FIN488 Data Science

Exam 1

Format and Guidance

• Open book exam including access to the internet.
• Attempts to communicate with any other person or plagiarism from class notes or the internet will be deemed potential violations of Academic Integrity and Honesty under UoA policies.
• Maximum time is 90 minutes.
• Upon completion email pcote@ualberta.ca from your ualberta account the firstname_lastname.Rmd and rendered html files.
• Answer where you see ... using your own words and insert R chunks where appropriate.

Grading

• The exam counts for 30 points.

Questions

• You will be performing analysis using a tidy workflow on the following stock tickers:

  • NASDAQ ETF (“QQQ”).
  • Gold ETF (“GLD”).
  • Albemarle, an industry leader in lithium (“ALB”).
• Use the time period from 2018-01-01 to 2019-10-01.

• Use daily log returns on adjusted prices (instead of daily settles).

Q1. Extract data and graph the stock prices. (2 points)

…

library(tidyquant); library(tidyverse)
ticker <- c("QQQ", "GLD", "ALB")
dfraw <- tq_get(ticker, get = "stock.prices", from = "2018-01-01", to = "2019-10-01")
dflong <- dfraw %>% dplyr::mutate(value = log(adjusted)- log(dplyr::lag(adjusted))) %>% na.omit() %>% select(date, symbol, value)

dflong %>% ggplot(aes(x = date, y = value, col = symbol))+ geom_line() + facet_wrap(.~symbol, ncol = 1, scale = "free")+ scale_y_continuous()
# to wide data
dfwide <- dflong %>% pivot_wider(names_from = symbol, values_from = value) %>% na.omit()

Q2. Compute their distribution moments. (4 points)

…

library(moments)
dflong %>% group_by(symbol) %>%  dplyr::summarize(mean=mean(value),
                 stdv=sd(value),
                 skew=moments::skewness(value),
                 kurt=moments::kurtosis(value))

## # A tibble: 3 x 5
##   symbol      mean   stdv    skew   kurt
##   <chr>      <dbl>  <dbl>   <dbl>  <dbl>
## 1 ALB    -0.00158  0.0222  -0.529   5.37
## 2 GLD    -0.000700 0.0208 -17.5   347.  
## 3 QQQ     0.000433 0.0131  -0.381   5.19

Q3. Generate Histograms including geom_rug(). (4 points)

…

library(scales)
dflong %>% group_by(symbol) %>% ggplot(aes(x = value))+
  geom_histogram(aes(y =..density..), bins = 300) + facet_wrap(.~symbol, ncol=1)+
  geom_rug(aes(x=value, y = 0), position = position_jitter(height = 0.5))

The remainder of the exam questions use ONLY ALB and QQQ. ### Q4. Correlation (10 points)

A) Compute the simple correlation during the whole period. (1 point)

…

cor <- stats::cor(dfwide$ALB,dfwide$QQQ)
cor

## [1] 0.4795587

B) Compute and chart a 60-day rolling correlation. (4 points)

library(tibbletime)
cor60 <- tibbletime::rollify(~cor(.x, .y),window=60)

dfroll <- dfwide %>% 
  dplyr::mutate(cor60= cor60(ALB, QQQ)) %>% 
  na.omit()
head(dfroll)

## # A tibble: 6 x 5
##   date            QQQ       GLD     ALB cor60
##   <date>        <dbl>     <dbl>   <dbl> <dbl>
## 1 2018-03-29  0.0181   0.000477  0.0349 0.359
## 2 2018-04-02 -0.0293   0.0116   -0.0416 0.381
## 3 2018-04-03  0.0112  -0.00757   0.0163 0.387
## 4 2018-04-04  0.0156   0.00119   0.0130 0.390
## 5 2018-04-05  0.00568 -0.00515   0.0445 0.392
## 6 2018-04-06 -0.0253   0.00468  -0.0423 0.412

cor60graph <- dfroll %>% ggplot(aes(x = date, y = cor60))+geom_line()
cor60graph

C) Chart the rolling 60-day standard deviations and choose the TRUE statement. (5 points)

sd60 <- tibbletime::rollify(sd,window=60)
dfrollsd <- tibbletime::as_tbl_time(dflong,index=date) %>% 
  group_by(symbol) %>% 
  dplyr::mutate(sd60=sd60(value)*sqrt(252)) %>% 
  ungroup() %>% na.omit()

dfrollsd %>% dplyr::filter(symbol == "QQQ" | symbol == "ALB" )%>% ggplot(aes(x = date, y = sd60, col = symbol))+ geom_line()

#The answer is A)ALB and QQQ both show heteroskedasticity. This is because their 60 days standard deviation are non-constant. 

1. ALB and QQQ both show heteroskedasticity.
2. Only ALB shows heteroskedasticity.
3. Only QQQ shows heteroskedasticity.
4. Neither ALB and QQQ show heteroskedasticity.

Q5. You now seek to understand how much of the variance in ALB returns is explained by QQQ returns. (10 points)

A) Do a scatter plot of the data consistent with the regression equation. (2 points)

dfwide %>% ggplot(aes(x = QQQ, y = ALB))+ geom_point()+ stat_smooth(method = "lm")

…

B) Perform a linear regression model. (2 points)

…

fit <- stats::lm(ALB ~ QQQ, data = dfwide)
summary(fit)

## 
## Call:
## stats::lm(formula = ALB ~ QQQ, data = dfwide)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.098294 -0.010557  0.001043  0.010968  0.079679 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.001737   0.000921  -1.886     0.06 .  
## QQQ          0.801556   0.070243  11.411   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.01926 on 436 degrees of freedom
## Multiple R-squared:   0.23,  Adjusted R-squared:  0.2282 
## F-statistic: 130.2 on 1 and 436 DF,  p-value: < 2.2e-16

C) Using regression summary, select the appropriate TRUE statement(s). (3 points)

1. More than 20% of the variance is explained.
2. Less than 20% of the variance is explained.
3. The regression and the beta are both significant.
4. The regression and the beta are both NOT significant.
5. The beta coefficient is > 0.5.
6. The beta coefficient is < 0.5.

Answers are

A/ More than 20% of the variance is explained. E/ The beta coefficient is > 0.5.

D) Print the residuals analysis plots and describe your interpretation. (3 points)

…be concise max a few bullet points…

library(ggfortify)
autoplot(fit,size = 0.5)

• Residuals vs. Fitted plot shows that residual are randomly and equally spread across a horizontal mean of 0. This indicates that ALB and QQQ don’t have a non-linear relationship.

• Normal Q-Q shows the residual follow a straight line, which indicates that the residuals are normally distributed

• Scale-locations have evenly spread points around a horizontal line. This means that the variance is constant.

• Overall, the residuals show that the linear relationship is significant between QQQ and ALB.