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?
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.
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 :-)
Understanding the BS equation is not needed. What is needed is an understanding of the bell curve.
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 $5, the chance is 2.3% per the chart below. The higher volatility makes the option more valuable as there's a higher chance of it being 'in the money.'
My answer is an oversimplification, per your request.
When volatility is higher, the option is more likely to end up in-the-money. Moreover, when it ends up in-the-money, it is likely to be over the strike price by a greater amount. Consider a call option. With high volatility, moves in the stock price are big - both up moves and down moves. If the stock moves up by a lot, the call option holder will benefit greatly. On the other hand, when the stock moves down, below a certain point the option holder does not care how big a down move the stock has. His downside is limited. Hence, the value of the option is increased by high volatility.
I know everyone who searches this is looking for this answer. Bump so people are able to get this concept instead of looking all over the web for it.
A few remarks that have not been highlighted yet in the other answers.
As vol goes higher, the value of an ATMF call and the value of an ATMF put will increase; initially pretty linear in vol, until they approach their limits (S for the call, present value of strike for the put), then they'll taper off towards said limit.
Since the value of both call and put go up, the reasoning that "it's more likely that the call will end up in the money" is fallacious. It's rather that when it ends up in the money, it'll be way in the money.
The probability that the call lands in the money will actually decrease as vol goes up. In fact, the value of a ATMF high digital (paying 1$ 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!)
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:
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.
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
