While it makes sense the Sigma gets larger, the value of the security will

also increase. an easy way to see that perhaps is

imagine that S0, the initial stock price, is much less than in, than the strike of

K. Well, in that case, if Sigma is very very

small, the chances of the stock price growing enough, so that the option ends

up in the money, will be zero, or approximately zero.

On the other hand, as Sigma gets sufficiently large, the probability that

st will be greater than K will actually increase, in which case, the option value

will be non-zero. And so it makes sense that the call

price, the initial price of the option C zero, should be increasing in Sigma.

And we will see that that is indeed the case.

The Vega of an option is a partial derivative of the option price with

respect to the volatility parameter Sigma.

Vega, therefore, measures the sensitivity of the option price to Sigma.

And using the Black-Scholes formula, it can easily be calculated.

Vega is equal to Delta C Delta Sigma, which turns out to be e to the minus eT,

s square root of time to majority times phi of d1, where phi is the probability

density function of a standard normal random variable.

Now, we can also compute the Vega for a European put option by using put-call

parity. So, this is put-call parity here.

Remember, so it actually implies. So, it implies that the put price is

equal to the the call plus e to the minus rt times K minus e to the minus cT times

S. And so, therefore, we can actually

compute Delta P, Delta Sigma, we see it's equal to, well Delta C, Delta Sigma.

That's the first term here. And then these other two terms don't

depend on Sigma at all. So, it's plus zero minus zero.

And so, we see Delta P, Delta Sigma equals Delta c, Delta Sigma.

And so, the Vega of a European put option is the same as the Vega of a European

call option. Here's a question for us.

Is the concept of Vega inconsistent in any way with the Black-Scholes model?

And the answer is yes. If you recall, the Black-Scholes mo-,

model assumes that St, the stock price of any time t, is equal to S zero e to the

Mu minus Sigma squared over 2 times t, plus Sigma times Wt, where Wt is a

standard Brownian motion. Mu and Sigma are constants in this model.

They are not assumed to change. And that indeed was the assumption of the

Black-Scholes model. They assumed continuous trading.

They assumed that there were no transactions, costs, and that short sales

were allowed, and that borrowing or lending at the risk free interest rate,

or was also possible. Using these assumptions, they constructed

a self-financing trading strategy that replicated the payoff of the option, and

that is indeed how they are paying the Black-Scholes formula.

Nothing in their model allowed Sigma to change, Sigma was a known constant.

And yet when we're talking about Delta C, Delta Sigma, we're implicitly recognizing

the fact that Sigma can change. And indeed, in the marketplace, Sigma

does change. So, in that sense, the, While,

mathematically, one can always define Delta C, Delta Sigma, there's no problem

with that. Within the economics of the Black-Scholes

model, it isn't consistent to talk about Sigma changing.

Because we ob-, we obtained the Black-Scholes option price under the

assumption that Sigma could not change. Here are some plots of Vega for options,

for European options, as a function of the stock price at time t equal to zero,

and as the time to maturity varies. So, we've got three different times to

maturity T equals 0.05 years, T equals 0.25 years, and T equals 0.5 years.

There are probably two things to notice first.

The first observation is as follows. Note that if I pick any one of these

options, the Vega goes to 0 as the stock price moves away from the strike, which

was $100 here. So, what is going on here.

Well, it's very simple. So, again re-, returning to what we did

in previous modules, we know the following.

Let's take a call option as our example. .

We know that the call option price[BLANK_AUDIO] will be approximately

equal to, and again, ignoring interest rate factors and so on.

It would approximately be equal to S0 minus K, for S0 being very large.

And by very large, I mean much larger than K.

And sufficiently large, that I'm almost certain I'm going to be exercising the

option. It will be equal to zero for S0 being

very small. And very small here means much smaller

than K. And indeed, small enough that the chances

of exercising the option are approximately zero.

Well, we can see here that Delta C, Delta Sigma, therefore, must be equal to 0 in

this situation because the partial derivative of S0 minus K with respect to

sigma is 0. And also 0 down in this situation as

well. And therefore, for S0 very large, which

is up here, or S0 very small, which is down here, we see that Delta C, Delta

Sigma goes to 0. And indeed, that's what we see for each

of these three options. So, that's the first observation.