As per the Black-Scholes model, the value of a call option is directly proportional to the volatility. Without getting into the derivation of the BS equation, is it possible to intuitively understand why this is so?

High volatility just means the underlying stock is volatile, it does not imply if the stock is going up and down. But call options should go up in price only when the underlying stock goes up in price.

So how come high volatility always means a high price for call option?

  • 1
    Volatility = unpredictability. Steady, relatively slow trading at only minor price variations allows a more reliable prediction of future behavior, making the party selling the option more confident in the (low) chance you'll exercise it at a significant loss to him. A highly volatile stock, trading at high volume for wildly varying prices, reduces the predictability of the spot price as of the option date, and therefore also reduces the confidence of the option seller that he won't lose his shirt on the deal. So, he'll want more money for the option to ensure he doesn't.
    – KeithS
    Commented Jan 24, 2014 at 20:59
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    Volatility = unpredictability, but that goes both for the call option buyer and the seller, but looks like the benefit in thsi case goes to the seller only.
    – Victor123
    Commented Jan 24, 2014 at 21:54
  • 2
    You're correct, but both have power over the option price; the seller explicitly offers the price for the option (which he'll want to be as high as possible to cover his uncertainty), and the buyer has the choice to accept that price or not (the buyer will want it low for the same ends). Supply and demand; they'll meet in the middle.
    – KeithS
    Commented Jan 24, 2014 at 21:58
  • This question was originally written wrongly. The option price depends on the actual volatility, whereas the question wrote about implied volatility. Implied volatility is what you get if you run the Black-Scholes equation in reverse - taking the current price and computing what volatility theoretically would have given it. I edited the question to correct this mistake.
    – user32479
    Commented Dec 23, 2015 at 1:47

9 Answers 9


The mathematics make it easier to understand why this is the case.

Using very bad shorthand, d1 and d2 are inputs into N(), and N() can be expressed as the probability of the expected value or the most probable value which in this case is the discounted expected stock price at expiration. d1 has two σs which is volatility in the numerator and one in the denominator. Cancelling leaves one on top. Calculating when it's infinity gives an N() of 1 for S and 0 for K, so the call is worth S and the put PV(K). At 0 for σ, it's the opposite.

More concise is that any mathematical moment be it variance which mostly influences volatility, mean which determines drift, or kurtosis which mostly influences skew are all uncertanties thus costs, so the higher they are, the higher the price of an option.

Economically speaking, uncertainties are costs. Since costs raise prices, and volatility is an uncertainty, volatility raises prices.

It should be noted that BS assumes that prices are lognormally distributed. They are not. The closest distribution, currently, is the logVariance Gamma distribution.

  • 2
    Thanks, but is it possible to understand the relation using economic principles (supply-demand) without the math?
    – Victor123
    Commented Jan 24, 2014 at 18:05
  • If you're gonna write an equation, then write an equation. Why break the equation into a sentence??? wtf
    – NoName
    Commented Jan 1, 2021 at 8:33

I agree that high volatility just means the underlying stock price fluctuates more, and it does not imply if the stock is going up or down.

But a high volatility in the price of an underlying also means that there is a higher chance that the underlying price could reach extreme prices (albeit in either direction). However, if you purchased a call option then if the underlying price reached an extremely high value, then you will be richly rewarded. But if the underlying price reached an extremely low value, you won't lose any more than the initial premium that you paid. There is no additional risk on your side, it's capped to the premium that you paid for the call option.

It's this asymmetric outcome (Heads - I win, Tails - I don't lose) combined with high volatility that means that call options will increase in value when the underlying price becomes more volatile.

If the optionality wasn't there then the price wouldn't be related to the volatility of the underlying. But that would be called a Future or a Forward :-)

  • 2
    Nice answer here. I fixed a typo, note that loose rhymes with moose. My 3rd grade teacher taught me this, and I'm sticking with it. Commented Apr 5, 2017 at 16:53
  • This is the most intuitive answer that I have found. Thanks.
    – James LT
    Commented Nov 26, 2017 at 1:41
  • Best answer. Without price accounting for volatility, you can just yolo with the most volatile option till you hit gold.
    – NoName
    Commented Jan 1, 2021 at 8:30

Understanding the BS equation is not needed. What is needed is an understanding of the bell curve.

enter image description here

You seem to understand volatility. 68% of the time an event will fall inside one standard deviation. 16% of the time it will be higher, 16%, lower.

Now, if my $100 stock has a STD of $10, there's a 16% chance it will trade above $110. But if the STD is $5, the chance is 2.3% per the chart below. The higher volatility makes the option more valuable as there's a higher chance of it being 'in the money.'

My answer is an oversimplification, per your request.

