Small-Time Approximations for Exponential Terms in Option Valuation

# Question on an approximation in pricing formula


# Question on an approximation in pricing formula












I am reading the book An Introduction to Financial Option Valuation. The following on page 58 makes me confused:

> For the formula: $\exp \left\{ -1.96\sigma \sqrt{t}+(\mu-0.5 \sigma^2)t \right\}$, if $t$ is small, then it is approximately equal to $\exp \left (-1.96 \sigma \sqrt{t} \right )$. Moreover, the second formula approximagely equals $1 - 1.96 \sigma \sqrt{t}$.

I don't understand how can we get the second and the third expression. If $t$ is very small, then $\sqrt{t}$ should be infinitesimal. Then, why has $(\mu-0.5 \sigma^2)t$ disappeared in the second formula, but not $-1.96 \sigma \sqrt{t}$?

## Answer by QuantIbex (score 5, accepted)

https://quant.stackexchange.com/a/8528

To simplify notations, let $a:= -1.96\sigma$ and $b := \mu - 0.5\sigma^2$. The development in the book could be justified if both $a\sqrt{t}$ and $bt$ are small (close to zero), and if we have that $|a\sqrt{t}| > |bt|$.

Recall that

- $\exp (x+y)= \exp(x)\exp(y)$,

- $\exp(x)\approx 1 + x,\quad \text{if } x\approx 0$.

Then, using these properties we have \begin{align} \exp (a\sqrt{t} + bt) &= \exp (a\sqrt{t}) \exp (bt)\\ &\approx \exp (a\sqrt{t}) (1 + bt) \\ &\approx \exp (a\sqrt{t}),\tag{1} \\ &\approx 1 + a\sqrt{t}, \end{align} where the approximation in (1) follows from the fact that $bt$ is (very) close to zero.

The fact that $bt$ dissapeared but not $a\sqrt{t}$ comes from the fact that they probably assume that $|a\sqrt{t}| > |bt|$. This assumption should be explicitly stated or obvious from the context (given the typical values for $\sigma$ and $\mu$).

## Answer by jens_bo (score 3)

https://quant.stackexchange.com/a/8539

I wanted to add this side note to Quantelbex' answer:

Both factors in $\exp(a\sqrt t)\exp(b t)$ go to one as $t$ goes to zero, but for small $t$, the $\exp(b t)$ term approaches one faster. For $t=\frac {a^2}{b^2}$ both factors are the same, if $t$ is smaller than $\frac {a^2}{b^2}$, we have $\exp(a\sqrt t) > \exp(bt)$. Thus the approximation that $\exp(a \sqrt t + b t) \approx \exp(a \sqrt t)$ for small $t$.

Using the Taylor expansion for the exponential we can calculate the error that is made with this approximation:\begin{align}\exp(a \sqrt t + b t) - \exp(a\sqrt t) &= 1+a\sqrt t +b t + \frac 12 a^2 t +\mathcal O(t^{3/2}) - (1+a\sqrt t + \frac 12 a^2 t + \mathcal O(t^{3/2}))\\ &= b t + \mathcal O(t^{3/2})\end{align}

The second approximation is just the Taylor expansion of the first one and the error is given by: \begin{align}\exp(a \sqrt t + b t) - 1-a\sqrt t &= 1+a\sqrt t +b t + \frac 12 a^2 t +\mathcal O(t^{3/2}) - 1-a\sqrt t\\ &= (b + \frac 12 a^2) t + \mathcal O(t^{3/2})\end{align}

So in both cases the error vanishes linear in $t$, but in order to understand the error one makes, one needs to know $a$ and $b$ to know if the approximations are acceptable for the values of $t$ you consider.