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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, ...


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 ...


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 ...


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.


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 ...


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 ...


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.

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