# ECON 203 Lecture Notes - Lecture 1: Statistical ModelPremium

8 pages36 viewsSpring 2017

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**preview**shows pages 1-2. to view the full**8 pages of the document.**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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