44
Here couple pointers that may make it clearer:
Drift can be replaced by the risk-free rate through a mathematical construct called risk-neutral probability pricing.
Why can we get away with that without introducing errors? The reason lies in the ability to setup a hedge portfolio, thus the market will not compensate us for the drift above and beyond the ...
41
The risk-neutral measure $\mathbb{Q}$ is a mathematical construct which stems from the law of one price, also known as the principle of no riskless arbitrage and which you may already have heard of in the following terms: "there is no free lunch in financial markets".
This law is at the heart of securities' relative valuation, see this very nice paper by ...
31
Being on the sell side and selling options you can intuitively think of it like this:
An option is like any other product that is being produced out of ingredients and because of the competitive situation of the producer this is done by the cheapest possible production process.
The ingredients are in a simple (Black Scholes) setting a stock and and a risk ...
24
Life Without a Risk-Neutral Measure
How would we price assets without the measure $\mathbb Q$? Well, we would start with some version of the Euler equation $P_t=\mathbb{E}_t[M_{t+1}P_{t+1}]$, where $M$ is the stochastic discount factor (SDF). This equation holds under very weak assumptions (law of one price) and uses real-world probabilities. So, we take the ...
14
You can price an asset paying $X_{t+1}$ in two ways:
$$P_t=\frac{1}{R_f}\sum_{\omega} Q(\omega)X_{t+1}(\omega)$$
$$P_t=\sum_{\omega} P(\omega)M_{t+1}(\omega)X_{t+1}(\omega)$$
As you can see, the price is making a joint statement (i.e. you can recover $Q(\omega)$) regarding both the probability of an event $P(\omega)$ and how much people dislike that event, i....
13
This is a loaded question. Ross' recovery theorem has both flaws and insights. The single answer thus far did a great job of addressing the flaws from an economics perspective. No one questions that the math is wrong: it is correct. Here is a mathematical insight from Ross' work. Abstracting from the finance and economics, the purely probabilistic content of ...
13
I'll outline how you can estimate the (implied) real-world density function from (observed) option prices. Having found this real-world density, you can then compute all sorts of probabilities and quantify the market's expectation of future prices.
Recall firstly that (European-style) options are priced as risk-neutral expectation of the discounted payoff. ...
12
In the derivatives context, "arbitrage free" means almost surely for the probability measure under consideration. This is in opposition with statistical arbitrage used at high frequencies for example.
More precisely the assumption is that there is no $T\geq 0$ and self-financed portfolio $V$ such that $V_0 = 0$, $P(V_T < 0) = 0$ and $P(V_T > 0) > ...
12
In order to estimate a probability of an event with small probability, you might want to try to estimate the probability for a changed random variable that allocates a larger probability mass for the event happening. So, in your case, you might want to change the original $N(0, 1)$ to $N(100, 1)$ because for the second r.v. the probability of it being higher ...
12
There is a deeper issue. Frequentist distributions are not probability distributions because they are designed to be minimax distributions rather than actual distributions. This ignores all of the other problems and this also ignores risk-neutral versus any other measure of risk aversion.
An even deeper issue is that these models presume that the ...
12
Recall that any traded asset divided by a numéraire is a martingale under the measure associated to that numéraire. For the 3 interest rates you mention, the natural measure (namely the one that makes those processes martingales) is deduced from the structure of the rate.
Always keep in mind that the value at $t_0$ of a cash flow $C$ paid at $T$ is equal to ...
11
You're missing the point of the risk-neutral framework.
The idea is as follows: assume the real probability measure called $\mathbb{P}$. The thing is, because investors are not risk-neutral, you cannot write that $v_0 = E_\mathbb{P} [ e^{-rT} V_T]$.
Using the Fundamental Theorem of Asset Pricing, you know that if the market is arbitrage-free, then there ...
11
A stochastic volatility model for a single risky asset can't be complete because you have two sources of randomness. But you can easily make it complete by adding a derivative whose value depends on the volatility. For example, if you add a variance swap in the Heston model then it becomes complete. This allows you to calibrate the model.
But your ...
11
Assume a constant risk-free rate $r$ and no dividends. Generalisation is straightforward.
To preclude arbitrage opportunities, under the risk-neutral measure $\Bbb{Q}$, the discounted asset price process should be a $\Bbb{Q}$-martingale i.e.
$$ S_0 = \Bbb{E}^\Bbb{Q}_0 \left[ e^{-rt} S_t \right] \iff \Bbb{E}^\Bbb{Q}_0 \left[ S_t \right] = S_0 \exp(rt) \...
11
Q: What does the risk-neutral price represent if the option is not replicable?
In an incomplete market, there is no unique martingale measure but instead a set $Q$ of equivalent martingale measures. Consequently, there is an interval of arbitrage-free prices:
$ \Big( inf_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX], sup_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX] \Big)...
11
$\mathbb{P}$ is the true probability measure. Measure $\mathbb{Q}$ is a measure of convenience that allows risk neutral pricing. Stochastic discount factor $M$ takes you between the two.
