Apply it to your data 13
How much should you expect from your $100 investment after 20 years?
Reed Wilson
2023-05-03
# Load packages
# Core
library(tidyverse)
library(tidyquant)
# Source function
source("../00_scripts/simulate_accumulation.R")1 Import stock prices
Revise the code below.
- Replace symbols with your stocks.
- Replace the from and the to arguments to date from 2012-12-31 to present.
symbols <- c("TSLA", "GOOG","MSFT", "AAPL")
prices <- tq_get(x = symbols,
get = "stock.prices",
from = "2012-12-31",
to = "2023-04-30")2 Convert prices to returns
asset_returns_tbl <- prices %>%
group_by(symbol) %>%
tq_transmute(select = adjusted,
mutate_fun = periodReturn,
period = "monthly",
type = "log") %>%
slice(-1) %>%
ungroup() %>%
set_names(c("asset", "date", "returns"))3 Assign a weight to each asset
Revise the code for weights.
- The vector weights should have a length equal to the number of assets in the portfolio.
- The values in the vector weights should sum to 1.
# symbols
symbols <- asset_returns_tbl %>% distinct(asset) %>% pull()
symbols## [1] "AAPL" "GOOG" "MSFT" "TSLA"# weights
weights <- c(0.30, 0.30, 0.20, 0.20)
weights## [1] 0.3 0.3 0.2 0.2w_tbl <- tibble(symbols, weights)
w_tbl## # A tibble: 4 × 2
## symbols weights
## <chr> <dbl>
## 1 AAPL 0.3
## 2 GOOG 0.3
## 3 MSFT 0.2
## 4 TSLA 0.24 Build a portfolio
portfolio_returns_tbl <- asset_returns_tbl %>%
tq_portfolio(assets_col = asset,
returns_col = returns,
weights = w_tbl,
rebalance_on = "months",
col_rename = "returns")
portfolio_returns_tbl## # A tibble: 124 × 2
## date returns
## <date> <dbl>
## 1 2013-01-31 -0.000977
## 2 2013-02-28 -0.000782
## 3 2013-03-28 0.0208
## 4 2013-04-30 0.111
## 5 2013-05-31 0.154
## 6 2013-06-28 -0.0180
## 7 2013-07-31 0.0706
## 8 2013-08-30 0.0670
## 9 2013-09-30 0.0298
## 10 2013-10-31 0.0508
## # … with 114 more rows5 Simulating growth of a dollar
# Get mean portfolio return
mean_port_return <- mean(portfolio_returns_tbl$returns)
mean_port_return## [1] 0.0212265# Get standard deviation of portfolio returns
stddev_port_return <- sd(portfolio_returns_tbl$returns)
stddev_port_return## [1] 0.069037836 Simulation function
No need
7 Running multiple simulations
# Create a vector of 1s as a starting point
sims <- 51
starts <- rep(100, sims) %>%
set_names(paste0("sims", 1:sims))
starts## sims1 sims2 sims3 sims4 sims5 sims6 sims7 sims8 sims9 sims10 sims11
## 100 100 100 100 100 100 100 100 100 100 100
## sims12 sims13 sims14 sims15 sims16 sims17 sims18 sims19 sims20 sims21 sims22
## 100 100 100 100 100 100 100 100 100 100 100
## sims23 sims24 sims25 sims26 sims27 sims28 sims29 sims30 sims31 sims32 sims33
## 100 100 100 100 100 100 100 100 100 100 100
## sims34 sims35 sims36 sims37 sims38 sims39 sims40 sims41 sims42 sims43 sims44
## 100 100 100 100 100 100 100 100 100 100 100
## sims45 sims46 sims47 sims48 sims49 sims50 sims51
## 100 100 100 100 100 100 100# 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 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 sims1 100
## 2 1 sims2 100
## 3 1 sims3 100
## 4 1 sims4 100
## 5 1 sims5 100
## 6 1 sims6 100
## 7 1 sims7 100
## 8 1 sims8 100
## 9 1 sims9 100
## 10 1 sims10 100
## # … 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, 0.25, 0.5, 0.75, 1)) %>%
round(2)## 0% 25% 50% 75% 100%
## 988.04 5859.61 12302.15 22761.59 58582.99Visualizing Simulations with ggplot
# Step 1: Summarize data into max, median, and min of the 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 58583. 12302. 988.# Step 2 Plot
monte_carlo_sim_51 %>%
# Filter for max, median, and 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 120 months",
subtitle = "Maximum, Median, and Minimum Simulation")
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 upon the results, the median growth expectation is $12,302 after 20 years. The best case scenerio and max growth was $58,583 and worst case min of $988. Given the near 11 year growth of these four technology stocks, the numbers are not surprising with all four seeing rapid growth over this time period. Limitations for this simulation would be that the Monte Carlo simulation assumes only the normal distributions of returns, but the distribution of returns is negatively skewed, making this simulation too optimistic. This is true especially considering the data used for the simulation. If the data came during a bullish period you would see more optimism compared to a bearish period, where you would see a more pessimistic outcome.