# confused about interest rate vs option price

I know it might be a stupid question but…

when I read schweser notes, it states that “as the risk free rate increases, the value of the call (put) will increase (decrease)”. think about it for a while and it makes sense to me.

Now when I start doing part 1 practice exam of 2012, I read the official explanation for one quesiton, it goes" when interest rate rise, stock price have a tendency to fall. This increases the value of a put option on a stock."

seems conflicting to me…can someone pls help me here?? thx a lot!

Think about the Black-Scholes formula for a minute. In the second term, when valuing a _ call _, you are subracting the discounted value of the strike price – and this is of course discounted by the risk free rate. The larger that rate, the smaller the subtracted term in the equation, and the larger the difference. In other words, the call value is higher because a higher risk-free rate made a lower PV of the strike price.

The order of the terms is conceptually reversed in the _ put _ valuation – here you are subtracting the probability-adjusted stock value from the discounted strike. Thus, a larger risk-free rate would mean a smaller first term in the put equation, shrinking the difference. Lower risk-free rates lead to larger first terms, meaning higher differences – voila, a higher put price.

There is no conflict if you think about it. Forget about the equations for a minute. Do you agree that a rising interest rate environment is bad for stock prices? Historically, it has been. Therefore, a rising rate environment would be good for put buyers (as you profit from the underlying stock decreasing in value). Conversely, a decreasing rate environment is good for stock prices, hence, good for a call buyer.

All of that is consistent with our walkthrough on the formulas – high interest rates decrease call values and increase put values.

Good luck in your final studies.

Another way to think about it,

put call parity:

C +Ke^-rt = S+P

C = S+P-Ke^-rt

Hence, r increases, C increases.

for P,

P = C+Ke^-rt-S

Hence, r increases P decreases

Hope we all pass!