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Apr
12
accepted A few questions about signs of the Greek letters
Apr
11
comment A few questions about signs of the Greek letters
I think what you meant is that under risk neutral probability, $F_T=E(S_T)=S_0{e}^{rT}$ is increasing in $r$. There's no dispute about that. But notice the ${e}^{-rT}$ term in $C={e}^{-rT}E[max(S_T-K,0)]$. So even if you can conclude that $E[max(S_T-K,0)]$ is increasing in $r$, can you really conclude ${e}^{-rT}E[max(S_T-K,0)]$ is increasing in $r$ from that?
Apr
11
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Apr
11
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Apr
7
revised A few questions about signs of the Greek letters
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Apr
7
revised A few questions about signs of the Greek letters
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Apr
7
comment A few questions about signs of the Greek letters
@Owe: I was not claiming that $C$ can go up or down (it can only go up in the BS model) following an increase in $r_f$. I meant when $r_f$ increases, there are two contradicting forces on the movement of $C$. But in the BS model the latter force always prevails. I want to know why, and I want to know whether it holds true for any distributions of prices of the underlying.
Apr
7
revised A few questions about signs of the Greek letters
Deleted the third question, because it doesn't make sense
Apr
6
comment A few questions about signs of the Greek letters
I understand your point. But as I emphasized in the question, this is not the whole story. As $r$ increases, present value of future payoff also gets discounted more. Why is the gain from the bond always outweighs the loss from discounted payoff?
Apr
6
comment A few questions about signs of the Greek letters
By put-call parity $C=S-K/{e}^{rT}+P$. But the value of the call is always equal to that of the leverage on the RHS. Why is buying the call more attractive? And what do you mean by "not paying for the underlying until a later date"?
Apr
6
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Apr
6
revised A few questions about signs of the Greek letters
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Apr
6
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Apr
6
asked A few questions about signs of the Greek letters