  • 1
    Thanks very much, but higher volatility also means the stock has a bigger chance (16%) of getting to 90. So what makes a call option more valuable for such a stock?
    – Victor123
    Commented Jan 24, 2014 at 20:09
  • 4
    The higher volatility provides a higher chance of getting past your target price. Both Puts and Calls are priced higher. Think about it. Commented Jan 25, 2014 at 0:21

When volatility is higher, the option is more likely to end up in-the-money. Moreover, when it ends up in-the-money, it is likely to be over the strike price by a greater amount. Consider a call option. With high volatility, moves in the stock price are big - both up moves and down moves. If the stock moves up by a lot, the call option holder will benefit greatly. On the other hand, when the stock moves down, below a certain point the option holder does not care how big a down move the stock has. His downside is limited. Hence, the value of the option is increased by high volatility.

I know everyone who searches this is looking for this answer. Bump so people are able to get this concept instead of looking all over the web for it.


A few remarks that have not been highlighted yet in the other answers.

  1. As vol goes higher, the value of an ATMF call and the value of an ATMF put will increase; initially pretty linear in vol, until they approach their limits (S for the call, present value of strike for the put), then they'll taper off towards said limit.

  2. Since the value of both call and put go up, the reasoning that "it's more likely that the call will end up in the money" is fallacious. It's rather that when it ends up in the money, it'll be way in the money.

  3. The probability that the call lands in the money will actually decrease as vol goes up. In fact, the value of a ATMF high digital (paying 1$ if S(T)>K) goes to zero as vol goes up, while the value of the low digital goes to present value of 1$. (When thinking about this, remember that the forward is kept constant!)

  4. Option pricing works by hedging, that is replicating the option value. Every time you re-hedge a call (or put), you lose a bit (because of gamma). The higher the vol, the further the stock will move typically, so the more you lose. Thus, it costs more to produce a call (or put) when vol is higher. That's why its BS price increases with vol (until the limits are approached - and notice that there's no more gamma then).


Well, the increase in the price of the call can be understood by the fact that with increasing volatility the profit form long gamma position hedging increases.

This is because from the point of view of no-arbitrage pricing, it is irrelevant how likely the stock is to go up or down because delta-neutral is a hedee against both the possibilities.

In a long gamma position, if the stock's price goes up or down, our portfolio always benefits. Hence, the higher the volatility, the greater the chance of the stock going up or down, more is our portfolio value, more is the price.


As per the Black-Scholes model, the value of a call option is directly proportional to the volatility. Without getting into the derivation of the BS equation, is it possible to intuitively understand why this is so?

No, you cannot disregard the BS equation and intuitively understand why the value of a call option is directly proportional to the volatility.

I'm dizzy from all of the quant-like attempts to answer your question. The answer is really quite simple. An option pricing formula has 5 inputs (strike price, underlying price, time until expiration, volatility, carry cost, and dividend if any). It's a formula. Period.

Let's try something a lot simpler. Let's pretend that the option pricing formula is:

  • Price = (1.6753) x Volatility

Now what happens to Price if Volatility increases? It increases. And conversely, it declines if Volatility decreases.

Now if you don't like 6th grade level explanations like this one, look at the formulas used to calculate d1 and d2 in the pricing model and therein lies your answer.

  • 2
    I think you are missing the point of the question. Yes, multiplying 1 number by a bigger number increases it. That is not the question. The question is really why are you multiplying by volatility, why not divide? Why is the formula this way? Reading the other answers, I now know why the equation does this. Due to the nature of call options the higher volatility means I can potentially make more money and I don't stand to loose much. I was confused too because high volatility just means it can be really low or really high so at first, it didn't make sense this increases the price
    – devo
    Commented Sep 16, 2020 at 20:44

Let's say a stock trades at $100 right now, and you can buy a $100 call option. When you buy the call option (and the money you paid is gone), one of two things can happen: The share price goes up, or the share price goes down.

If the share price goes up, you profit. If the share price goes down, you don't lose! Because once the shares are below $100, you don't exercise the call option, and you don't lose any money.

So if you have a share that is rock solid at $100, you don't make money. If you have a share where the company owner took some ridiculous risk, and the shares could go to $200 or the company could go bankrupt, then you have a 50% chance to make $100 and a 50% chance to not lose anything. That's much more preferable.


The entire premise of purchasing a call option is your expectation that the prices will rise. So even though there is a possibility of prices falling, you wouldn't mind paying higher premiums in a volatile market for a call option because you're bullish and are expecting the volatility to eventually turn out in your favour i.e. prices to rise

  • 1
    Not true - you can also buy a call option to set a maximum price for your purchase, which might be lower than the current price. And the answer doesn't explain how volatility affects option prices.
    – D Stanley
    Commented Sep 19, 2017 at 13:41

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