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1

it depends on asset class. In some classes, the future is more liquid than the underlying eg oil so it makes more sense to work with its volatility. Also, the forward price and the spot price only have the same volatility if we assume deterministic interest rates. For short-dated equity type products, this is reasonable. For long-dated products and ...


1

well you are really asking why is the ATM value so linear in $\sigma. $ If you take a Taylor about $\sigma =0$ when ATM you get the well-known expression $$ \frac{1}{\sqrt{2\pi}} \sigma \sqrt{T} S_t +{\cal O}(\sigma^3 T^{3/2}) $$ which gives the approximate linearity. For details of the derivation see for example my book Concepts etc.


0

Black--Scholes takes log-normal stock price movements whereas your model gives normal price movements.


0

In the BS model, everything is explicit. If your spot increases by $h\%$, the price will increase by $\Delta_{rel,h}\%$ where $$ \Delta_{rel,h} = \frac{C_{BS}(S(1+h),T,K,\sigma)}{C_{BS}(S,T,K,\sigma)} - 1 $$ That's high school math.


0

Try Finite Differences to calculate your Greeks, it will give all the greeks for that specific underlying moviment. In order to back out the dollar change in your pnl just multiply each greek by the amount held in that position.


3

you don't need $ud=1.$ In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have $ud=1.$ All you need is $$ d < e^{r \Delta t} < u $$ The tree recombines provided $u$ and $d$ don't change from step to step. See my book More Mathematical Finance for a comprehensive review and ...


0

These years, The new frontier has been around optimal trading, market impact and orderbook dynamics. They are plenty of sources around, one of them being this bi-yearly conference. You have all the slides on the web site: Market Microstructure: Confronting many Viewpoints. The next challenge is to link previous quant and economic knowledge with ...


-2

http://www.algorithmicfinance.org/ It's a free peer reviewed journal, depending on your definition of discovery you might find interesting tidbits there.


3

Ross had an interesting paper thats making the rounds: The Recovery Theorem. He claims that the physical measure can be recovered from option prices under certain conditions. I think that's getting a lot of academic interest recently.


3

The Asian option is cash settled, so the bank will transfer you $0.5. Delivering the shares and doing some trades is not possible. You can't buy the spot for the average price over a period, you just pay the spot price. Since you're into Asian options, I assume asian option valuation is useful for you to assess whether you're not paying too much.


1

You need to make a distinction between reality and the model you are considering. 1) In your model, the conclusion is valid: in your model holding the stock is equivalent to holding the zero strike call. This is because you make many implicit assumptions (basically these with zero risk free rate). 2) Yes these assumptions i.e. your model are very ...


1

Buying a call at time $t=0$ with strike $K=0$ on a stock whose value is $S_0$ will produce the following cash flows ensure a cash flow at time $t=T$ of $S_T$, because as you mentioned $(S_T - 0, 0 )^{+}=S_T$ because $S_t \geq 0 ~ \forall t$ by definition. This cash flow is replicable by buying the stock for $S_0$ at $t=0$. By the law of one price, when ...


2

Based on the example you gave, it seems that indeed your inputs are inconsistent. The intrinsic call value is $S-e^{-rT}K = 286.52355\dots$, which is higher than the market value, implying that there exists an arbitrage. Instead, one of your inputs is probably wrong. Even if the interest rate is set to $0$, the intrinsic call value is still above your bid, ...


0

That is the answer, $S-e^{-r}K$. It depends on the values $S$ of the security at time $0$, on $K$ and on the interest rate $r$. All of these you can assume know at time $0$, i.e. now.


2

You should always think: I buy the one which is to cheap and sell the one that is too expensive and figure it out. The figuring out in this case is noting that: $C\geq 0$ since it will never cost you money The option is strictly better than $S-K$ so has a higher price. Now to your strategy: You buy $C(T_2)$ (the cheap) and sell $C(T_1)$ (the ...


0

you need to check implied volatility... Calendars must be opened when IV is low, hopping that will increase, that way you will capture the volatility increase.


1

I'll use a European call option as an example, I think you can easily generalize it for a put option. Given underlying $S(t) = S_t$, maturity $T$, strike $K$ and risk-free rate $r$, the price of a call option as time $t$ under the rik-neutral measure $Q$ is $$ \begin{align} C(t, S_t) & = \mathbb{E}^Q \left[ e^{-r(T-t)} \max (S_T - K, 0) \right] \\ ...


-1

When the stockprice follows a GBM, the arbitrage-free value of an EU call is given by the Black Scholes model: \begin{align} C(S, t) &= N(d_1)S_0 - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S_0}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= d_1 - \sigma\sqrt{T - t} ...


2

You could simulate many (100000) 3 day price paths for the stock using the geometric brownian motion. Then for each simulated path, calculate the option value and store them. Then calculate the return difference for each of the calls and order them from smallest to largest. The 5% cutoff is your 3 day 95% VaR.


2

puts can be decreasing in time to maturity. This is why you sometimes early exercise an American put. This tends to happen deeply in the money with large r and zero dividend rate.


1

You can find the solution here: http://www.wiley.com/legacy/wileychi/pwiqf2/supp/c02.pdf For all solutions see my answer here: http://quant.stackexchange.com/a/16061/12


0

The concept of replication is indeed applied to IR products, after all they are also hedged in practice. However, in the equity world we start with the replicating portfolio and then arrive to the pricing formula. In contrast, for IR products we employ a convenient numeraire which helps us to arrive at the pricing formula directly (in a non-constructive and ...


2

I think you need to ask your question differently to get better answers than mine. Your Black Scholes part has two problems. First positive infinity should be negative infinity. Second, you are assuming zero dividends in Black Scholes but you are assuming a possibly positive div yield q in the CEV part. If the div yield q is sufficiently positive in the ...



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