Generally, we assume that an interest rate increase makes the call price more expensive. From my understanding it is because the expected return on the stock price increases. However the interest rate increase makes also the discount factor lower, and we know that the price today is the discounted expected payoff. Does that mean that we assume that the expected return increase on the stock always outweigh the discount factor decrease, so that the price today increases?
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1$\begingroup$ Maybe this is relevant to you: quant.stackexchange.com/questions/39801/…. $\endgroup$– Daneel OlivawMar 28, 2019 at 16:54
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No, they're actually the same, kind of. The risk-free rate that we assume the underlying stock price grows at is the same as the rate we discount once we have determined if that path exceeds the strike price (the discount factor we multiply the stock price by uses the same $r$ as the one used in $N(d_1)$. Accordingly, if you have a theoretical zero-strike call option on a non-dividend paying stock worth \$100 and the risk free rate is 3%, the value of that call option would be \$100 - the same as the price of the stock since you will always exceed the strike price and it's guaranteed to be worth the stock at the future point in time.
Now, if the strike was \$100, the probability of exceeding the strike price goes down quite a bit. One thing that helps the stock hit the \$100 is that the stock is assumed to drift higher at a risk-free rate of 3%. So, the value of the call option is worth more because it increases the probability of receiving payment, even though it doesn't increase the value of that payment if the payment is certain to be made.
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$\begingroup$ Please note that my explanation assumes the assumptions underlying the Black Scholes option pricing hold. $\endgroup$– RandyFMar 28, 2019 at 22:17