# 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 $-1.96 \sigma \sqrt{t}$?

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I agree with the assumption QuantIbex stated at the very end. Not knowing the text but if the authors did not elaborate why they made such assumption then I find it a pretty weak treatise and without specific reasons (especially without establishing a relationship between sigma, mu, and t) such assumption could potentially even be wrong. –  Matt Wolf Jul 19 '13 at 11:40

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

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

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 $\mu$).

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Answer downvoted, so deemed not useful. How could it be improved to become useful? –  QuantIbex Jul 19 '13 at 23:56
Yesterday,I saw two answers for my question.But now,only one left.I didn't vote any answer since it requires 15 reputation but I have only 1.Both your answers are helpful.Because the key point I didn't understand before reading your answer is that exp(x) approximates 1+x if x is infinitesimal.And I also made a mistake about x and the square root of x.If x is within (0,1),then,x must be smaller than square root of x. –  Hebe Jul 20 '13 at 3:33
The approximation $\exp (x) = 1 + x$ when $x$ is near $0$ comes from the Taylor expansion of the exponential function. Another useful approximation is $\log (1 + x) = x$. –  QuantIbex Jul 20 '13 at 8:57
@Hebe, feel free to come back and vote once you've reach the required reputation threshold. As the author of the question you also have the option to accept one of the answers (quant.stackexchange.com/help/someone-answers). –  QuantIbex Jul 20 '13 at 9:08
This is very basic maths, how can this be downvoted ? –  lmorin Jul 22 '13 at 12:46

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 $t$.
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.
(+1) for the nice and clear explanation. The condition $t<a^2/b^2$ is more intuitive formulation of the condition on $t$ than $|a\sqrt{t}|>|bt|$. –  QuantIbex Jul 20 '13 at 12:57
x ~ y does NOT imply $e^{x}$ ~ $e^{y}$ –  lmorin Jul 24 '13 at 9:24