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Can someone explain this mathematical "trick"? (This comes from a long derivation from pricing European options in Finance).

$$I = {1 \over {\sqrt {2\pi } }}\int\limits_{ - \infty }^{ - {x \over {\sqrt {2\tau } }}} {{e^{ - {1 \over 2}{{\left( {x' - {{(k - 1)\sqrt {2\tau } } \over 2}} \right)}^2}}}{e^{{1 \over 2}(k - 1)x + {1 \over 4}{{(k - 1)}^2}\tau }}} dx'$$

$$ = {{{e^{{1 \over 2}(k - 1)x + {1 \over 4}{{(k - 1)}^2}\tau }}} \over {\sqrt {2\pi } }}\int\limits_{ - \infty }^{ - {x \over {\sqrt {2\tau } }} - {1 \over 2}(k - 1)\sqrt {2\tau } } {{e^{-{1 \over 2}{s^2}}}ds} $$

$$={e^{{1 \over 2}(k - 1)x + {1 \over 4}{{(k - 1)}^2}\tau }}N( - d2)$$

I don't get how he can change the integral boundary? Also, how is that $N(-d2)$; when looking at the formula below, I don't see it, but help me please.

note: $$N(d) = {1 \over {\sqrt {2\pi } }}\int\limits_{ - \infty }^d {{e^{ - {1 \over 2}{s^2}}}ds} $$

$$d2 = {x \over {\sqrt {2\tau } }} + {1 \over 2}(k - 1)\sqrt {2\tau } $$

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  • $\begingroup$ I can't tell if this is the property he's using just by looking at the problem (so many symbols!) but it could be the property that $$\int_a^b g'(x)*f(g(x)) \,dx=\int_{g(a)}^{g(b)} f(x) \,dx $$ $\endgroup$ Commented Apr 22, 2017 at 17:05

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For the first part it looks like they made the substitution

$$ s = x' - \frac{1}{2}(k-1)\sqrt{2\tau} $$

Then, to get the new upper limit of integration you let $x' = -\frac{x}{\sqrt{2\tau}}$, which was the old limit. The lower limit does not change because $s \to -\infty$ as $x \to -\infty$. Note that the integrand after the substitution should actually be $e^{-\frac{1}{2}s^2}$.

As for the part with $N(-d2)$ , just replace $d$ with $-d2$ in the formula for $N(d)$ and you should see how the result matches your expression for $I$.

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  • $\begingroup$ thank you for the rply, the last point was fairly obvious and i understand it now. However the first point i still dont understand how $$\int\limits_{ - \infty }^{ - {x \over {\sqrt {2\tau } }}} $$ goes to this or becomes this $$\int\limits_{ - \infty }^{ - {x \over {\sqrt {2\tau } }} - {1 \over 2}(k - 1)\sqrt {2t} } $$ even if you make the substitution which you where right on $\endgroup$
    – asdf
    Commented Apr 22, 2017 at 17:35
  • $\begingroup$ The original limit is in terms of $x'$, because the integral has a $dx'$. Thus, at the upper limit we have $x' = -\frac{x}{\sqrt{2\tau}}$. We need to change this to be a limit in terms of the new variable of integration $s$. Since $s = x'-\frac{1}{2}(k-1)\sqrt{2\tau}$ and at the upper limit $x' = -\frac{x}{\sqrt{2\tau}}$ we have that at the upper limit $s = -\frac{x}{\sqrt{2\tau}}-\frac{1}{2}(k-1)\sqrt{2\tau}$. $\endgroup$
    – wgrenard
    Commented Apr 22, 2017 at 17:47

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