Refik Turkeli MFIN 809 HW1 Rmarkdown

refik_turkeli_hw1.Rmd

---
title: "MFIN 809 Homework 1"
author: "Refik Türkeli"
date: "11/11/2018"
output: pdf_document
---

```{r setup, include = FALSE}
library(tidyverse)
library(moments)
library(scales)
library(kimisc)
library(knitr)
```

# Exercise 1

Portfolio deviation from benchmark return is given as follows.

```{r echo = FALSE}
portfolio_return <- tribble(
    ~year, ~deviation,
    1992L,    -0.0714,
    1993L,     0.0162,
    1994L,     0.0248,
    1995L,    -0.0259,
    1996L,     0.0937,
    1997L,    -0.0055,
    1998L,    -0.0089,
    1999L,    -0.0919,
    2000L,    -0.0511,
    2001L,    -0.0049,
    2002L,     0.0684,
    2003L,     0.0304
)

portfolio_return %>% 
    mutate(deviation = percent(deviation)) %>% 
    kable()
```

## 1) Binning the returns

Dividing into 4 intervals.

```{r}
binned <- portfolio_return %>% 
    mutate(bins = cut(deviation * 100, breaks = 4)) %>% 
    group_by(bins) %>% 
    summarise(freq = n()) %>% 
    mutate(cum_freq = cumsum(freq),
           rel_freq = freq / sum(freq),
           cum_rel_freq = cumsum(rel_freq))
```

```{r echo = FALSE}
binned %>% 
    mutate_at(c("rel_freq", "cum_rel_freq"), percent) %>% 
    kable()
```

\newpage

## 2) Constructing a histogram

```{r fig.height = 3}
ggplot(binned, aes(x = bins, y = freq)) +
    geom_bar(stat = "identity")
```

## 3) Modal interval of the data

Modal interval is the interval with the highest frequency. 

```{r}
binned %>% 
    filter(freq == max(freq)) %>% 
    select(bins, freq) %>% 
    kable()
```

## 4) Tracking error

```{r results = "asis"}
sd(portfolio_return$deviation) %>% 
    percent()
```

\newpage

# Exercise 2

Annual returns for MSCI Germany Index.

```{r echo = FALSE}
msci <- tribble(
    ~year, ~return,
    1993L,  0.4621,
    1994L, -0.0618,
    1995L,  0.0804,
    1996L,  0.2287,
    1997L,   0.459,
    1998L,  0.2032,
    1999L,   0.412,
    2000L, -0.0953,
    2001L, -0.1775,
    2002L, -0.4306
)

msci %>% 
    mutate(return = percent(return)) %>% 
    kable()
```

## 1) Frequency table

```{r}
binned <- msci %>% 
    mutate(bins = cut(return * 100, breaks = 5)) %>% 
    group_by(bins) %>% 
    summarise(freq = n()) %>% 
    mutate(cum_freq = cumsum(freq),
           rel_freq = freq / sum(freq),
           cum_rel_freq = cumsum(rel_freq))
```

```{r echo = FALSE}
binned %>% 
    mutate_at(c("rel_freq", "cum_rel_freq"), percent) %>% 
    kable()
```

\newpage

## 2) Histogram

```{r fig.height = 3}
ggplot(binned, aes(x = bins, y = freq)) +
    geom_bar(stat = "identity")
```

## 3) Modal interval of the data

```{r}
binned %>% 
    filter(freq == max(freq)) %>% 
    select(bins, freq) %>% 
    kable()
```

## 4) Symmetry

The frequency distribution is not symmetric. Higher returns have a higher frequency compared to lower returns.

\newpage

## 5-10) Statistical summaries

```{r}
stat_list <- list(
    mean = mean(msci$return),
    median = median(msci$return),
    compound = prod(msci$return + 1) ^ (1 / length(msci$return)) - 1,
    perc30 = quantile(msci$return, probs = 0.3),
    range = max(msci$return) - min(msci$return),
    MAD = mean(abs(msci$return - mean(msci$return))),
    var = var(msci$return),
    sd = sd(msci$return),
    skewness = skewness(msci$return),
    kurtosis = kurtosis(msci$return) - 3
) %>% 
    map_at(c("mean", "median", "compound", "perc30", "range", "MAD", "sd"), percent) %>% 
    map_at(c("skewness", "kurtosis", "var"), partial(round, digits = 2))

tibble(stat = names(stat_list), value = stat_list) %>% 
    kable()
```

## Comments about Skewness and Kurtosis

The distribution is negatively skewed. It is skewed to the left. This means that the median is larger than the mean. 

