Tag Info

Hot answers tagged

6

the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work. The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value. This is a well-known problem. As the bump size goes to zero, the ...


3

If you're asking what the FX Outright for 1M EUR/PLN is, given that table, then yes the answer is just outright = spot + fwd points, which is 3.4550 + 0.0079 = 3.4629 (you had the wrong column for your 1M value). Usually fwd points are quoted directly (i.e. not as an outright), using a divisor set by market convention. I expect EUR/PLN divisor to be 10,000, ...


3

Options have an asymmetric payoff profile: The payoffs are zero for almost all cases and positive else (as we well know). If the option is OTM, most of its payoffs are zero. A rise in volatility will hence increase the likelihood for instead positive payoffs from a change in the underlying price (i.e. delta increases). If the option is already ITM, ...


3

The time to expiry is required, but it's included in the inputs: the two discounts $e^{-rT}$ and $e^{-qT}$ and the standard deviation $\sigma\sqrt{T}$. You might argue it could be documented more clearly, and I might agree with you.


2

Given that by delta means that if the price goes up by 0.01% i.e. one basis point, you gain 15 and vice versa if the price goes down by one basis point. You know that the daily standard deviation is 2.2%, than again you know that $ 220*15 = 3300$ is the standard deviation of your portfolio. So, since we are using a normal distribution you can look at a table ...


1

The "right" thing to do is to treat the options as derivative contracts. Let's say for simplicity that you are using Monte Carlo to compute VaR. Then you would simulate the equity prices on each iteration, and then apply an option-pricing formula to get the corresponding option prices on that iteration. This lets you obtain an accurate simulated portfolio ...


1

The Black-Scholes delta: $$\partial_SC=N\left(\dfrac{\ln\left(\frac{S_0}{K}\right) +(r - q + \frac{1}{2}\sigma^2)(T - t)}{\sigma\sqrt{T - t}}\right)$$ As you can see this delta would go to$1$ if $\sigma\to\infty$ (and $t<T$).


1

It's a combination of too few sample paths and/or too small an increment. Your estimation error on the price is magnified by the $dS^2$. Try using a larger sample or a larger increment. Alternatively, you can use a multiplier instead of a fixed increment; in my experience, it usually yields better results.


1

http://www.theoptionsguide.com/in-the-money-covered-calls.aspx Look at the chart on this site. Do you see how the line is upward sloping for a covered call until the stock reaches a price of 45? To simplify things, you have a delta of 1 until 45 and then a delta of 0 after 45 onwards. Delta is the slope of the line. Delta of a covered call changes depending ...


1

First, I think you made a mistake in your computations above. Where you wrote $(30-20)$, I think you really meant $(30-(-20))$ i.e. $30+20$, yielding a gamma P&L of $1000$ instead of $200$. Your total P&L over $[90,170]$ would then be $110$ instead of $-690$. It doesn't matter for my answer either way, just thought I'd point it out for confused ...


1

When pricing FX options, the underlying is the spot or forward exchange rate. The foreign currency is analogous to a stock where the owner of the foreign currency receives a "dividend yield" equal to the risk-free rate in the foreign currency.


1

I am assuming you are talking about probability of becoming at least $\Delta = 30$, otherwise probability is zero. Hard to give a complete answer as quite some information is missing. As you are seeking for the probability, the outcome definitely depends on which model of underlying you are using. Moreover, even if you are using the BS model, some parameter ...


1

I've started thinking about this, too. My gedanken conclusion turned out to be too simple once I found what I was after: http://www.investment-and-finance.net/derivatives/o/option-beta.html, which I've confirmed in Black & Scholes (1973) p10 (eq 15). In short: $$ \beta_{\text{option}} = \frac{S\cdot\Delta}{O}{\beta_S} $$ where $S$ is the underlying ...


1

For Black-Scholes, $\Delta_C=\partial_{S} C=N(d_1)$, $d_1= \frac{\ln\left(\frac{S_t}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)}{\sigma\sqrt{T - t}}$ You may fit the volatility $\sigma$ to this term by $$\Delta_C({\hat{\sigma}})=0.25$$Note that $\Delta_P=1-\Delta_C$ by Put-Call-Parity.



Only top voted, non community-wiki answers of a minimum length are eligible