Tag Info

New answers tagged

4

The time-$t$ price of a European call on a non-dividend paying stock with spot price $S_t$, when the strike is $K$ and the time to maturity is $\tau = T − t$, is the discounted expected value of the payoff under the risk-neutral measure $Q$ $$C(t,{{S}_{t}},K,T)={{e}^{-r(T-t)}}\mathbb{E}_{t}^{Q}\,[{{({{S}_{T}}-K)}^{+}}]={{e}^{-r\tau ...


3

Note that, for the first term, it is \begin{align*} e^{-rT} \mathbb{E}(S_T \mathbb{I}_{S_T >X}), \end{align*} which is not equal to $e^{-rT}S\,\mathbb{P}(S_T >X)$. Here, $\mathbb{P}$ is the risk-neutral measure and $\mathbb{E}$ is the corresponding expectation operator. Let $\tilde{\mathbb{P}}$ be the probability measure with the stock price process ...


1

Take a look at Hull's Appendix of the Volatility Smiles chapter. (Chapter 16 in my version). It gives a method to calculate the probability density function based on option prices: $$ g(K) = e^{rT} \frac{\partial ^2 c}{\partial K^2} $$ This result comes from the Breeden Litzenberger 1978 paper.


0

How to use the stock as Numeraire: $$\mathbb{\tilde{E}}[e^{-rT}(S_T-K)^+]=\mathbb{\tilde{E}}\left[e^{-rT}S_T\left(1-\frac{K}{S_T}\right)^+\right]$$ $$=S_0\mathbb{\tilde{E}}\left[\frac{e^{-rT}S_T}{S_0} \left(1-\frac{K}{S_T}\right)^+\right]$$ $$=S_0\mathbb{\hat{E}}\left[\left(1-\frac{K}{S_T}\right)^+\right]$$ Where under $\mathbb{\hat{P}}$ the stock follows ...


0

Risk-neutral is just one of the many possible measures. It's the most common because we can discount an asset by the risk-free rate under this measure. Of course, we can use any other measure, such as pricing under the stock measure. The mathematics will be very similar, you'll still try to form a martingale under the measure. In @Farahvartish's answer, you ...


1

If $\{N_t\}_t$ be a numerair,and $S_t$ any asset price process, then there exists a measure $Q^S$ which is equivalent to the risk-neutral measure $Q$ such that process $$\frac{S_t}{N_t}$$ is a martingale under $Q^S$ and Derivative pricing under the new measure $Q^S$ is similar as risk-neutral pricing. In particular, the time-t price of a derivative that ...


1

If $\mu$ is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for ...


0

First your equation for returns is false. Forgetting about the jump, it does not reduce to Gbm returns. The variance term from Ito's formula is missing. In the case of a jump, a similar term should appear. Secondly, the distribution is obviously not "what market things are the real probabilities": you chose to impose specific sizes for the jump, the market ...


0

See this resource at github , it uses different methods to calculate IV programmatically


2

It is. Note that the interval I is open on both ends. Moreover, \begin{align*} C_{BS}(\sigma) = S\Phi(d_1)-Ke^{-rT}\Phi(d_2), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable, and \begin{align*} d_{1,2} = \frac{\ln\frac{S}{K e^{-rT}} \pm \frac{1}{2}\sigma^2T}{\sigma \sqrt{T}}. \end{align*} Note that ...


1

For the terminal distributions, I don't have the closed-form solution to hand, but it's computable, since we can price power options (with payoffs like $(S_T^n-K)^+$). You need to find $$ E[S_T C_{K,T}] = \int_K^\infty x(x-K) \cdot p_{BS}(x) dx \\=-Ke^{(r-q)T} C_{K,T} + \int_K^\infty x^2 \cdot p_{BS}(x) dx $$ The latter formula is just a power-option ...



Top 50 recent answers are included