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08-29-2007, 04:15 PM   #21
shinichi9htv

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ặc ặc 3hoa bốc phét linh tinh thế thôi ạ :P, nhưng bài táo hay phết mấy đại ca cũng thử làm xem

Thực ra bài semimartingale cũng intuitive phết mà. Có thể thấy là cái B_t nó quét hết cả R nên qua cái điểm gián đoạn, cái conditional expectation trước và sau stopping time của điểm gián đoạn khác nhau (tất nhiên nếu muốn chặt chẽ thì phải consider 1 dãy stopping time -> + infty....)

Em đang "en vacances", rỗi hơi nên ngồi post linh tinh. Mấy đại ca có gì bỏ qua cho em ngựa non háu đá nhá
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08-29-2007, 08:14 PM   #22
Khoa Tran

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Quote:
Originally Posted by shinichi9htv View Post
12. (from quantfinancejob.com) trong rổ có 10 quả táo xanh, 20 quả táo vàng và 30 quả táo đỏ. Nhắm mắt lấy táo lần lượt từ trong rổ ra. Tính xác suất để khi lấy ra hết táo xanh thì vẫn còn ít nhất 1 quả táo vàng và 1 quả táo đỏ.
Is it 1/3?
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08-29-2007, 08:35 PM   #23
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hình như chưa đúng rồi. Khoa làm thế nào?
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08-29-2007, 09:13 PM   #24
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Giả sử táo xanh cuối cùng được bốc ở lần thứ x, táo vàng cuối cùng y, và táo đỏ cuối cùng z (1\leq x,y,z\leq 30). There are 6 possibilities: x<y<z, y<z<x, etc. Lấy ra hết táo xanh vẫn còn ít nhất 1 táo vàng và 1 táo đỏ tương đương với x<y<z or x<z<y. By symmetry, xác suất của 2 possibilities out of 6 là 1/3.

Không biết có pitfall nào không?

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08-29-2007, 11:22 PM   #25
shinichi9htv

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sao lại symmetry được, có nhiều táo đỏ nên khả năng lấy hết táo xanh trước táo đỏ nhiều hơn mà (ví dụ thế)
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08-30-2007, 03:12 AM   #26
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Quote:
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sao lại symmetry được, có nhiều táo đỏ nên khả năng lấy hết táo xanh trước táo đỏ nhiều hơn mà (ví dụ thế)
Arrgh, my damn mistake from the beginning. Somehow I assumed that there were 10 each color.

But then we can use the same approach. We split the yellow ones into 2 groups and the red ones into 3 groups with the last one in each group respectively numbered y1, y2, z1, z2, z3 to obtain symmetry.

Now, there are 6!=72 possibilities in total and a few special ones that we can count.

Saying another way, the problem can be scaled down to 1 green apple + 2 yellow apples + 3 red apples.

Last edited by Khoa Tran; 08-30-2007 at 07:05 AM.
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08-30-2007, 05:00 AM   #27
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let's consider the problem of 13 green apples, 37 yellow apples and 41 red apples :-?
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08-30-2007, 10:51 AM   #28
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Quote:
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10. (ML) T_1 < T_2. Pricing forward-start option (cliquet?) E(\frac{S_{T_2}}{S_{T_1}}-K)^{+}
Assuming log-normal stock price and deterministic volatility during [T1,T2] ("Deterministic" means that the vol is a given function of time in the interval [T1, T2] but not changing as time goes by!], the cliquet price is the call option price with moneyness K. [Too lazy to write down the details, though; too late in the night. The calculations are straightforward.]

How to hedge it? Simply take the full premium to buy share(s) at S_0. Wait until T1, sell the whole amount of shares at S_1 then buy the call option(s) struck at moneyness K. You will be then fully covered.

The challenge is with smile and/or time-dependent volatility.

Why are cliquets interesting? I was once told (sigh, never seen or carried out the proof myself) that every contingent claim is (statically) hedgeable by using the full set of cliquets (for every strike and every expiry date T1 and maturity date T2) together with the full set of vanillas (again, at every strike and every expiry date).

In other words, IF this statement holds true, cliquets and vanillas will complete the market. This is where their beauty is. [Bruno Dupire's dream would be fulfilled! The entire arduous assumptions about stock price dynamics would be gone for good...]

[I have some doubt about its validity, though... But I might be wrong.]
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08-30-2007, 04:39 PM   #29
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Carr formula shows us how to replicate every contingent claim statitically by a set of fwd, out the money calls and puts:
<br />
E(H(F_T)) = H(F_0) + \int_{0}^{F_0}H''(K)(F_T-K)^{+}dK + \int_{F_0}^{\infty}H''(K)(K-F_T)^{+}dK<br />
(smt like that, perhaps you could find an exact formula with "static hedging")

I think the cliquets give us a "forward smile" - a prediction of smile.
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08-30-2007, 04:44 PM   #30
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Well, every european contingent claim (convex payoff, whatever...) can be replicated by a bunch of calls (or puts), now if you throw in the cliquets (a call is nothing but a cliquet with T_1 = 0), you can cover all path-dependent options.
Regarding the hedge, does vol need to be deterministic ? I think this semi-static hedge always works no matter what the vol is.
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