Current value of option risk free rate=$r$
volatility of stock price=$\sigma$
continuous dividend rate=$q$
$a>0,K>0$
If your stock price S becomes below $K$ at maturity T,
the option A pays you $aS_T$.
Otherwise, this option pays you zero.
I have to prove that the curret value($v_0$) of this option A is
$v_0=aS_0e^{-qT}\phi(-d)$
where $d=\frac{(ln\frac{S_0}{K}+(r-q+\sigma^2/2))}{\sigma\sqrt{T}}$ and
$\phi$ is the cumulative distribution function of standard normal distribution.
I learned Black-Scholes formula but I can't even figure out how to start.
Any suggestions please?
 A: if you know how to derive B&s formula, you're halfway there. The basic principle of math. finance is that the price of a contingent claim is just the expctation of discounted payoff, i.e. $\mathbb{E}[e^{-rT}\text{Payoff}]$.
Here $\text{Payoff} = aS_T \mathbb{I}_{\{S_T < K\}}$ and so we have to compute the following, where $z \sim N(mT,\sigma^2 T), m = (r-q-\frac{\sigma^2}{2})$:
$$
\begin{align}
\mathbb{E}[e^{-rT} aS_T \mathbb{I}_{\{S_T < K\}}] &= ae^{-rT} \int_{- \infty}^{+\infty}\frac{S_0 e^{z}e^{-\frac{(z-mT)^2}{2 \sigma^2 T}}}{\sqrt{2 \pi \sigma^2 T}} \mathbb{I}_{\{S_0 e^z < K\}}dz\\ 
&= \text{Substitute} \;  y=\frac{z-mT}{\sigma T} \; \text{and take into account the carachteristic function} \\
&= aS_0e^{-rT}\int_{-\infty}^{\frac{\ln \frac{K}{S_0}-mT}{\sigma T}}e^{mT}\frac{e^{-\frac{1}{2}(y^2 -2\sigma \sqrt{T} y)}}{\sqrt{2 \pi}}dy \\
&= \text{Complete the square in the exponent} \\
&= aS_0e^{-rT} e^{mT}\int_{-\infty}^{\frac{\ln \frac{K}{S_0}-mT}{\sigma T}}e^{\frac{\sigma^2T}{2}}\frac{e^{-\frac{1}{2}(y-\sigma \sqrt{T})^2}}{\sqrt{2 \pi}}dy \\
&= \text{Substitute} \; t = y-\sigma \sqrt{T} \\
&= aS_0e^{-rT} e^{mT}\int_{-\infty}^{-d}\frac{e^{-\frac{t^2}{2}}}{\sqrt{2 \pi}}dt \\
&= \text{Simplify the exponentials and notice the normal CDF} \\
&= aS_0 e^{-qT} \Phi(-d)
\end{align}
$$
