# Pricing a riskless asset in the Black & Scholes market

Consider a Black&Scholes Market where a risky asset evolves according to: $$\frac{dS_t}{S_t}=\mu dt+\sigma dB_t$$ $$S_o=s$$ Riskless asset is associated with risk free rate r. I want to represent the value of the payoff given only by the riskless asset using the B&S formula.

Is it correct to simply use the B&S formula by eliminating the underlying part $S_0$ to obtain the payoff of the riskless assets? $$P_o=S_0 \phi (d1)-Ke^{-rT} \phi (d2)$$ $P_o=S_0 \phi (d1)-Ke^{-rT} \phi (d2)$ (Price of a call) $P_o=-Ke^{-rT} \phi (d2)$ ( Price of the riskless asset)

• Please remove the "graph" as it is completely useless as it is. Commented Jan 19, 2015 at 19:30
• A riskless asset has no variance in price (it is deterministic) by definition, so you're simply solving the ODE $dS_t/S_t = r dt$ and you get the answer you expect. Commented Jan 19, 2015 at 19:51

You get the value of the pay-off (assuming it is a european option of expiray $T$) by looking at $P_T$, as by definition, this is the pay-off. Indeed, for a european call of strike $K$ and expiry $T$ you have: $$P(t,S_t) = S_t \varphi( d_{+} ) - Ke^{-r(T-t) } \varphi( d_{-} )$$ where $\varphi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{t^2}{2}}dt$ and $d_{\pm} = \frac{\ln\left(\frac{S_t}{Ke^{-r(T-t)}}\right) \pm \frac{\sigma^2}{2} (T-t)}{\sigma\sqrt{T-t}}$. And this $P(t,S_t)$ tends to $\left( S_T - K\right)_{+}$ when $t$ tends to $T$.
Now, first of all, you're totally wrong if think a quantity existing at time $t=0$ could be a pay-off, which is a time $t=T$ living quantity. But even if you keep the second half of the price, and look its limit as $t$ tends to $T$, this will have nothing to do with any riskless-asset. Wrong again.