Apply it to your data 13
How much should you expect from your $100 investment after 20 years?
Sondre Asheim
2023-12-04
# Load packages
# Core
library(tidyverse)
library(tidyquant)
# Source function1 Import stock prices
Revise the code below.
- Replace symbols with your stocks.
- Replace the from and the to arguments to date from 2012-12-31 to present.
symbols <- c("AAPL", "CL=F", "ROKU")
prices <- tq_get(x = symbols,
get = "stock.prices",
from = "2012-12-31",
to = "2023-10-31")2 Convert prices to returns
asset_returns_tbl <- prices %>%
group_by(symbol) %>%
tq_transmute(select = adjusted,
mutate_fun = periodReturn,
period = "monthly",
type = "log") %>%
slice(-1) %>%
ungroup() %>%
set_names(c("asset", "date", "returns"))3 Assign a weight to each asset
Revise the code for weights.
- The vector weights should have a length equal to the number of assets in the portfolio.
- The values in the vector weights should sum to 1.
# symbols
symbols <- asset_returns_tbl %>% distinct(asset) %>% pull()
symbols## [1] "AAPL" "CL=F" "ROKU"# weights
weights <- c(0.5, 0.3, 0.2)
weights## [1] 0.5 0.3 0.2w_tbl <- tibble(symbols, weights)
w_tbl## # A tibble: 3 × 2
## symbols weights
## <chr> <dbl>
## 1 AAPL 0.5
## 2 CL=F 0.3
## 3 ROKU 0.24 Build a portfolio
portfolio_returns_tbl <- asset_returns_tbl %>%
tq_portfolio(assets_col = asset,
returns_col = returns,
weights = w_tbl,
rebalance_on = "months",
col_rename = "returns")
portfolio_returns_tbl## # A tibble: 130 × 2
## date returns
## <date> <dbl>
## 1 2013-01-31 -0.0598
## 2 2013-02-28 -0.0300
## 3 2013-03-28 0.0178
## 4 2013-04-30 -0.0117
## 5 2013-05-31 0.00626
## 6 2013-06-28 -0.0483
## 7 2013-07-31 0.0913
## 8 2013-08-30 0.0476
## 9 2013-09-30 -0.0261
## 10 2013-10-31 0.0280
## # ℹ 120 more rows5 Simulating growth of a dollar
# Get mean portfolio return
mean_port_return <- mean(portfolio_returns_tbl$returns)
mean_port_return## [1] 0.009936184# Get standard deviation of portfolio returns
stddev_port_return <- sd(portfolio_returns_tbl$returns)
stddev_port_return## [1] 0.078339576 Simulation function
simulate_accumulation <- function(initial_value, N, mean_return, sd_return) {
# Add a dollar
simulated_returns_add_1 <- tibble(returns = c(initial_value, 1 + rnorm(N, mean_return, sd_return)))
# Calculate the cumulative growth of a dollar
simulated_growth <- simulated_returns_add_1 %>%
mutate(growth = accumulate(returns, function(x, y) x*y)) %>%
select(growth)
return(simulated_growth)
}
simulate_accumulation(initial_value = 100, N = 120, mean_return = 0.005, sd_return = 0.01) %>%
tail()## # A tibble: 6 × 1
## growth
## <dbl>
## 1 186.
## 2 186.
## 3 188.
## 4 190.
## 5 193.
## 6 191.dump(list = c("simulate_accumulation"),
file = "../00_scripts/simulate_accumulation.R")7 Running multiple simulations
# Create a vector of 1s as a starting point
sims <- 51
starts <- rep(100, sims) %>%
set_names(paste0("sim", 1:sims))
starts## sim1 sim2 sim3 sim4 sim5 sim6 sim7 sim8 sim9 sim10 sim11 sim12 sim13
## 100 100 100 100 100 100 100 100 100 100 100 100 100
## sim14 sim15 sim16 sim17 sim18 sim19 sim20 sim21 sim22 sim23 sim24 sim25 sim26
## 100 100 100 100 100 100 100 100 100 100 100 100 100
## sim27 sim28 sim29 sim30 sim31 sim32 sim33 sim34 sim35 sim36 sim37 sim38 sim39
## 100 100 100 100 100 100 100 100 100 100 100 100 100
## sim40 sim41 sim42 sim43 sim44 sim45 sim46 sim47 sim48 sim49 sim50 sim51
## 100 100 100 100 100 100 100 100 100 100 100 100# Simulate
# for reproducible research
set.seed(1234)
monte_carlo_sim_51 <- starts %>%
# Simulate
map_dfc(simulate_accumulation,
N = 240,
mean = mean_port_return,
sd_return = stddev_port_return) %>%
# Add the column, month
mutate(month = seq(1:nrow(.))) %>%
# Arrange column names
select(month, everything()) %>%
set_names(c("month", names(starts))) %>%
pivot_longer(cols = -month, names_to = "sim", values_to = "growth")
monte_carlo_sim_51## # A tibble: 12,291 × 3
## month sim growth
## <int> <chr> <dbl>
## 1 1 sim1 100
## 2 1 sim2 100
## 3 1 sim3 100
## 4 1 sim4 100
## 5 1 sim5 100
## 6 1 sim6 100
## 7 1 sim7 100
## 8 1 sim8 100
## 9 1 sim9 100
## 10 1 sim10 100
## # ℹ 12,281 more rows# Calculate the quantiles for simulated values
probs <- c(0, 0.25, 0.5, 0.75, 1)
monte_carlo_sim_51 %>%
group_by(sim) %>%
summarise(growth = last(growth)) %>%
ungroup() %>%
pull(growth) %>%
# Find the quantiles
quantile(probs = probs) %>%
round(2)## 0% 25% 50% 75% 100%
## 41.39 320.47 754.03 1517.09 4490.518 Visualizing simulations with ggplot
Line plot with max, median, and min
# Step 1 Summarize data into max, median, and min of last value
sim_summary <- monte_carlo_sim_51 %>%
group_by(sim) %>%
summarise(growth = last(growth)) %>%
ungroup() %>%
summarise(max = max(growth),
median = median(growth),
min = min(growth))
sim_summary## # A tibble: 1 × 3
## max median min
## <dbl> <dbl> <dbl>
## 1 4491. 754. 41.4monte_carlo_sim_51 %>%
group_by(sim) %>%
filter(last(growth) == sim_summary$max |
last(growth) == sim_summary$median |
last(growth) == sim_summary$min) %>%
ungroup() %>%
# Plot
ggplot(aes(x = month, y = growth, col = sim)) +
geom_line() +
theme(legend.position = "none") +
theme(plot.title = element_text(hjust = 0.5)) +
theme(plot.subtitle = element_text(hjust = 0.5)) +
labs(title = "Simulating Growth of $100 over 240 months",
subtitle = "Max, Median, and Minimum Simulation")
Based on the Monte Carlo simulation results, how much should you expect from your $100 investment after 20 years? The median results were $754, which is what I should expect What is the best-case scenario? The best-case scenario is $4,491! What is the worst-case scenario? The worst case scenario is $41, which is a loss of almost $60. What are limitations of this simulation analysis? The limitations to this analysis is that it is probably more optimistic than real. The second limitation is about the data set we used. During the first five years, we were in a raising bull market, which means that the mean is probably on the high side for the first half of the graph, but since we have results until present time, the analysis is a bit more accurate.