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answer a financial math question
why would a stock have value if it doesn't pay dividends
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Partial ownership/etc?
JM |
Oh I figured it out.
from Pt = (K - St)*D(T-t) + Ct D(T-t) = 1 => Pt > K - St = return from exercising option => you will never exercise option => European and American options are identical |
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stock = stock option?
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Obviously the price of an American option is at least the price of the European option. The point is that if the interest rate is 0 and there are no dividends, it turns out that you never exercise the American option before the expiration so they are the same.
No offense, but I'm pretty sure this math is beyond your training and you couldn't have actually helped me. |
I know the difference between an option and what it was the option of (I do some agri products' futures), its just that the OP didn't mention option actually it did, I missed it.
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That would depend if it's a call or put option.
Regardless, someone might sell a call option based purely on the timing of their investments rather than the movement of the stock, forcing them to exercise prior to the expiration date. |
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Obviously, several of these assumptions aren't actually true. They are useful mathematical constructs. |
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This post is utterly ridiculous. I suggest you read a basic financial math book. Real investors, if they wish to take a market view, will resell the option for >= the intrinsic value in the market (strictly greater if implied volatility is anything other than 0). If the price was ever less than the intrinsic value then there exists a simple arbitrage. Exercising early is always suboptimal, modulo transaction costs and some minor difficulties like shorting at a retail level. |
That's physician.
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Ok i've now ADBLOCKED ben kenobi to avoid any temptation to view his posts
jesus christ I don't understand how someone so stupid can even be literate adblock http://www.discussworldissues.com/im...ons/icon14.gif |
Two years? Did he get an associate's degree or something?
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This is like watching two paraplegics run a marathon.
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A long uniform slab of with mass M is sliding on a "frictionless" horizontal surface with (constant) speed S before a small, dense block with mass 2M is placed on top of the front of it by a person nearby. The small block has an initial velocity of ZERO with respect to the fixed horizontal surface so it immediately starts skidding along the surface of the moving slab. The coefficient of kinetic friction between the block and slab is Mu . As the block skids the slab slows down until the velocities of both objects with respect to the fixed surface are identical; after which they move together. ( The block does NOT reach the back of the slab and fall off!)
a) Develop an expression for the time it takes the block and slab to reach the same velocity. Express your answer in terms of g, Mu, and S. b) Find an expression for the final speed of both objects with respect to the fixed surface. c) Find the magnitude of the frictional force between the objects once the slab and block are both gliding with the same speed. |
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I think this is to be expected. Sorry http://www.discussworldissues.com/fo...ies/tongue.gif At least I'm not like ben--I don't think I'm a physicist. |
It's not even a very hard problem...basic mechanics
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