Styjun

I have a dataframe and want to use uniroot on each row to solve Implied Volatility based on Black-Scholes formula. What is the correct way to use uniroot.all to solve for each row? It should produce a new column vector of results.

The code below has this error

"Error in S/K : non-numeric argument to binary operator"

. I suspect the trouble comes when uniroot is trying to solve multiple rows instead of each row one-by-one.

I've tried to modify to a vectorized version of the bscall function but this doesn't seem to best way to do it.

df <- data.frame( strike = c(80,120,100,100), 
 type = c("C", "C", "C","C"),
 optionPrice = c(22,3,7,9),
 futurePrice = c(100, 100,100,100),
 time_to_expiry = c(0.1, 0.1,1,1.2))

bscall <-
 function (S,K,r,T,sig) 
 d1 <- (log(S/K)+(r+0.5*sig^2)*T) / (sig*sqrt(T)) 
 d2 <- d1 - sig*sqrt(T)
 price <- S*pnorm(d1) - K*exp(-r*T)*pnorm(d2)
 return(price)
 


apply(df, 1,
 function(z) uniroot.all( function(x) bscall(z[4],z[1],r,z[5],x) - z[3], interval = c(0,1) ))

The first row of df is 80, "C", 22, 100, 0.1. It contains a character string. Therefore, when you do apply(df, 1, ....), the rows are coerced to character vectors:

df <- data.frame( strike = c(80,120,100,100), 
 type = c("C", "C", "C","C"),
 optionPrice = c(22,3,7,9),
 futurePrice = c(100, 100,100,100),
 time_to_expiry = c(0.1, 0.1,1,1.2))

apply(df, 1, function(z) z)
# [,1] [,2] [,3] [,4] 
# strike " 80" "120" "100" "100"
# type "C" "C" "C" "C" 
# optionPrice "22" " 3" " 7" " 9" 
# futurePrice "100" "100" "100" "100"
# time_to_expiry "0.1" "0.1" "1.0" "1.2"

First: you should include the library statement you used for package rootSolve.
Where are variables r and sig defined?

There is no need for uniroot.all as you can see by inspecting function bscall. There is no need for iteration; price is calculated directly from the input.

Define df as follows taking into account the first answer.

df <- data.frame( strike = c(80,120,100,100), 
 # type = c("C", "C", "C","C"),
 optionPrice = c(22,3,7,9),
 futurePrice = c(100, 100,100,100),
 time_to_expiry = c(0.1, 0.1,1,1.2))

Column type has been commented out.

Define your function bscall as

bscall <-
 function (S,K,r,T,sig) 
 d1 <- (log(S/K)+(r+0.5*sig^2)*T) / (sig*sqrt(T)) 
 d2 <- d1 - sig*sqrt(T)
 price <- S*pnorm(d1) - K*exp(-r*T)*pnorm(d2)
 return(price)

Give r and sig a value

r <- .05
sig <- 1

Apply your function to rows of df (indices into z changed because of change to df)

apply(df, 1,
 function(z) bscall(z[3],z[1],r,z[4],sig) - z[2])

giving

[1] 2.258107 3.168623 32.840162 34.366636

as answer.

The rest is up to you.