Risk free rate vs call option price
- Thread starter nsivakr
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Hi nsivakr, a way to look at it is, a higher risk-free rate decreases the PV of the (fixed) exercise price. This is found in the minimum value of the option, which is the value of the option if the asset were to grow at the risk free rate.
Minimum value call = S(0) - K*exp(-rT), which is the BSM "stripped" of its N(.) functions that essentially capture the volatility: S(0)*N(d1) - K*exp(-rT)*N(d2). So, as (r) goes up, the minimum value goes up, because K*(-rT) goes down, which we may think of as the PV of the exercise strike. Put another way, if (r) goes up, we need to deposit less cash today to receive back the same fixed strike price in the future. I hope that helps, thanks,
Minimum value call = S(0) - K*exp(-rT), which is the BSM "stripped" of its N(.) functions that essentially capture the volatility: S(0)*
little bit of calculus to prove this,
value of call ,
c(r)=S(0)*N(d1) - K*exp(-rT)*N(d2) where call option value is the function of r that is the interest rate
differentiating w.r.t r above equation,
dc/d(r)=d/dr[S(0)*N(d1) - K*exp(-rT)*N(d2)]
dc/d(r)=d/dr[S(0)*N(d1)] - K*d/dr[exp(-rT)*N(d2)]
dc/d(r)=S(0)*d/dr[N(d1)] - exp(-rT)*K*d/dr[N(d2)]
N(d1) and N(d2) aree functions of d1 and d2 we can write above to be approximate,
dc/d(r)=S(0)*d/dr[(d1)] - exp(-rT)*K*d/dr[(d2)]
d1=[ln(S(0)/K)+T(r+.5*vol^2)]/[vol*sqrt(T)]=>dd1/dr=d[[ln(S(0)/K)+T(r+.5*vol^2)]/[vol*sqrt(T)]]/dr=[sqrt(T)/vol]
d2=[ln(S(0)/K)+T(r-.5*vol^2)]/[vol*sqrt(T)]=>dd2/dr=d[[ln(S(0)/K)+T(r-.5*vol^2)]/[vol*sqrt(T)]]/dr=[sqrt(T)/vol]
dc/d(r)=S(0)*[sqrt(T)/vol] - exp(-rT)*K*[sqrt(T)/vol]
dc/d(r)=[sqrt(T)/vol]*[S(0) - exp(-rT)*K]
as r increases keeping all others factors constant exp(-rT) decreases=>exp(-rT)*K decreases=>[S(0) - exp(-rT)*K] increases=>[sqrt(T)/vol]*[S(0) - exp(-rT)*K] increases which is nothing but dc/d(r) so as r increases dc/d(r) increases that means change in call is more than change in interest rate and is on positive side ie. if dr=+change in r than dc changes positively
and as first derivative of function>0 wrt any variable than the function is an increasing one in that variable
=>thus the function c(r) is an increasing function w.r.t the value of r thus as r increase the value of option increases.
thanks
value of call ,
c(r)=S(0)*N(d1) - K*exp(-rT)*N(d2) where call option value is the function of r that is the interest rate
differentiating w.r.t r above equation,
dc/d(r)=d/dr[S(0)*N(d1) - K*exp(-rT)*N(d2)]
dc/d(r)=d/dr[S(0)*N(d1)] - K*d/dr[exp(-rT)*N(d2)]
dc/d(r)=S(0)*d/dr[N(d1)] - exp(-rT)*K*d/dr[N(d2)]
N(d1) and N(d2) aree functions of d1 and d2 we can write above to be approximate,
dc/d(r)=S(0)*d/dr[(d1)] - exp(-rT)*K*d/dr[(d2)]
d1=[ln(S(0)/K)+T(r+.5*vol^2)]/[vol*sqrt(T)]=>dd1/dr=d[[ln(S(0)/K)+T(r+.5*vol^2)]/[vol*sqrt(T)]]/dr=[sqrt(T)/vol]
d2=[ln(S(0)/K)+T(r-.5*vol^2)]/[vol*sqrt(T)]=>dd2/dr=d[[ln(S(0)/K)+T(r-.5*vol^2)]/[vol*sqrt(T)]]/dr=[sqrt(T)/vol]
dc/d(r)=S(0)*[sqrt(T)/vol] - exp(-rT)*K*[sqrt(T)/vol]
dc/d(r)=[sqrt(T)/vol]*[S(0) - exp(-rT)*K]
as r increases keeping all others factors constant exp(-rT) decreases=>exp(-rT)*K decreases=>[S(0) - exp(-rT)*K] increases=>[sqrt(T)/vol]*[S(0) - exp(-rT)*K] increases which is nothing but dc/d(r) so as r increases dc/d(r) increases that means change in call is more than change in interest rate and is on positive side ie. if dr=+change in r than dc changes positively
and as first derivative of function>0 wrt any variable than the function is an increasing one in that variable
=>thus the function c(r) is an increasing function w.r.t the value of r thus as r increase the value of option increases.
