Apply it to your data 13
How much should you expect from your $100 investment after 20 years?
Aalaya Elgohery
2024-10-04
# Load packages
# Core
library(tidyverse)
library(tidyquant)
# Source function
source("../00_scripts/simulate_accumulation.R")1 Import stock prices
Revise the code below.
- Replace symbols with your stocks.
- Replace the from and the to arguments to date from 2012-12-31 to present.
symbols <- c("GOOG", "GME", "NVDA", "V")
prices <- tq_get(x = symbols,
get = "stock.prices",
from = "2012-12-31")2 Convert prices to returns
asset_returns_tbl <- prices %>%
group_by(symbol) %>%
tq_transmute(select = adjusted,
mutate_fun = periodReturn,
period = "monthly",
type = "log") %>%
slice(-1) %>%
ungroup() %>%
set_names(c("asset", "date", "returns"))3 Assign a weight to each asset
Revise the code for weights.
- The vector weights should have a length equal to the number of assets in the portfolio.
- The values in the vector weights should sum to 1.
# symbols
symbols <- asset_returns_tbl %>% distinct(asset) %>% pull()
symbols## [1] "GME" "GOOG" "NVDA" "V"# weights
weights <- c(0.25, 0.25, 0.25, 0.25)
weights## [1] 0.25 0.25 0.25 0.25w_tbl <- tibble(symbols, weights)
w_tbl## # A tibble: 4 × 2
## symbols weights
## <chr> <dbl>
## 1 GME 0.25
## 2 GOOG 0.25
## 3 NVDA 0.25
## 4 V 0.254 Build a portfolio
portfolio_returns_tbl <- asset_returns_tbl %>%
tq_portfolio(assets_col = asset,
returns_col = returns,
weights = w_tbl,
rebalance_on = "months",
col_rename = "returns")
portfolio_returns_tbl## # A tibble: 144 × 2
## date returns
## <date> <dbl>
## 1 2013-01-31 0.00716
## 2 2013-02-28 0.0451
## 3 2013-03-28 0.0484
## 4 2013-04-30 0.0804
## 5 2013-05-31 0.0311
## 6 2013-06-28 0.0607
## 7 2013-07-31 0.0398
## 8 2013-08-30 -0.00126
## 9 2013-09-30 0.0418
## 10 2013-10-31 0.0666
## # ℹ 134 more rows5 Simulating growth of a dollar
# Get mean portfolio return
mean_port_return <- mean(portfolio_returns_tbl$returns)
mean_port_return## [1] 0.02174311# Get standard deviation of portfolio returns
stddev_port_return <- sd(portfolio_returns_tbl$returns)
stddev_port_return## [1] 0.095143316 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("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)
my_data_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")
my_data_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# Find quantiles
my_data_sim_51 %>%
group_by(sim) %>%
summarize(growth = last(growth)) %>%
ungroup() %>%
pull(growth) %>%
quantile(probs = c(0, 0.25, .50, 0.75, 1)) %>%
round(2)## 0% 25% 50% 75% 100%
## 290.31 3419.56 9649.18 22035.42 81099.588 Visualizing simulations with ggplot
Line Plot of Simulations with Max, Median, and Min
my_data_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 $100 over 240 months")
Line plot with max, median, and min
# Step 1, Summarize data into max, median, and min of the last value
sim_summary <- my_data_sim_51 %>%
group_by(sim) %>%
summarize(growth = last(growth)) %>%
ungroup() %>%
summarize(max = max(growth),
median = median(growth),
min = min(growth))
sim_summary## # A tibble: 1 × 3
## max median min
## <dbl> <dbl> <dbl>
## 1 81100. 9649. 290.# Step 2 Plot
my_data_sim_51 %>%
# Filter for max, median, and min sim
group_by(sim) %>%
filter(last(growth) == sim_summary$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 240 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 on the graphs created and the Monte Carlo Simulation results, the very best case scenario for these is approximately 60,000 growth after 20 years, the worst-case scenario is approximately 0, and the expected growth is around 10,000. Some of the limitations of this simulation analysis include uncertainty, you never know what you’re actual growth is going to be, the next limitation is assumptions, we are going based off of assumptions but you could end up with the lowest growth, seeing the median is around 10,000, there is, there is not many that will be near the 60,000 mark. Another limitation is a restricted range, there are several limitations and obstacles that this simulation does not take into account meaning the actual growth may be lower for the best and worst-case scenarios. Overall, you should expect 10,000 from a $100 investment after 20 years.