Why Call Prices Rise with Maturity Even at Negative Rates

# Call option prices in terms of maturity with negative interest rates


# Call option prices in terms of maturity with negative interest rates












let's assume that interest rates are constant, $r$. When $r\geq 0$, we can see that if $T_1<T_2$ and $C_1$ (resp. $C_2$) is the price of a call option on a non-dividend paying stock with maturity $T_1$ (resp. $T_2$), then $C_1<C_2$.

I am trying to understand what happens when $r<0$. But all I have are the optimal bounds $S_0-KB_i < C_i < S_0$, where $K$ is the strike price and $B_i=e^{-rT_i}$. I can't seem to deduce any consequences from this alone.

## Answer by Cettt (score 1)

https://quant.stackexchange.com/a/37776

the short answer: we have $C_1 \leq C_2$ for all $r \in \Bbb R$. Here it is important to note that $C_1$ and $C_2$ have the same discounted strike, i.e. $C_1$ has strike $K_1 := K \cdot \exp(rT_1)$ and $C_2$ has strike $K_2 := K \cdot \exp(rT_2)$

The longer answer: one way to reason why option prices are increasing with maturity is the fact that the discounted underlying is a martingale (with respect to the pricing measure) and the fact that the function $x \mapsto (x-K)^+$ is convex. To be more precise let's fix a strike $K \in \Bbb R$ and denote the time-$T$ price of the underlying by $S_T$. Then the process $\{\exp(-rT)S_T\}_{T \geq 0}$ is a martingale. A well known result (which is a simple application of Jensen's formula) states that the process $\{(\exp(-rT)S_T - K)^+\}_{T \geq 0}$ is a submartingale. In particular, submartingales have increasing expectations which implies \begin{align} C_1 = & \exp(-rT_1) \cdot \Bbb E \Bigl[\Bigl(S_{T_1} - K \cdot \exp(rT_1) \Bigr)^+ \Bigr] = \Bbb E \Bigl[\Bigl(\exp(-rT_1)S_{T_1} - K \Bigr)^+ \Bigr] \\ \leq & \Bbb E \Bigl[\Bigl(\exp(-rT_2)S_{T_2} - K \Bigr)^+ \Bigr] = \exp(-rT_2) \cdot \Bbb E \Bigl[\Bigl(S_{T_2} - K \cdot \exp(rT_2) \Bigr)^+ \Bigr] = C_2. \end{align}

Note that I did not assume any model or particular distribution for $S$, so this result for 'every' model.