!)s 443 with s ~nO ="D', I " I , Again, don’t worry about and variance I yet. We know that if ~n+ then a standardization of (by subtracting the mean and dividing by the standard deviation) is 9"D'- . distributed n ei'4"o44 in part Now, let’s fill in the pieces that we need. We have already computed b.i. The tricky part is I , but the Central Limit Theorem gives us guidance. Because is an f f average of i.i.d. random variables, its variance is I . The standard deviation of is I . C**F A98>_*f='-2 C**F=8HA4_ We have the elements that we need: I Before we proceed with the analytic answer, this is a good opportunity for some estimation. The standard deviation of average damage is slightly under $500. How often will average damage 4 4 F 4 )? This should ]B^ s 44C43*F=L8HA4'-_ 4 %]B^ s =8>_*B`']B^ sm=8>_*B=8>9'D4_98 T Indeed, this is a very rare event: average damage exceeds $2,000 around 1 percent of the time. (Consider how important independence of each was for this result.) exceeds its average ($1,000) by $1,000, or about two standard deviations ( not happen often, definitely less than 2.5 percent of the time. 5