The kurtosis we calculate is the excess kurtosis. In this case it is negative, meaning that the distrbution is less peaked than the normal distribution.

\newpage

# Exercise 3

Annual returns for the MSCI Germany Index and JP Morgan government bonds index is provided as follows.

```{r, echo = FALSE}
index_return <- tribble(
    ~year,   ~msci,    ~jpm,
    1993L,  0.4621,  0.1574,
    1994L, -0.0618,  -0.034,
    1995L,  0.0804,   0.183,
    1996L,  0.2287,  0.0835,
    1997L,   0.459,  0.0665,
    1998L,  0.2032,  0.1245,
    1999L,   0.412, -0.0219,
    2000L, -0.0953,  0.0744,
    2001L, -0.1775,  0.0555,
    2002L, -0.4306,  0.1027
)

index_return %>% 
    mutate_at(c("msci", "jpm"), percent) %>% 
    kable()
```

## 1) Calculate portfolio return

For a porfolio of MSCI 60% and JPM 40%, returns are as follows.

```{r}
portfolio_return <- index_return %>% 
    mutate(portfolio = msci * 0.6 + jpm * 0.4) %>% 
    select(year, portfolio)
```

```{r echo = FALSE}
portfolio_return %>% 
    mutate(portfolio = percent(portfolio)) %>% 
    kable()
```

The expected value of the portfolio is:

```{r, results = "asis"}
percent(mean(portfolio_return$portfolio))
```

\newpage

## 2-3) Coefficient of variation and Sharpe

```{r}
portfolio <- portfolio_return$portfolio

cv_sharpe <- list(
    cv_msci = sd(index_return$msci) / mean(index_return$msci),
    cv_jpm = sd(index_return$jpm) / mean(index_return$jpm),
    cv_portfolio = sd(portfolio) / mean(portfolio),
    sharpe_msci = (mean(index_return$msci) - 0.0433) / sd(index_return$msci),
    sharpe_jpm = (mean(index_return$jpm) - 0.0433) / sd(index_return$jpm),
    sharpe_portfolio = (mean(portfolio) - 0.0433) / sd(portfolio)
)

tibble(name = names(cv_sharpe), value = as.numeric(cv_sharpe)) %>% 
    kable(digits = 2)
```

Eventhough the best return per unit of risk is with JPM, portfolio manages to keep the expected returns relatively higher with less risk. 

\newpage 

# Exercise 4

Ratios for common stock in an equally weighted portfolio are given as follows.

```{r echo = FALSE}
stock <- tribble(
                  ~stock,  ~P_E, ~P_S,   ~P_B,
           "Aber.&Fitch", 13.67, 1.66,   3.43,
             "Albemarle", 14.43, 1.13,   1.96,
                  "Avon", 28.06, 2.45, 382.72,
    "Berkshire Hathaway", 18.46, 2.39,   1.65,
               "Everest", 11.91, 1.34,    1.3,
             "FPL Group",  15.8, 1.04,    1.7,
      "Johnson Controls", 14.24,  0.4,   2.13,
          "Tenneco Auto",  6.44, 0.07,  41.31
)

kable(stock)
```

## 1) Mean and medians

```{r}
stock %>% 
    gather(ratio, value, P_E, P_S, P_B) %>% 
    group_by(ratio) %>% 
    summarise(mean = mean(value), median = median(value)) %>% 
    kable(digits = 2)
```

## 2) Comments

Price to Book ratios mean and median values have a large difference because the mean is highly susceptible to outliers. In this case, Avon and Tenneco's large P/B values distort the mean. Median gives a better sense of P/B.

It may not be a good idea to directly take the mean or median values of these ratios. Ratios don't give any information about the price and price determines the amount of shares (weight) of a stock in the portfolio even when it is equally weighted. A better measure for finding the mean P/E can be 

``Total Portfolio Value / Total EPS``.