How to find r in put-call Parity?

[Steve Jobs]

Assume it's the put-call parity and all variables are given except the risk free rate (r). The answer in the book does not provide the detail. Is there a quick way to find the r in this equation by the calculator or can you guide step by step what to do:

-23.79 = -24 e^(-0.25 r)

 

[ShaktiRathore]

hi there,
p+S=c+Xe^(-rT)
p+S-c=Xe^(-rT)
e^(-rT)=p+S-c/X
-rT=ln(p+S-c/X)=>rT=ln(X/p+S-c)=>r=(1/T)*ln(X/p+S-c) use ln function in the calculator find ln of X/p+S-c and divide by T the result to get the final result.
to be more approximate: use ln(1+x)~1+x
so that r~(1/T)*[1+(X/p+S-c)]
thanks

 

[Steve Jobs]

Thanks a lot Shakti, really helpful.

 

[David Harper CFA FRM]

btw, this is a thematic building block; liberating the exponent, because continuous compounding of returns is common. For example,
If p = f*exp[-(r+s)*t], need to be able to similarly solve for implied credit spread, s, where P = bond PV price, r = riskfree rate and t = time to maturity.
\(if P=F\cdot {{e}^{-(r+s)T}}\text{ what is s as function of the others?} \)

 

[Steve Jobs]

Hi David,

My math skills are not that good, so I really don't know what you mean by FUNCTION in regard to that equation!!

Any way I understand that for the equation: x = e^(r*t), to get the (r*t) => Ln(x)=r*t

Thanks David,

 

[ShaktiRathore]

hi,
P=F*e^-(r+s)T
=>P/F=e^-(r+s)T
=>ln(P/F)=-(r+s)T
=>ln(F/P)=(r+s)T
=>(1/T)ln(F/P)=(r+s)
s=(1/T)ln(F/P)-r where s is the function of others T,F,P,r similarly you can calculate r as a function of other params based on the same theme.

thanks

 

[Steve Jobs]

Thanks Shakti, It's clear.

 

[David Harper CFA FRM]

Thanks Shakti, I maybe could have just said, "solve for s" in terms of the others variables, where I did mean to give an example of Shakti's step, in particular, where he goes from P/F=e^-(r+s)T to ln(P/F)=-(r+s)T, by taking the natural log (LN) of both sides. This requires using the fact that ln[exp(x)] = x. As ln() and exp() are inverse functions, it's also true that exp[ln(x)] = x, so when continuous compounding is involved it's often the case we get to a step where we want to take exp() or ln() of both sides in pursuit of a goal to get one variable isolated on one side of the equation (e.g., "solving for spread, s, as a function of the other variables"), thanks for raising it!