r <- c(0.10, -0.10, 0.05) # three daily returns as decimals
print(r) # -> [1] 0.10 -0.10 0.05[1] 0.10 -0.10 0.05print(1 + r) # -> [1] 1.10 0.90 1.05[1] 1.10 0.90 1.05How do I build an equity curve?
Turn a column of returns into the growth of one unit of money with cumprod(1 + r), and check the last value against the prices.
An equity curve is what one unit of money is worth after each return in turn. Add 1 to every return to get a growth factor, then multiply the growth factors together as you go: cumprod(1 + r).
In this lesson I build the curve on three returns I can check by hand, then on a column of daily returns from a price file, and I show that the last value of the curve is the last close over the first.
Step 1. Three returns
cumprod() is the running product. Element 1 is the first growth factor, element 2 is the first times the second, element 3 is the first times the second times the third.
Up 10%, down 10%, up 5%. Those three returns become the growth factors 1.10, 0.90 and 1.05.
equity <- cumprod(1 + r) # running product down the growth factors
print(equity) # -> [1] 1.1000 0.9900 1.0395[1] 1.1000 0.9900 1.0395A gain of 10% followed by a loss of 10% leaves 0.99, not 1.0. The 10% loss comes off 1.10, so it takes away 0.11 while the gain only put on 0.10.
print(equity[2]) # -> [1] 0.99[1] 0.99print(1.10 * 0.90) # -> [1] 0.99 <- the same two factors by hand[1] 0.99print(equity[2] == 1.10 * 0.90) # -> [1] TRUE[1] TRUEThe whole curve is that multiplication carried on. Here is the last value against the product written out longhand.
by_hand <- 1.10 * 0.90 * 1.05 # the three growth factors written out
last <- equity[length(equity)] # the final level of the curve
print(by_hand) # -> [1] 1.0395[1] 1.0395print(last) # -> [1] 1.0395[1] 1.0395print(all.equal(last, by_hand)) # -> [1] TRUE[1] TRUEIn Lesson 8 prod(1 + r) - 1 gave the total over the whole stretch as one number. cumprod() gives the same multiplication, but keeps every step.
print(prod(1 + r) - 1) # -> [1] 0.0395 <- Lesson 8, the single total[1] 0.0395print(round((last - 1) * 100, 4)) # -> [1] 3.95 <- the curve's last value, minus 1[1] 3.95One unit of money became 1.0395, a gain of 3.95%.
Step 2. A curve from a file of closing prices
prices.csv sits next to this lesson and holds simulated closes: 40 business days for three tickers. read.csv() reads dates as character, so as.Date() turns the Date column into real dates. The dates here are written YYYY-MM-DD, so text order and calendar order already agree and arrange(Date) returns the same rows either way. The conversion is for what real dates give you: subtracting two of them, printing them as dates, and matching them against dates in another table.
library(dplyr)
prices <- read.csv("prices.csv") # 120 rows, three tickers
prices$Date <- as.Date(prices$Date) # character to Date
print(head(prices, 3)) Date Ticker Close Volume
1 2026-01-02 AAA 186.66 2183494
2 2026-01-02 CCC 416.98 3117814
3 2026-01-02 DDD 125.18 7447373# -> Date Ticker Close Volume
# -> 1 2026-01-02 AAA 186.66 2183494
# -> 2 2026-01-02 CCC 416.98 3117814
# -> 3 2026-01-02 DDD 125.18 7447373
print(nrow(prices)) # -> [1] 120[1] 120I want one ticker. Base R does it with a logical row index, the same idea as Lesson 5.
aapl_base <- prices[prices$Ticker == "AAA", ] # keep the rows where Ticker is AAA
aapl_base <- aapl_base[order(aapl_base$Date), ] # oldest first, what arrange(Date) does below
print(head(aapl_base, 3)) Date Ticker Close Volume
1 2026-01-02 AAA 186.66 2183494
4 2026-01-05 AAA 186.38 8463199
7 2026-01-06 AAA 188.51 5299573# -> Date Ticker Close Volume
# -> 1 2026-01-02 AAA 186.66 2183494
# -> 4 2026-01-05 AAA 186.38 8463199
# -> 7 2026-01-06 AAA 188.51 5299573
print(nrow(aapl_base)) # -> [1] 40[1] 40The row names are 1, 4, 7: base R keeps the numbering from the full table. filter() renumbers from 1, which is why the printed tables below start at row 1.
