Apply it to your data 13
How much should you expect from your $100 investment after 20 years?
Jason Zink
2022-12-08
# Load packages
# Core
library(tidyverse)
library(tidyquant)
# Source function
source("../00_scripts/simulate_accumulation.R")1 Import stock prices
Revise the code below.
- Replace symbols with your stocks.
- Replace the from and the to arguments to date from 2012-12-31 to present.
symbols <- c("WMT", "AMZN", "TGT")
prices <- tq_get(x = symbols,
get = "stock.prices",
from = "2012-12-31",
to = "2017-12-31")2 Convert prices to returns
asset_returns_tbl <- prices %>%
group_by(symbol) %>%
tq_transmute(select = adjusted,
mutate_fun = periodReturn,
period = "monthly",
type = "log") %>%
slice(-1) %>%
ungroup() %>%
set_names(c("asset", "date", "returns"))3 Assign a weight to each asset
Revise the code for weights.
- The vector weights should have a length equal to the number of assets in the portfolio.
- The values in the vector weights should sum to 1.
# symbols
symbols <- asset_returns_tbl %>% distinct(asset) %>% pull()
symbols## [1] "AMZN" "TGT" "WMT"# weights
weights <- c(0.25, 0.2, 0.1)
weights## [1] 0.25 0.20 0.10w_tbl <- tibble(symbols, weights)
w_tbl## # A tibble: 3 × 2
## symbols weights
## <chr> <dbl>
## 1 AMZN 0.25
## 2 TGT 0.2
## 3 WMT 0.14 Build a portfolio
portfolio_returns_tbl <- asset_returns_tbl %>%
tq_portfolio(assets_col = asset,
returns_col = returns,
weights = w_tbl,
rebalance_on = "months",
col_rename = "returns")
portfolio_returns_tbl## # A tibble: 60 × 2
## date returns
## <date> <dbl>
## 1 2013-01-31 0.0208
## 2 2013-02-28 0.00943
## 3 2013-03-28 0.0250
## 4 2013-04-30 -0.00233
## 5 2013-05-31 0.00955
## 6 2013-06-28 0.00544
## 7 2013-07-31 0.0317
## 8 2013-08-30 -0.0457
## 9 2013-09-30 0.0301
## 10 2013-10-31 0.0443
## # … with 50 more rows5 Simulating growth of a dollar
# Get mean portfolio return
mean_port_return <- mean(portfolio_returns_tbl$returns)
mean_port_return## [1] 0.008098315# Get standard deviation of portfolio returns
stddev_port_return <- sd(portfolio_returns_tbl$returns)
stddev_port_return## [1] 0.0234029No need
6 Simulation function
No need
7 Running multiple simulations
# Create a vector of 1s as a starting point
sims <- 51
starts <- rep(1, sims) %>%
set_names(paste0("sim", 1:sims))
starts## sim1 sim2 sim3 sim4 sim5 sim6 sim7 sim8 sim9 sim10 sim11 sim12 sim13
## 1 1 1 1 1 1 1 1 1 1 1 1 1
## sim14 sim15 sim16 sim17 sim18 sim19 sim20 sim21 sim22 sim23 sim24 sim25 sim26
## 1 1 1 1 1 1 1 1 1 1 1 1 1
## sim27 sim28 sim29 sim30 sim31 sim32 sim33 sim34 sim35 sim36 sim37 sim38 sim39
## 1 1 1 1 1 1 1 1 1 1 1 1 1
## sim40 sim41 sim42 sim43 sim44 sim45 sim46 sim47 sim48 sim49 sim50 sim51
## 1 1 1 1 1 1 1 1 1 1 1 1# Simulate
# For reproducible research
set.seed(1234)
monte_carlo_sim_51 <- starts %>%
# Simulate
map_dfc(.x = .,
.f = ~simulate_accumulation(initial_value = .x,
N = 240,
mean_return = mean_port_return,
sd_return = stddev_port_return)) %>%
# Add a column month
mutate(month = 1:nrow(.)) %>%
select(month, everything()) %>%
# Rearrange column names
set_names(c("month", names(starts))) %>%
# Transform to long form
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 1
## 2 1 sim2 1
## 3 1 sim3 1
## 4 1 sim4 1
## 5 1 sim5 1
## 6 1 sim6 1
## 7 1 sim7 1
## 8 1 sim8 1
## 9 1 sim9 1
## 10 1 sim10 1
## # … with 12,281 more rows# Find quantiles
monte_carlo_sim_51 %>%
group_by(sim) %>%
summarise(growth = last(growth)) %>%
ungroup() %>%
pull(growth) %>%
quantile(probs = c(0, 0.5, 1)) %>%
round(2)## 0% 50% 100%
## 3.06 7.06 12.448 Visualizing simulations with ggplot
monte_carlo_sim_51 %>%
ggplot(aes(x = month, y = growth, color = sim)) +
geom_line() +
theme(legend.position = "none") +
theme(plot.title = element_text(hjust = 0.5)) +
labs(title = "Simulating growth of $100 over 240 months")
# Step 1 Summarize data into max, median, min of last value
sim_summary <- monte_carlo_sim_51 %>%
group_by(sim) %>%
summarise(growth = last(growth)) %>%
ungroup() %>%
summarise(max = max(growth),
median = median(growth),
min = min(growth))
sim_summary## # A tibble: 1 × 3
## max median min
## <dbl> <dbl> <dbl>
## 1 12.4 7.06 3.06# Step 2 Plot
monte_carlo_sim_51 %>%
# Filter for max, median, min sim
group_by(sim) %>%
filter(last(growth) == sim_summary$max |
last(growth) == sim_summary$median |
last(growth) == sim_summary$min) %>%
ungroup() %>%
# Plot
ggplot(aes(x = month, y = growth, color = 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, Minimum Simulation")
Line Plot of Simulations with Max, Median, and Min
Based on the Monte Carlo simulation results, how much should you expect from your $100 investment after 20 years? What is the best-case scenario? What is the worst-case scenario? What are limitations of this simulation analysis?
Based on the Monte Carlo simulation results the best-case scenario that you could be making is a max of 12.4. While the worst case scenario would be a return of 3.06, with an average return being 7.06. Limitations would be that we are not able to take into account events that are unforeseeable. This portfolio is based off the fact that we are looking at a normal distribution of overall returns.