Level l Schweser notes for derivatives

Hi, I am looking at question 10 from the Schweser study notes for derivatives. Chapter. - basics of derivative pricing and valuation.

Question - a decrease in risk free rate of interest will

A. Increase put and call option prices

B. Decrease put option prices and increase call option prices

C. Increase put option prices and decrease call option prices

Should the answer be B instead of option C? They are talking about price here and not the value. Am I understanding it correctly?

Look at put-call parity:

P0 + S0 = C0 + X / (1 + rrf)

A decrease in the risk-free rate increases the present value of the strike price (X); to maintain equality, either C0 has to decrease or P0 has to increase.

Looks like C to me.

Sure, thanks. This helps to understand the concept. Just to clarify, do they distinguish between the value and its price? (I assumed they are different as with forwards) Or is value and price used interchangeably, both meaning price for options?

I was using the following equation and hence getting it wrong s-x, x being the strike price, which I was discounting, and the difference between s-x was going up, which I assumed was the price, but that is the payoff.

The assumption is that options are priced fairly, so price and value are the same number.

Understood…Thanks a lot… :slight_smile:

My pleasure.