Now the dplyr version, with the returns from Lesson 15 in the same pipeline. lag(Close) is yesterday’s close on today’s row.
aapl <- prices |>
filter(Ticker == "AAA") |> # one ticker
arrange(Date) |> # oldest first, so lag reaches back a day
mutate(ret = Close / lag(Close) - 1) # today's close over yesterday's, minus 1
print(head(aapl[, c("Date", "Close", "ret")], 3)) Date Close ret
1 2026-01-02 186.66 NA
2 2026-01-05 186.38 -0.001500054
3 2026-01-06 188.51 0.011428265# -> Date Close ret
# -> 1 2026-01-02 186.66 NA
# -> 2 2026-01-05 186.38 -0.001500054
# -> 3 2026-01-06 188.51 0.011428265The first row has no day before it, so lag() puts NA there. cumprod() would carry that NA all the way down the column, so the row goes.
aapl <- aapl |> filter(!is.na(ret)) # drop the leading NA row
print(nrow(aapl)) # -> [1] 39 <- 40 closes give 39 returns[1] 39With 39 clean returns, the curve is one mutate().
aapl <- aapl |> mutate(equity = cumprod(1 + ret)) # one unit of money, day by day
print(head(aapl[, c("Date", "ret", "equity")], 4)) Date ret equity
1 2026-01-05 -0.001500054 0.9984999
2 2026-01-06 0.011428265 1.0099111
3 2026-01-07 -0.006737043 1.0031073
4 2026-01-08 0.010841700 1.0139826# -> Date ret equity
# -> 1 2026-01-05 -0.001500054 0.9984999
# -> 2 2026-01-06 0.011428265 1.0099111
# -> 3 2026-01-07 -0.006737043 1.0031073
# -> 4 2026-01-08 0.010841700 1.0139826The first day of the curve is just 1 + ret, because there is nothing yet to multiply by. Every day after that takes yesterday’s level times today’s growth factor.
Step 3. Check the end against the prices
Multiplying every growth factor together is the same as dividing the last close by the first, since each close cancels the one after it. So the final value of the curve has to match.
last_eq <- aapl$equity[nrow(aapl)] # final level of the curve
closes <- aapl_base$Close # the 40 AAA closes, oldest first
print(round(last_eq, 6)) # -> [1] 0.90764[1] 0.90764print(round(closes[length(closes)] / closes[1], 6)) # -> [1] 0.90764 <- last close over first[1] 0.90764print(c(closes[1], closes[length(closes)])) # -> [1] 186.66 169.42[1] 186.66 169.42One unit of money became 0.90764. Ten thousand becomes 9076.40, because the curve scales.
print(round(10000 * last_eq, 2)) # -> [1] 9076.4[1] 9076.4Step 4. cumprod against cumsum
cumsum() is the running total instead of the running product. Both run down the same 39 returns.
aapl <- aapl |> mutate(added = cumsum(ret)) # running total of the returns
print(tail(aapl[, c("Date", "equity", "added")], 3)) Date equity added
37 2026-02-24 0.9248902 -0.07536057
38 2026-02-25 0.9183542 -0.08242729
39 2026-02-26 0.9076396 -0.09409454# -> Date equity added
# -> 37 2026-02-24 0.9248902 -0.07536057
# -> 38 2026-02-25 0.9183542 -0.08242729
# -> 39 2026-02-26 0.9076396 -0.09409454
print(round((last_eq - 1) * 100, 4)) # -> [1] -9.236[1] -9.236print(round(sum(aapl$ret) * 100, 4)) # -> [1] -9.4095[1] -9.4095The price went from 186.66 to 169.42, which is -9.236%. The two final figures are not the same number: simple returns compound, they do not add, which is the point Lesson 8 made with prod() against sum().
Your turn
Build the equity curve for CCC from prices.csv. What is one unit of money worth at the end, and does it equal the last close over the first?
TipShow answer
library(dplyr)
prices <- read.csv("prices.csv")
prices$Date <- as.Date(prices$Date)
msft <- prices |>
filter(Ticker == "CCC") |> # one ticker
arrange(Date) # oldest first
closes <- msft$Close # all 40 closes, kept before any row is dropped
msft <- msft |>
mutate(ret = Close / lag(Close) - 1) |> # daily returns
filter(!is.na(ret)) |> # drop the leading NA
mutate(equity = cumprod(1 + ret)) # the curve
last_eq <- msft$equity[nrow(msft)] # final level
print(round(last_eq, 6)) # -> [1] 1.04547
print(round(closes[length(closes)] / closes[1], 6)) # -> [1] 1.04547
print(round((last_eq - 1) * 100, 4)) # -> [1] 4.547Lesson 19 groups the table by ticker, so a curve like this one can be built for all three at once.