Deriving the Black–Scholes Put Price Under the Risk-Neutral Measure

# B-S Put Option Formula: Derivation using expected value under Q


# B-S Put Option Formula: Derivation using expected value under Q












I have been working on an old problem in one of my finance classes and, since no solution has been provided and I won't be able to contact my teacher anytime soon, I was hoping I could ask you guys to give me some feedback on my solution.

Here's the problem:

Consider a measure Q under which the dynamics of St are:

$\frac{dS_{t}}{S_{t}}=rdt+\sigma dW_{t}^{Q}$

where $W_{t}^{Q}$ is a Brownian motion under Q.Solve the boundary value problem you found above via an equivalent martingale method under the measure Q, i.e.

$P(t,S_{t})=e^{-r(T-t)}E{_{t}}^{Q}[P(T,S_{t})]$

Here is my attempt at a solution:

$E{_{t}}^{Q}[P(T,S_{t})]=E{_{t}}^{Q}[K\times 1_{S_{t}\leq K}]-E{_{t}}^{Q}[S_{T}\times 1_{S_{t}\leq K}]$ where 1 is an indicator function

$E{_{t}}^{Q}[P(T,S_{t})]=K\times Prob^{Q}(S_{T}\leq K)-E{_{t}}^{Q}[S_{t}e^{(r-\frac{\sigma ^{2}}{2})(T-t)+\sigma (W_{T}^{Q}-W_{t}^{Q}))}\times 1_{S_{t}\leq K}]$

$E{_{t}}^{Q}[P(T,S_{t})]=K\times Prob^{Q}(S_{T}\leq K)-S_{t}e^{r(T-t))}E{_{t}}^{Q}[e^{(-\frac{\sigma ^{2}}{2})(T-t)+\sigma (W_{T}^{Q}-W_{t}^{Q}))}\times 1_{S_{t}\leq K}] $

Now, here comes the part where I am not sure whether how I proceeded is correct.

Fact: $E[e^{yZ-0.5y^2}\times 1_{Z\geq -a}]=\Phi (y+a)$, for $Z\sim N(0,1)$

Proof (provided to us):

$E[e^{yZ-0.5y^2}\times 1_{Z\geq -a}]=\int_{-\infty }^{\infty }e^{yZ-0.5y^2}\times 1_{Z\geq -a}[\frac{1}{\sqrt{2\pi }}e^{-0.5Z^2}]dZ$ $=\frac{1}{\sqrt{2\pi }}\int_{-a}^{\infty }e^{-0.5(Z-y)^2}dZ=\Phi (y+a)$

From this I went on to rewrite $1_{S_{t}\leq K}$ :

$S_{t}\leq K$

$S_{t}e^{(r-\frac{\sigma ^{2}}{2})(T-t)+\sigma (W_{T}^{Q}-W_{t}^{Q}))}\leq K$

$lnS_{t}+(r-\frac{\sigma ^2}{2})(T-t)+\sigma (W_{T}^{Q}-W_{t}^{Q})\leq lnK$

$\sigma (W_{T}^{Q}-W_{t}^{Q})\leq lnK -lnS_{t}-(r-\frac{\sigma ^2}{2})(T-t)$

$\sigma \sqrt{T-t}Z\leq ln\frac{K}{S_{t}}-(r-\frac{\sigma ^2}{2})(T-t)$

$Z \geq\frac{ln\frac{S_{t}}{K}+(r-\frac{\sigma ^2}{2})(T-t)}{\sigma \sqrt{T-t}}$

${ \frac{ln\frac{S_{t}}{K}+(r-\frac{\sigma ^2}{2})(T-t)}{\sigma \sqrt{T-t}}}\equiv a$

and so

$1_{Z\geq a}$

Therefore,

$E[e^{-\frac{\sigma ^{2}}{2}(T-t)+\sigma\sqrt{T-t}Z}\times 1_{Z\geq a}]=\Phi (y-a)$

is it correct here to subtract a, instead of adding it as given in the proof?

Further:

$\Phi (y-a)=\sigma \sqrt{T-t}-\frac{ln\frac{S_{t}}{K}+(r-\frac{\sigma ^2}{2})(T-t)}{\sigma \sqrt{T-t}}$

$\Phi (y-a)=\frac{\sigma^2 (T-t)}{\sigma\sqrt{T-t}}\ - \frac{ln\frac{S_{t}}{K}+(r-\frac{\sigma ^2}{2})(T-t)}{\sigma \sqrt{T-t}}$

giving:

$\Phi (y-a)= -\frac{ln\frac{S_{t}}{K}+(r+\frac{\sigma ^2}{2})(T-t)}{\sigma \sqrt{T-t}}=\Phi (-d_{1})$

Finally:

$P(t,S_{t})=e^{-r(T-t)}[K \times Prob^{Q}(S_{T}\leq K)-S_{t}e^{r(T-t)}\Phi (-d_{1})]$

$=e^{-r(T-t)}K\Phi (-d2)-S_{t}\Phi (-d_{1})$$

Which is the B-S put option formula, if I'm not mistaken. Again, as mentioned above, I am not sure if everything I did was correct, especially the application of the fact I've given above. I'd be grateful, if someone more knowledgable than could quickly go over my solution and let me know if I went wrong anywhere :)

Thanks for any help!