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Lecture 1

ECON 203 Lecture 1: Analysis (1)
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8 Pages
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Department
Economics
Course Code
ECON 203
Professor
Strow

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Regression for Options 25% ITM
for_reg=subset(results, (UnderlyingPrice-Strike)/UnderlyingPrice>(0.0)
& (UnderlyingPrice-Strike)/UnderlyingPrice<(0.25)
& Last>0.01
& exp_payoff<20
& Last<20
& actual_payoff<25)
Benchmark Model- How well does the market predict payoff (Last vs. Actual)
Call:
lm(formula = actual_payoff ~ Last, data = for_reg)
Residuals:
Min 1Q Median 3Q Max
-21.950 -2.417 -0.426 2.455 14.989
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.99303 0.17989 16.64 <2e-16 ***
Last 0.99776 0.02684 37.17 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.292 on 1860 degrees of freedom
Multiple R-squared: 0.4262, Adjusted R-squared: 0.4259
F-statistic: 1381 on 1 and 1860 DF, p-value: < 2.2e-16
R2 = 42.62% - as we can see from the graph, the market does not predict the
payoff accurately based on the last price of the option.
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How well the expected payoff mimics the Last price (Exp_payoff vs Last)
Call:
lm(formula = Last ~ exp_payoff, data = for_reg)
Residuals:
Min 1Q Median 3Q Max
-7.6133 -0.6888 -0.0075 0.6858 10.5237
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.717374 0.074559 9.622 <2e-16 ***
exp_payoff 0.780406 0.009946 78.461 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.786 on 1860 degrees of freedom
Multiple R-squared: 0.768, Adjusted R-squared: 0.7678
F-statistic: 6156 on 1 and 1860 DF, p-value: < 2.2e-16
R2 = 76.8% The data points fit the trendline much better in this example. The
option price generated by the model fits the market price quite well.
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Does expected payoff add anything to the predictive power of Last?
Call:
lm(formula = actual_payoff ~ Last + exp_payoff, data = for_reg)
Residuals:
Min 1Q Median 3Q Max
-21.5852 -2.1865 -0.3599 2.2896 14.5518
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.43017 0.17170 14.15 < 2e-16 ***
Last 0.25014 0.05211 4.80 1.72e-06 ***
exp_payoff 0.75973 0.04641 16.37 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.014 on 1859 degrees of freedom
Multiple R-squared: 0.4985, Adjusted R-squared: 0.4979
F-statistic: 923.8 on 2 and 1859 DF, p-value: < 2.2e-16
with a p-value of 2e-16 (or 2 with 15 zeros in front of it), the exp_payoff
variable is significant in predicting payoffs.
When we include expected payoff, the value of our R2 increases by 7.23% (from
42.62% to 49.85%) This roughly translates into expected payoff being able to
predict payoffs about 7% better than the market.
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Description
Regression for Options 25% ITM for_reg=subset(results, (UnderlyingPrice-Strike)/UnderlyingPrice>(0.0) & (UnderlyingPrice-Strike)/UnderlyingPrice0.01 & exp_payoff<20 & Last<20 & actual_payoff<25) Benchmark Model- How well does the market predict payoff (Last vs. Actual) Call: lm(formula = actual_payoff ~ Last, data = for_reg) Residuals: Min 1Q Median 3Q Max -21.950 -2.417 -0.426 2.455 14.989 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.99303 0.17989 16.64 <2e-16 *** Last 0.99776 0.02684 37.17 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 4.292 on 1860 degrees of freedom Multiple R-squared: 0.4262, Adjusted R-squared: 0.4259 F-statistic: 1381 on 1 and 1860 DF, p-value: < 2.2e-16 R2 = 42.62% - as we can see from the graph, the market does not predict the payoff accurately based on the last price of the option. How well the expected payoff mimics the Last price (Exp_payoff vs Last) Call: lm(formula = Last ~ exp_payoff, data = for_reg) Residuals: Min 1Q Median 3Q Max -7.6133 -0.6888 -0.0075 0.6858 10.5237 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.717374 0.074559 9.622 <2e-16 *** exp_payoff 0.780406 0.009946 78.461 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.786 on 1860 degrees of freedom Multiple R-squared: 0.768, Adjusted R-squared: 0.7678 F-statistic: 6156 on 1 and 1860 DF, p-value: < 2.2e-16 R2 = 76.8% The data points fit the trendline much better in this example. The option price generated by the model fits the market price quite well. Does expected payoff add anything to the predictive power of Last? Call: lm(formula = actual_payoff ~ Last + exp_payoff, data = for_reg) Residuals: Min 1Q Median 3Q Max -21.5852 -2.1865 -0.3599 2.2896 14.5518 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.43017 0.17170 14.15 < 2e-16 *** Last 0.25014 0.05211 4.80 1.72e-06 *** exp_payoff 0.75973 0.04641 16.37 < 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 4.014 on 1859 degrees of freedom Multiple R-squared: 0.4985, Adjusted R-squared: 0.4979 F-statistic: 923.8 on 2 and 1859 DF, p-value: < 2.2e-16 with a p-value of 2e-16 (or 2 with 15 zeros in front of it), the exp_payoff variable is significant in predicting payoffs. When we include expected payoff, the value of our R2 increases by 7.23% (from 42.62% to 49.85%) This roughly translates into expected payoff being able to predict payoffs about 7% better than the market. 10% ITM Last v Actual Call: lm(formula = actual_payoff ~ Last, data = for_reg) Residuals: Min 1Q Median 3Q Max -21.2685 -2.3220 -0.4358 2.3252 14.2027 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.50991 0.23450 10.70
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