Finance Solutions to Problem Set #6: Demand Estimation and Forecasting

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Finance 30210 Soluions o Problem Se #6: Demand Esimaion and Forecasing 1) Consider he following regression for Ice Cream sales (in housands) as a funcion of price in dollars per pin. My daa is aken from muliple locaion a one poin in ime. SUMMARY OUTPUT Sales p Regression Saisics R Square 0.26 Sandard Error 0.65 ANOVA df SS MS Regression 1 0.845 0.845 Residual 28 11.7 0.42 Toal 29 12.5 Coefficiens Sandard Error Sa Inercep 70.64 47.14 1.49 Price -20.47 14.3-1.42 a) Calculae a forecas for he demand for ice cream a he sample mean price of $3.25 per pin. So, our regression equaion is Sales 70.64 20.47 p Plug in 3.25 o ge a forecas of: Sales 70.64 20.47 3.25 4.11 So, 4,110 pins of ice cream sold. b) Calculae a 95% confidence inerval for his forecas. We need o calculae he sandard error for he forecas 2 p p 1 2 1 Var sales / p ˆ 1 N N Var p

We are forecasing a he sample mean, so p p 2 0. 2 1 1 p ˆ Var sales / 1.42 1.434 N 29 SE.434.66 So a 95% confidence inerval would be 4.11 +/- 2*(.66) = (2.79, 5.43) c) Calculae he elasiciy of demand a a price of $3.25 Sales 70.64 20.47 p p % Q Q p 3.25 20.47 16.2 % p p q 4.11 2) Now, suppose we ran a differen regression. Now, we have SUMMARY OUTPUT ln Sales p Regression Saisics R Square 0.23 Sandard Error 0.05 ANOVA df SS MS Regression 1 0.049 0.049 Residual 28 0.866 0.030 Toal 29 0.916 Coefficiens Sandard Error Sa Inercep 17.48 12.82 1.36 Price -4.95 3.91-1.26 a) Calculae a forecas for he demand for ice cream a he sample mean price of $3.25 per pin.

Now, we have he regression equaion ln Sales 17.48 4.95 p Plug in price. ln Sales 17.48 4.953.25 1.39 Now, we need o conver back o level from logs 1.39 e 4.014 so, 4, 014 pins sold. b) Calculae a 95% confidence inerval for his forecas. As from before. 2 1 1 p ˆ Var sales / 1.03 1.031 N 29 SE.031.176 Because sales is in logs, he above sandard error is read as 17.6%, so a 94% confidence inerval is 4.014 +/- 2*(17.6% ) (2.601, 5.426) c) Calculae he elasiciy of demand a a price of $3.25 p % Q ln Q p 4.953.25 16.1 % p p

3) Le s go back o he original regression. Sales p We could have a misspecificaion here. I seems reasonable ha ice cream sales will increase on a ho day. So Temperaure will be posiively correlaed wih ice cream sales. If emperaure is negaively correlaed wih price, in wha direcion is my esimaed coefficien on price biased. Explain. The problem here is ha he coefficien is picking up he effec on sales of emperaure as well as price. When price is high, demand is low, bu when price is high, emperaure is low (negaive correlaion beween price and emperaure) and a low emperaure lowers ice cream sales. So our esimaed coefficien is oo negaive (i.e. biased down). In general, we have ˆ TCorr p, Temp We know ha he correlaion beween price and emperaure is negaive and ha he esimaed effec beween emperaure and sales should be posiive, so he esimaed coefficien is equal o he rue coefficien plus somehing negaive (biased downward 4) Suppose I run a differen regression for ice cream sales. I include Income per capia. Sales p I 1 2 I ge a saisically significanly posiive coefficien for income. However, I discover ha here is a highly posiive correlaion beween price and income. a) Should I be worried abou his? Explain. Wih income having a posiive correlaion wih price, we have wha s called mulicollineariy. My coefficiens will no be biased, bu my sandard errors are all off. The issue here is ha because income and price always move in he same direcion, i s hard for he regression o separae each variable s effec on sales.

b) How could I correc his problem? I could simply remove income from he regression, bu hen I creae misspecificaion bias (see quesions 3). The bes soluion is o fine a variable ha is highly correlaed wih income, bu uncorrelaed wih price and use ha as a proxy for income in he regression. 5) Coninuing wih my regression. Sales p I 1 2 I discover ha he correlaion beween my regression residuals and price is posiive. Why is his and why is his a problem? The reason for his is because he price of ice cream is deermined joinly by supply and demand. Suppose ha he error erm is posiive in he above regression. This is an increase in demand.bu an increase in demand will lead o an increase in supply.and he rise in supply is due o a rise in price. There s he posiive correlaion beween he errors and price. The issue here is ha if we believe ha price is deermined simulaneously by supply and demand, is difficul o separae supply effecs and demand effecs. 6) Now, suppose ha raher han using a cross secion, I use a ime series. Sales p I 1 2 I run a Durbin-Wason es and ge a Durban-Wason saisic of.25. Should I be worried abou his? Explain. A Durban Wason saisic lies beween 0 and 4. A value of 2 indicaes no auocorrelaion (auocorrelaion means ha oday s error erm is correlaed wih yeserday s error erm). Greaer han 2 is negaive auocorrelaion and less han 2 is posiive auocorrelaion. Wih a value of.25, here is significan posiive correlaion across ime in he error erms. One soluion is o run a regression in differences raher han levels. For example, we could ry Sales p I 1 2

