I was wondering why the price of an option increases with Rho (price change for a derivative relative to a change in the risk-free rate of interest). I found this explanation on a website:

"Each standard equity options contract represents 100 shares of the underlying stock. Because it’s much cheaper to buy a call options contract than it is to buy 100 shares of stock, call buyers are willing to pay more for call options when rates are relatively high because they can invest the difference. A call seller, on the other hand, would want additional incentive to sell a call option (versus selling the stock outright) if interest rates are high in order to compensate for forgoing the cash from the stock sale. In other words, the higher call options premium when interest rates are high is the “opportunity cost” of forgone interest. "

This makes sense, however, this contradicts in my opinion the knowledge that:

  • Higher interest rates tend to negatively affect stock prices
  • (Call) Option price decreases with decreases stock price (how much depends on Delta)

So shouldn't the (call) option price go down with increasing interest rates?

Can someone give me more insights? Thanks!


Your arguments, though correct, have no implication on option price because they are facts about how the market would behave in the future. However, the fact that the option can be replicated today by a dynamic strategy all but constrains the option price, so that real world probabilities/opinions on what would happen have no way to influence the market price.

See this answer for why real world speculation does not matter. All analysis therefore should somehow rest on the argument that the option can be replicated, and this replication controls the price. Equivalently, one can analyse risk neutral dynamics.

Coming to the direction of price change. Let's see what happens in the degenerate case where there's no stock price volatility. Assume that the strike is sufficiently low so that the option is ITM. In this case, price of the option is $S-K*exp(-rt)$, which increases with rates.

Note: If I change the payoff to something else, say $Max(0,S-K*exp(2rt))$, the sign of rho might be different. Infact this thing should be treated as an equity-rates hybrid, with a 2 factor model.

  • $\begingroup$ Thank you for the quick answer! $\endgroup$ May 24 at 10:51
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    $\begingroup$ Nice simplification, to assume $\sigma=0$, which then makes $d_1$ & $d_2$ go to infinity, which makes $N(d_1)$ & $N(d_2)$ go to 1. Saves a lot of nasty algebra, having to work with $r$ within the $N(d_1)$ & $N(d_2)$ terms! +1 from me. $\endgroup$ May 24 at 11:15
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    $\begingroup$ @JanStuller This is what tens of thousands of multiple choice questions does to you! Boundary cases are your friend. $\endgroup$
    – Arshdeep
    May 24 at 17:42
  • $\begingroup$ Lol, how did you end up doing tens of thousands of multiple choice questions? :) $\endgroup$ May 24 at 17:47
  • $\begingroup$ The top engineering entrance exam in my country operates on multiple choice questions. It's hideously competitive so it's not uncommon for the top students to go through those many questions in 2-4 years. $\endgroup$
    – Arshdeep
    May 24 at 17:49

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