If you care about prices you can either: (1) work under $\mathbb{Q}$ or (2) work under $\mathbb{P}$ with a stochastic discount factor $M$. There's an isomorphic ...
11
The risk neutral probability measure $Q$ is the true probability measure $P$ times the stochastic discount factor $M$ but rescaled so $Q$ sums to 1.
Simple derivation
For maximum simplicity, I'll work in a discrete probability space with $n$ possible outcomes. Everything goes through under measure theory in more general, infinite number of outcome ...
10
It depends on the purpose of your simulation.
If you want to model the asset price path for pricing some derivative then you need the risk-neutral measure (thus you take the risk-less rate as drift).
Why? Because the risk-neutral measure makes your pricing compatible with the pricing of other contracts in the market. It makes the prices consistent.
If ...
10
$$dS / S = \mu dt + \sigma dW \\
\\ dS / S -r dt= \mu dt - rdt + \sigma dW \\
\\ dS / S -r dt= [\frac{(\mu - r)}{\sigma}dt + dW]\sigma \\$$
Then, Girsanov tells us that, as long as the risk premium is bounded from below, we can write
$[\frac{(\mu - r)}{\sigma}dt + dW]\sigma$ as $\sigma d\tilde{W}$ where $\tilde{W}$ is simply another brownian motion with ...
10
Recently I came across an interesting intuitive explanation:
Suppose driftless market. Market price is 105, strike price is 100. Call option costs 8, put option 3. (intrinsic value of call is 5, time value of both is 3)
Now the market starts drifting upwards massively. You say, that you would probably price call higher, e.g. at 10. Would you also price put ...
10
Very simply, Ross' framework assumes a great deal to extract the true pricing kernel. Time homogeneity, additively separable state dependent utility, (discrete time Markovian structure - though these have been relaxed.) In particular, there are two schools of criticism, one is that time homogeneity makes little sense in the real market. In fact, the Recovery ...
10
The short answer is:
As long as a derivative can be perfectly replicated via hedging in the underlying asset then the price of the derivative should be independent of investors' risk aversion and hence the application of risk-neutral probabilities and discounting of the future expected payoff under risk neutral probability leads to the same price of the ...
10
this is probably the most asked question in quantitative finance... There are many answers. One nice example to consider is what if the calls were struck at zero.
The call then pays the stock price at time $T$ and so it's value today must the stock price today since we can replicate by holding one unit of stock. This will be true regardless of the drift of ...
10
Your butterfly is short a straddle and long a strangle. If you add a long straddle with the same strike/notional you are now just long a strangle.
The payoff for a strangle is zero if the terminal price is between the two strikes and positive otherwise. Once you take the premium into account you will see that you make a loss if the terminal price is between (...
9
"Intuitively, everything else being equal, if a stock has higher drift, shouldn't it have higher probability of finishing in-the-money (and higher probability of having higher payoff), and the call option should be worth more?"
All these other answers are focusing on the wrong aspect of the question - it is true that the maths makes the drift drop out from ...
9
You may want to consider splitting two important, yet very different concepts:
Pricing a derivative security with contingent payoff and forecasting an asset.
Pricing a derivative can be achieved through setting up a hedge portfolio and track its evolution and "value" at any point in time before the derivative security pays off. Risk-neutral pricing is a ...
9
Risk-neutrality isn't really a property of a model. Instead, it describes a certain calibration of a model (almost always represented by an SDE).
We say a model has been calibrated to risk-neutral probabilities if
model parameters can be inferred from traded security prices, and
there's some anti-arbitrage assumption and hedging scheme available for those ...
9
It is a very interesting question. There is a brief explanation in the book Martingale methods in financial modelling. Basically, it says that, the interest short rate $r_t$ can be modeled in any martingale measure $Q$, however, as long as the zero-coupon bond price $P(t, T)$ is defined by
\begin{align*}
P(t, T) = E^{Q}\Big(e^{-\int_t^T r_s ds}
\mid \...
9
The risk-neutral probability density function $q(.)$ is indeed given by
$$ q(S_T=s) = \frac{1}{P(0,T)} \frac{ \partial^2 C }{\partial K^2} (K=s,T) $$
where $P(0,T)$ figures the relevant discount factor. This is known as the Breeden-Litzenberger identity.
Because you do not observe a continuum of call prices in practice, you can use a finite difference ...
9
The result you're looking for is
$$ \left. \frac{d\Bbb{P}}{d\Bbb{Q}}\right\vert_{\mathcal{F}_t} = \left( \left. \frac{d\Bbb{Q}}{d\Bbb{P}}\right\vert_{\mathcal{F}_t} \right)^{-1} $$
This is a result from measure theory but since you mention it, let's see how we can show it based on Girsanov theorem.
Starting from the definitions you provide and introducing ...
Only top voted, non community-wiki answers of a minimum length are eligible