7) Suppose ha you are in he consrucion business. You are ineresed in forecasing oal consrucion spending in he US for June of 2018. You run he following regression: Spending 1 Spending is in millions of dollars and your daa is monhly from Jan. 1993. SUMMARY OUTPUT Regression Saisics R Square 0.55 Sandard Error 135815 ANOVA df SS MS Regression 1 6.477E+12 6.47E+12 Residual 280 5.16E+12 18445763725 Toal 281 1.16E+12 Coefficiens Sandard Error Sa Inercep 591453.8 16132 36.66 Price 1861.7 99.35 18.79 a) By how much does consrucion spending increase per year? We have he following equaion Spending 591, 493 1,861 So, spending increases by 1,861 per monh or 1,861(12) = 22,332 per year. This is in millions, so spending increases by around $22.3B per year. b) Calculae your forecas for June 2018 ( = 305) Spending 591, 493 1,861 305 1,159, 098 (approx. $1.159T)

c) Calculae a 95% confidence inerval for your forecas. (The average of ime is 141 and he variance of he ime variable is 6650.5) 2 2 1 Var Spending / ˆ 1 N ( N 1) Var 305 141 2 1 Var Spending / 305 18,445,763,725 1 18,777,026,009 281 280*6650.5 SE 18, 777, 026, 009 137, 029 So, a 95% confidence inerval is 1,159,098 +/- 2*137,029 8) Now, you run an exponenial regression. ln SUMMARY OUTPUT Spending 1 Regression Saisics R Square 0.59 Sandard Error.161 ANOVA df SS MS Regression 1 10.88 10.88 Residual 280 7.29.026 Toal 281 18.1 Coefficiens Sandard Error Sa Inercep 13.28 0.019 692.8 Price 0.0024 0.00018 20.43 a) By how much does consrucion spending increase per year? Consrucion spending rises by.24 percen per monh or 2.88 percen per year.

b) Calculae your forecas for June 2018 ( = 305) Spending ln 13.28.0024 13.28.0024 305 14.012 e 14.012 1, 217,122 c) Calculae a 95% confidence inerval for your forecas. (The mean of ime is 141 and he variance of he ime variable is 6650.5) 305 141 2 1 Var Spending / 305.026 1.0264 281 280*6650.5 SE.0264.162 So, we have 1,217,122 +/- 32.4% 9) Now, suppose ha you include dummy variables for he firs hree quarers. ln SUMMARY OUTPUT Spending 1 2D1 3D2 4D3 Regression Saisics R Square 0.59 Sandard Error.161 Coefficiens Sandard Error Sa Inercep 13.28 0.019 692.8 Price 0.0024 0.00018 20.43 D1 -.0128 0.0273 -.47 D2 -.00447.02733 -.16 D3 -.00094.027621 -.03 Wha does his regression ell you abou he seasonaliy of consrucion? A firs pass, I could say ha relaive o he 4 h quarer (ha s he dummy no included), Consrucion spending is 1.28% lower in he firs quarer Consrucion spending is.447% lower in he second quarer Consrucion spending is.094% lower in he hird quarer However, if I look a he sandard errors and he -sas, I see ha none of hese variables are significan, so here is no seasonaliy in oal consrucion spending!

10) Suppose ha you have he following daa for he daily dollar Euro Exchange rae: Dae Exchange Rae ($/Euro) Augus 4, 2016 1.1134 Augus 5, 2016 1.1080 Augus 8, 2016 1.1078 Augus 9, 2016 1.1110 Augus 10, 2016 1.1171 Augus 11, 2016 1.1168 Augus 12, 2016 1.1172 a) Make a forecas for Augus 14, 2016 using a moving average of lengh 3. For a MA(3), he forecas is x x x so, we have 3 1 2 x 1 Dae Exchange Rae ($/Euro) Forecas Augus 4, 2016 1.1134 --- Augus 5, 2016 1.1080 --- Augus 8, 2016 1.1078 --- Augus 9, 2016 1.1110 1.1097 Augus 10, 2016 1.1171 1.1089 Augus 11, 2016 1.1168 1.1120 Augus 12, 2016 1.1172 1.1150 Augus 13, 2017 1.1170

b) Make a forecas for Augus 14, 2016 using an exponenial smoohing model wih a smoohing parameer of.4. So, we have xˆ.4 x.6xˆ 1 We need an iniial forecas o kick hings off.les use he sample average of 1.1130 Dae Exchange Rae ($/Euro) Forecas Augus 4, 2016 1.1134 1.1130 Augus 5, 2016 1.1080 1.1132 Augus 8, 2016 1.1078 1.1111 Augus 9, 2016 1.1110 1.1097 Augus 10, 2016 1.1171 1.1102 Augus 11, 2016 1.1168 1.1130 Augus 12, 2016 1.1172 1.1145 Augus 13, 2017 1.1156