How does an option's time value (also known as extrinsic or instrumental value) depend on how far it is in the money or out of the money? In other words, how does the time value change as the underlying price changes?

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    $\begingroup$ I'm not sure if you're asking about the volatility smile or how Black Scholes works, but your question is impossible to parse. The option's price isn't derived from volatility. It's derived from volatility, spot, strike, risk-free rate, and time to expiration. So yes, the moneyness impacts option value. That's why quotes listed on an option chain are different at each price level. $\endgroup$ Commented Dec 31, 2011 at 2:23
  • $\begingroup$ @kamikaze_pilot Your question has been rewritten for clarity based on what I believe is your intended question. Please re-edit if I have erred. $\endgroup$ Commented Dec 31, 2011 at 23:46
  • $\begingroup$ I'm still not sure what the original poster was asking, but this edited question is now pretty decent. More over, the answer from @SpeedBoots is a solid solution. $\endgroup$ Commented Jan 5, 2012 at 23:33

4 Answers 4


By definition, an option's premia is the sum of intrinsic value and time value. The time value of premia declines as the option goes more ITM ("in-the-money") or OTM ("out-of-the-money"), ceteris paribus.

An intuitive explanation for this can be found by thinking of time value as the expected P&L of a long option position dynamically hedged by going short (long) Delta units of a call (put). Delta is the sensitivity of the option premia to the price of the underlying. It is a value between 0 and 1 and increases with moneyness. Delta of an ATM ("at-the-money") option is approximately 0.5. Suppose that you are long 100 ATM calls struck at X and delta-hedged by being short 50 units of the underlying at price S=X. If S goes up to S', your option is ITM and delta goes up as well--let's say to 0.6. Now, to be hedged, you have to sell 10 more units of the underlying at S'. Now suppose that the price goes back from S' to S. Delta decreases back to 0.5 and you have to buy back 10 units of the underlying to be hedged again. You have made a S' - S profit on 10 units by holding a market neutral position. Over the interval of time between re-hedges, your delta-hedged position gets long in an up market and short in a down market, and as you re-hedge you are realising little profits from the re-hedges as the underlying goes up and down in price.

This is a valuable property to have in a position that by construction takes no market risk. But there is no free lunch, and time value can be thought of as the price you have to pay in order to have exposure to this delta-hedging P&L over the life of your option position. (It is also why time value increases the more volatile the market expects the underlying to be--Implied Volatility--and time value erodes as the option's expiry gets closer. The sensitivity of option's premia--and therefore time value--to those two factors are called Vega and Theta, but that is another discussion.)

So why does time value decline the more ITM or OTM the option is? The reason is that the change in Delta given a change in underlying price is not linear, but is rather a convex function of price. The change in delta is greatest when the option is ATM, but as the option goes more and more ITM (OTM), Delta gets closer and closer to 1 (0) and the changes in Delta get smaller and smaller. Therefore, so do the little re-hedging profits and so does time value. Delta-hedging a deep ITM or OTM option doesn't generate much P&L for the simple reason that Delta doesn't change very much.

The sensitivity of Delta to a change in underlying price is called Gamma, and Gamma is the reason why time value is worth so little in very ITM option but worth allot in an ATM option. A more formal answer can be given to your question, but given the nature of the question, I hope that this intuitive sketch will be more illuminating.

  • $\begingroup$ Hi Robert, welcome to quant.SE and thanks for contributing your answer. I may be wrong, but my impression was that the question asks about the absolute level of time value, not the time value as a percentage of premium. Do you have thoughts on this point? $\endgroup$ Commented Jan 3, 2012 at 15:26
  • $\begingroup$ Quite right. The same intuition applies, but it's more clear to say that the absolute value of time value is greatest when the option is ATM and declines as it goes OTM or ITM. $\endgroup$ Commented Jan 5, 2012 at 11:47

Here is a plot of the time value for a typical call option as a function of spot. The image below was produced for a call option with the following parameters: Strike = 100, risk free rate = 0, volatility = 30%, Time to maturity = 0.5.

time value vs Spot

Here, by the way, is theta, the time decay (derivative of total value with respect to time) versus spot using the same parameters:

theta vs. spot

R Code for theta:

    BSOption <- function(S,K,r,v,T,type){
      d1 = (log(S/K)+(r+.5*v*v)*T)/(v*sqrt(T))
      d2 = d1 - v*sqrt(T)
      if(type == "call"||type == "Call"||type == "C"||type == "c"||type == 1)
        delta = pnorm(d1)
        theta = (-S*dnorm(d1)*v/(2*sqrt(T))) - 
        price = S*pnorm(d1)- K*exp(-r*T)*pnorm(d2)
        delta = -pnorm(-d1)
        theta = (-S*dnorm(d1)*v/(2*sqrt(T))) + 
        price = K*exp(-r*T)*pnorm(-d2)-S*pnorm(-d1)
      gamma = dnorm(d1)/(S*v*sqrt(T))
      vega = S*dnorm(d1)*sqrt(T)
      theta = theta/365
      BSoption = data.frame(price, delta, gamma, vega, theta)

S = seq(80,120,1); 
plot(S,BSOption(S,100,0,.3,.5,1)$theta,ylab="Theta",xlab="Spot"); title("Theta vs Spot")

R code for time value:

S = seq(80,120,1); 
intrinsic = array(0,c(length(S),1))
for(i in 1:length(intrinsic)){
    intrinsic[i] = max(S[i] - 100,0)
plot(S,BSOption(S,100,0,.3,.5,1)$price-intrinsic,ylab="Time Value",xlab="Spot");
title("Time Value vs Spot")
  • $\begingroup$ Great graph, I think this pretty conclusively answers the question. I just switched up the formatting to put the answer to the question at hand first. $\endgroup$ Commented Jan 5, 2012 at 15:46
  • $\begingroup$ When the deep ITM time value turns positive slightly, what happens? $\endgroup$
    – Wei Wu
    Commented Oct 16, 2020 at 9:06
  • $\begingroup$ This might be an obvious question, but would it be fair to say the first figure for a put option is just a mirror about an imaginary line in the middle of the figure? $\endgroup$
    – KaiSqDist
    Commented May 2 at 20:38

The deeper an option is in the money, the less likely it is that a long position in the option will result in a higher payoff than a (financed) long postion in the underlying (which has zero time value).

In other words: the value of the protection the call provides in cases when the underlying price falls below the strike, decreases with a rising underlying price (because the probability that the underlying price is at expiry below the strike decreases).


The options value is dependent on the moneyness of the option. Taking the partial derivative of Theta w.r.t. term S/K (whole term) will give the sensitivity of Theta to moneyness.

S/K is the moneyness.

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    $\begingroup$ I don't see how linking back to a wikipedia page mentioned in the question helps. We are looking for something more than what is on that wiki page. $\endgroup$ Commented Jan 3, 2012 at 21:54
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    $\begingroup$ Did you learn nothing from your suspension? You were put on a time-out because of your low-quality "contributions". $\endgroup$ Commented Jan 4, 2012 at 2:47

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