Black Scholes Formula and continuous compounding
$begingroup$
So i read that
In the black scholes formula, the term Ke^-rt does a 'backward' calculation..
like if the strike price is 500 dollars, to be exercised t years from now, at r%, then this term calculates what the value is today (the current price)
I know that this is derived from continuous compounding, but why? like why is the price of the stock said to increase like that?
finance
asked Jan 8 at 16:08
VanessaVanessa
727
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add a comment |
$begingroup$
So i read that
In the black scholes formula, the term Ke^-rt does a 'backward' calculation..
like if the strike price is 500 dollars, to be exercised t years from now, at r%, then this term calculates what the value is today (the current price)
I know that this is derived from continuous compounding, but why? like why is the price of the stock said to increase like that?
finance
asked Jan 8 at 16:08
VanessaVanessa
727
$endgroup$
add a comment |
$begingroup$
So i read that
In the black scholes formula, the term Ke^-rt does a 'backward' calculation..
like if the strike price is 500 dollars, to be exercised t years from now, at r%, then this term calculates what the value is today (the current price)
I know that this is derived from continuous compounding, but why? like why is the price of the stock said to increase like that?
finance
asked Jan 8 at 16:08
VanessaVanessa
727
$endgroup$
So i read that
In the black scholes formula, the term Ke^-rt does a 'backward' calculation..
like if the strike price is 500 dollars, to be exercised t years from now, at r%, then this term calculates what the value is today (the current price)
I know that this is derived from continuous compounding, but why? like why is the price of the stock said to increase like that?
finance
finance
asked Jan 8 at 16:08
VanessaVanessa
727
asked Jan 8 at 16:08
VanessaVanessa
727
asked Jan 8 at 16:08
VanessaVanessa
727
asked Jan 8 at 16:08
VanessaVanessa
727
asked Jan 8 at 16:08
VanessaVanessa
727
727
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
It's actually a pricing model for options, not shares (which is what stock markets trade). But either way, the reason compounding may be taken as continuous is because prices respond to supply and demand very rapidly. Suppose in the split-second $dt$ the price takes to update, it multiplies by $1+r dt$. Then in a time $t$, this becomes $(1+r dt)^{t/dt}approx e^{rt}$ (for backward rates, just change the sign of $r$).
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
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1 Answer
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1 Answer
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$begingroup$
It's actually a pricing model for options, not shares (which is what stock markets trade). But either way, the reason compounding may be taken as continuous is because prices respond to supply and demand very rapidly. Suppose in the split-second $dt$ the price takes to update, it multiplies by $1+r dt$. Then in a time $t$, this becomes $(1+r dt)^{t/dt}approx e^{rt}$ (for backward rates, just change the sign of $r$).
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
$endgroup$
add a comment |
$begingroup$
It's actually a pricing model for options, not shares (which is what stock markets trade). But either way, the reason compounding may be taken as continuous is because prices respond to supply and demand very rapidly. Suppose in the split-second $dt$ the price takes to update, it multiplies by $1+r dt$. Then in a time $t$, this becomes $(1+r dt)^{t/dt}approx e^{rt}$ (for backward rates, just change the sign of $r$).
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
$endgroup$
add a comment |
$begingroup$
It's actually a pricing model for options, not shares (which is what stock markets trade). But either way, the reason compounding may be taken as continuous is because prices respond to supply and demand very rapidly. Suppose in the split-second $dt$ the price takes to update, it multiplies by $1+r dt$. Then in a time $t$, this becomes $(1+r dt)^{t/dt}approx e^{rt}$ (for backward rates, just change the sign of $r$).
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
$endgroup$
It's actually a pricing model for options, not shares (which is what stock markets trade). But either way, the reason compounding may be taken as continuous is because prices respond to supply and demand very rapidly. Suppose in the split-second $dt$ the price takes to update, it multiplies by $1+r dt$. Then in a time $t$, this becomes $(1+r dt)^{t/dt}approx e^{rt}$ (for backward rates, just change the sign of $r$).
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
answered Jan 8 at 16:14
J.G.J.G.
27.7k22843
27.7k22843
add a comment |
add a comment |
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