Are Synthetic Options Truly Equivalent Options?

Here are some simple graphical illustrations provided by Cohen in "The Bible of Options Strategies", publsihed 2005 by FT Prentice Hall.

A long call (page 7) mostly shows positive Rho values (indicating that "higher interest rates would increase the value of the calls and therefore help the position; whereas a synthetic call (page 248) mostly shows negative Rho values (indicating that "higher interest rates are generally unhelpful to the position").

A long put (page 15) mostly shows negative Rho values; whereas a synthetic put (page 252) mostly shows positive Rho values.
 
Quote from OddTrader:

Here are some simple graphical illustrations provided by Cohen in "The Bible of Options Strategies", publsihed 2005 by FT Prentice Hall.

A long call (page 7) mostly shows positive Rho values (indicating that "higher interest rates would increase the value of the calls and therefore help the position; whereas a synthetic call (page 248) mostly shows negative Rho values (indicating that "higher interest rates are generally unhelpful to the position").

A long put (page 15) mostly shows negative Rho values; whereas a synthetic put (page 252) mostly shows positive Rho values.
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OddTrader, you're odd!

Why, oh why, do you have to keep posting these quotes? Why do you have to bring 'rho' into the discussion? What does rho have to do with anything? Please, I implore you, STOP, for goodness sake!
 
Here are my observations:

Synthetic long call/put options are not exactly equivalent of simple long call/put options.

Appears to me that they have nothing specially to do with whether they are American or European options.

I'm done with this thread now. Bye! :)
 
I would save about $100k in margin requirements this month, if I swapped all of my synthetic put for actual put positions.

But that's not by design.

That's only because my soon to be former FCM (RCG) has an incompetent, uninformed risk department.
 
Quote from OddTrader:

According to Weert:

Q

Since both the assumption c-p>S-Ke^-r(T-t) as well as c-p < S-Ke^-r(T-t) lead to arbitrage possibility, the put-call parity, c-p=S-Ke^-r(T-t), has to hold.
...


UQ
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Typo correction.
 
Quote from Martinghoul:

Why do you have to bring 'rho' into the discussion? What does rho have to do with anything?
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May I ask you one question:
Can you actually Read properly, other than you know how to write whatever you like to post? :p

When evaluating two supposedly equivalent options structures:
Why do you Not to bring 'rho' into the discussion? What does Not rho have to do with anything (within this context)? :D

Bye again, anyway!
 
Quote from OddTrader:

May I ask you one question:
Can you actually Read properly, other than you know how to write whatever you like to post? :p

When evaluating two supposedly equivalent options structures:
Why do you Not to bring 'rho' into the discussion? What does Not rho have to do with anything (within this context)? :D

Bye again, anyway!
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A parting shot, I see :)...

I can read properly, amico. It's just that it's my view that we have established, once and for all, the answer to your original put-call parity question. The answer will not change, no matter how many quotes from pundits/textbooks you post.

Specifically, in this case, rho actually doesn't matter in the slightest. I can explain why if you like, but we've agreed that this thread is over, so let's just leave it at that, shall we?
 
Quote from OddTrader:

A long call (page 7) mostly shows positive Rho values (indicating that "higher interest rates would increase the value of the calls and therefore help the position; whereas a synthetic call (page 248) mostly shows negative Rho values (indicating that "higher interest rates are generally unhelpful to the position").

A long put (page 15) mostly shows negative Rho values; whereas a synthetic put (page 252) mostly shows positive Rho values.
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I give you a grade of A+ for OBFUSCATION and a grade of F for ACCURACY and RELEVANCE.

Whatever you gain (or save) on the options due to interest rate change (rho) is offset by the change in carry cost on the underlying.
 
Quote from Martinghoul:

I can read properly, amico. It's just that it's my view that we have established, once and for all, the answer to your original put-call parity question. The answer will not change, no matter how many quotes from pundits/textbooks you post.

Specifically, in this case, rho actually doesn't matter in the slightest. I can explain why if you like, but we've agreed that this thread is over, so let's just leave it at that, shall we?
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I'm actually curious as to whether the author got it wrong or whether Oddball had no clue about what he was reading

:)
 
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