thanks
Thanks Shakti! I suppose your final should be equal to, is solving for, greek rho: K*T*exp(-rT)*N(d2)? Not sure I follow how you were able to drop N(.) ... thanks,
hi david,
as per your assumption of N(d1)=N(d2) for calculating impact of r on call price i asssumes that N(d1)=N(d2) diiferentiating both sides w.r.t the variable r d/drN(d1)=d/drN(d2)=>dd1/dr*N'(d1)=dd2/dr*N'(d2) now since dd1/dr=dd2/dr=[sqrt(T)/vol] =>N'(d1)=N'(d2) =e^(-.5*d1^2)
dc/d(r)=S(0)*d/dr[N(d1)] - exp(-rT)*K*d/dr[N(d2)]
dc/d(r)=S(0)*d/dd1[N(d1)] *dd1/dr- exp(-rT)*K*d/dd2[N(d2)]*dd2/dr
dc/d(r)=S(0)*[N'(d1)] *dd1/dr- exp(-rT)*K*[N'(d2)]*dd2/dr
dc/d(r)=S(0)*[e^(-.5*d1^2)] *dd1/dr- exp(-rT)*K*[e^(-.5*d1^2)]*dd2/dr
dc/d(r)=S(0)*[e^(-.5*d1^2)] *dd1/dr- exp(-rT)*K*[e^(-.5*d2^2)]*dd2/dr
as dd1/dr=dd2/dr=[sqrt(T)/vol]
dc/d(r)=[sqrt(T)/vol]*[e^(-.5*d1^2)]*[S(0)- exp(-rT)*K]=>as r increases=> exp(-rT)*K decreases=>[S(0)- exp(-rT)*K] increases so that if a t time t=0 dc/d(r)=0 than after time t=0 the dc/dr>0 as dc/dr increases with increase in r =>dc/dr>0 for increasing values of r so that c is an increasing function for increasing values of rand dc/dr<0 for decreasing values of r that is option value is a decreasing function for decreasing values of r.
thanks
as per your assumption of N(d1)=N(d2) for calculating impact of r on call price i asssumes that N(d1)=N(d2) diiferentiating both sides w.r.t the variable r d/drN(d1)=d/drN(d2)=>dd1/dr*N'(d1)=dd2/dr*N'(d2) now since dd1/dr=dd2/dr=[sqrt(T)/vol] =>N'(d1)=N'(d2) =e^(-.5*d1^2)
dc/d(r)=S(0)*d/dr[N(d1)] - exp(-rT)*K*d/dr[N(d2)]
dc/d(r)=S(0)*d/dd1[N(d1)] *dd1/dr- exp(-rT)*K*d/dd2[N(d2)]*dd2/dr
dc/d(r)=S(0)*[N'(d1)] *dd1/dr- exp(-rT)*K*[N'(d2)]*dd2/dr
dc/d(r)=S(0)*[e^(-.5*d1^2)] *dd1/dr- exp(-rT)*K*[e^(-.5*d1^2)]*dd2/dr
dc/d(r)=S(0)*[e^(-.5*d1^2)] *dd1/dr- exp(-rT)*K*[e^(-.5*d2^2)]*dd2/dr
as dd1/dr=dd2/dr=[sqrt(T)/vol]
dc/d(r)=[sqrt(T)/vol]*[e^(-.5*d1^2)]*[S(0)- exp(-rT)*K]=>as r increases=> exp(-rT)*K decreases=>[S(0)- exp(-rT)*K] increases so that if a t time t=0 dc/d(r)=0 than after time t=0 the dc/dr>0 as dc/dr increases with increase in r =>dc/dr>0 for increasing values of r so that c is an increasing function for increasing values of rand dc/dr<0 for decreasing values of r that is option value is a decreasing function for decreasing values of r.
thanks
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