has Kim left the building?
- Thread starter traderlux
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You got it all wrong.. I'm very grateful for any feedback or advice. Unlike other "gurus", I fully admit that I'm still learning every day (even after 15 years in the market).
That doesn't mean that I have to agree with 100% of what you say.
Thank you again.
That doesn't mean that I have to agree with 100% of what you say.
Thank you again.
Even after 15 years in the market you're "still learning"
...that basically... means you're not profitable, or Consistently or fruitfully profitable. -- and are searching for your place and/or formula in the market.
Lost, wandering soul.
But this is 2018...Make Trading Great Again...High-Five`
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Even after 15 years in the market you're "still learning"
...that basically... means you're not profitable, or Consistently or fruitfully profitable. -- and are searching for your place and/or formula in the market.
Lost, wandering soul.
But this is 2018...Make Trading Great Again...High-Five`
No. That means that I'm always looking to improve the strategies I'm using and fine tune them, and also to learn new strategies to diversify my trading arsenal.
That means that I'm always looking to become a better trader, and this will never stop.
As for my profitability - you can see all our history on our performance page. 1,000+ trades going back to 2011.
Y
Yes.
Thank you for your coaching sir.
Do you remember the theme to the Batman series??
Gamma gamma gamma gamma
Gamma gamma gamma gamma
Gamm-maaaaaa!
Hi JackRab, I started options trading only a few months ago, and I was not majored in finance or relative fields, but I am doing OK with high school math... Could you help me understand, analytically or quantitatively, why ITM strangle [+70C/+110P] is identical to OTM strangle [+70P/+110C]?
Part One:
Suppose we have strikes K1=70 & K2=110, underlying price S0=100, 60 days until maturity T=60, risk-free interest rate r, dividends present value D. Let C1 P1 C2 P2 be the price of 70C 70P 110C 110P respectively. And let us assume, for simplicity, that it is an European option. According to put-call parity on the two strikes, we have:
Eq1: C1 + D + K1 exp(-r T) = P1 + S0
Eq2: C2 + D + K2 exp(-r T) = P2 + S0
Subtracting the two yields:
Eq3: (C2 + P1) + (K2 - K1) exp(-r T) = (P2 + C1)
that is,
Eq4: Value[+70P/+110C] + 40exp(-60r) = Value[+70C/+110P]
I see the difference is 40exp(-60r), the interest rate component. Is the derivation above correct?
You said "the interest rate ... doesn't make a difference. Because both dividend and high lending fee cause the intrinsic value to drop, not the extrinsic. And by intrinsic value is meant the difference between strike price and underlying stock price including the interest rate component." I have some hard time understanding this part. Could you elaborate, analytically or quantitatively, on this a little bit?
Part Two:
Let's relax the assumption about European options; let it be an American option, with the possibility of early assignments. Put-call parity again:
Eq5: S0 - K1 <= C1 - P1 + D <= S0 - K1 exp(-r T)
Eq6: S0 - K2 <= C2 - P2 + D <= S0 - K2 exp(-r T)
Subtracting Eq6 from Eq5 yields:
Eq7: K2 exp(-r T) - K1 <= (C1 + P2) - (C2 + P1) <= K2 - K1 exp(-r T)
that is, Value[+70C/+110P] - Value[+70P/+110C] is bounded by range
[-K1 + K2 exp(-r T), K2 - K1 exp(-r T)]
or [-70 + 110exp(-60r), 110 - 70exp(-110r)]
If the above derivations correct? If so, is this range tight enough that you deem insignificant?
Sorry I am only savvy on high school math. Please correct me if anything is wrong...
The ATM options is really where the value of the future is... that means when there is no dividend or interest... the ATM is spot level.
Say spot = 100, r=0.
ATM strike = 100
both 100-call and 100-put only have extrinsic value...
If 100-call = 4.50... put call parity gives 100-put = 4.50
If 105-call = 2.50... 105 put = 7.50 (intrinsic = 5, extrinsic = same as the call)
Now you add interest... say the interest component = 1%
That means the ATM strike is 101....
Now, the 100-call is 1.00 ITM and the 100-put is 1.00 OTM.... and the ATM is the 101 strike. Both 101-call and -put have only extrinsic value.
If 101-call = 4.50, the 101-put = 4.50
If 100 call = 5.00, the 100 put = 4.00 (extrinsic is the same for both call and put).
So interest/dividend causes the In-The-Moneyness to shift...
Vega/Gamma is the same for both call and put with the same strike/expiry. Theta is different, but that's only because the interest is +/+ for the call and -/- for the put. You see, if the spot stays at 100 until expiry, the call loses 5 and the put 4... so there's 1.00 more theta in the call than the put. However, that's a financing thing.
If you buy the 100-put and hedge with delta 100, you basically have the synthetic 100-call. Now, you're theta is the lower one of the two.. since you bought the put... however the difference gets now paid in interest of 1.00, since you've bought the stock... (or lack of receiving interest, basically an opportunity loss, you could've put that money in interest bearing account and receive the 1.00 in interest). So in the end... you still pay the same amount of theta... only it's now split into interest paying.
Because of put/call-parity... put=call... which means they both have the same characteristics when hedged. That's how market makers hedge their risks... the close position risks by creating synthetic futures out of a call or a put and then hedging with 100 delta in the underlying makes basically a flat position.