# Asymptotics of binary call option in Black–Scholes model in the large deviations sense

Consider Black–Scholes model where asset log-price is given by

$$X _ { t } = \sigma W _ { t } + \mu t$$

for $$W_t$$ – Brownian motion.

I want to show that

$$P \left[ X _ { t } > k \right] \approx \exp \left( - \frac { k ^ { 2 } } { 2 \sigma ^ { 2 } t } \right) \quad \text { as } t \downarrow 0$$

in the large deviations sense, i.e.

$$\lim _{t\downarrow 0} t\cdot \log P[X_t>k] = \frac{-k^2}{2\sigma^2}$$

After rescaling and getting rid of vanishing prefactor $$\frac{1}{\sqrt{2\pi}}$$ we get

\begin{aligned} \lim _{t\downarrow 0} t\cdot \log P[X_t>k] &= \lim _{t\downarrow 0} t\cdot \log \int_{\frac{k-\mu t}{\sigma \sqrt t}}^{\infty}\exp\left(\frac{-x^2}{2}\right)\mathrm dx \\&=\lim _{t\downarrow 0} t\cdot \log \int_{\frac{k}{\sigma \sqrt t}}^{\infty}\exp\left(\frac{-x^2}{2}\right)\mathrm dx \end{aligned}

whereas in the second equality we applied dominated convergence.

Now I was looking into the mean-value theorem, but I am not sure how to rewrite the term in order to apply it directly.

Another idea would be to reformulate the problem in order to be able to apply Cramer theorem, since $$\varphi ^* (z) = \frac { (z - \mu) ^ { 2 } } { 2 \sigma ^ { 2 } }$$ is Legendre transform of the logarithm of the moment-generating function of the normal distribution, and Cramer theorem relates probability of the large deviations events in terms of Legendre transform of the logarithm of the mgf.

• The integral is the complementary error function. Its asymptotic behavior at infinity can be obtained by integrating by parts: $$\int_x^\infty e^{-t^2/2} dt = \frac {e^{-x^2/2}} x - \int_x^\infty \frac {e^{-t^2/2}} {t^2} dt = \dots \,.$$ Commented Mar 27, 2019 at 23:33
You could use the standard techniques of Cramer's theorem by rewriting the problem as a an average of normals by approximating $$1/t$$ with its nearest integer.
However, you can also evaluate the integral directly, as it's major contribution should come from values near $$k/\sigma\sqrt{t}$$. This is a direct consequence of Laplace's method. First let $$u=x\sigma\sqrt{t}/k$$. Rewrite the integral on $$[1,\infty]$$ with integrand $$\exp(-f(u)/t)$$, where $$f(u):=k^2u^2/2\sigma^2$$. Apply Laplace's method, computing $$f''(u)$$, to finish the proof.
• Cool, thanks for suggesting Laplace method. I guess you missed a square after $u$ in your definition of $f(u)$. Commented Mar 27, 2019 at 20:24