Quantitative Finance Interviews
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3.3 Partial Derivatives and Multiple Integrals

Partial derivative: w=f(x,y)⇒∂f∂x(x0,y0)=lim⁡Δx→0f(x0+Δx,y0)−f(x0,y0)Δx=fxw=f(x,y)\Rightarrow\displaystyle\frac{\partial f}{\partial x}(x_0,y_0)=\lim_{\Delta x\to0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=f_x

Second order partial derivatives: ∂2f∂x2=∂∂x(∂f∂x),∂2f∂x∂y=∂∂x(∂f∂y)=∂∂y(∂f∂x)\displaystyle\frac{\partial^2 f}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right),\quad\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)

The general chain rule: Suppose that w=f(x1,x2,⋯ ,xm)w=f(x_1,x_2,\cdots,x_m) and that each of variables x1, x2, ⋯ , xmx_1,\ x_2,\ \cdots,\ x_m is a function of the variables t1, t2, ⋯ , tnt_1,\ t_2,\ \cdots,\ t_n. If all these functions have continuous first-order partial derivatives, then ∂w∂ti=∂w∂x1∂x1∂ti+∂w∂x2∂x2∂ti+⋯+∂w∂xm∂xm∂ti\displaystyle\frac{\partial w}{\partial t_i}=\frac{\partial w}{\partial x_1}\frac{\partial x_1}{\partial t_i}+\frac{\partial w}{\partial x_2}\frac{\partial x_2}{\partial t_i}+\cdots+\frac{\partial w}{\partial x_m}\frac{\partial x_m}{\partial t_i} for each i,1≤i≤ni,1\leq i\leq n.

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Changing Cartesian integrals into polar integrals: The variables in two-dimension plane can be mapped into polar coordinates: x=rcos⁡θ, y=rsin⁡θx=r\cos\theta,\ y=r\sin\theta. The integration in a continuous polar region RR is converted to

∬Rf(x,y) dx dy=∬Rf(rcos⁡θ,rsin⁡θ)r dr dθ.\iint_R f(x,y)\,dx\,dy=\iint_R f(r\cos\theta,r\sin\theta)r\,dr\,d\theta.

Calculate ∫0∞e−x2/2 dx\displaystyle\int_0^\infty e^{-x^2/2}\,dx.

Solution: Hopefully you happen to remember that the probability density function (pdf) of the standard normal distribution is f(x)=12πe−x2/2\displaystyle f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}. By definition, we have

∫−∞∞f(x) dx=∫−∞∞12πe−x2/2 dx=2∫0∞12πe−x2/2 dx=1⇒∫0∞e−x2/2 dx=π2.\int_{-\infty}^{\infty}f(x)\,dx=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx=2\int_0^{\infty}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx=1\Rightarrow\int_0^{\infty}e^{-x^2/2}\,dx=\sqrt{\frac{\pi}{2}}.

If you’ve forgotten the pdf of the standard normal distribution or if you are specifically asked to prove ∫−∞∞12πe−x2/2 dx=1\displaystyle\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx=1, you will need to use polar integrals to solve the problem:

∫−∞∞e−x2/2 dx∫−∞∞e−y2/2 dy=∫−∞∞∫−∞∞e−(x2+y2)/2 dx dy=∫0∞∫02πe−(r2cos⁡2θ+r2sin⁡2θ)/2r dr dθ=∫0∞∫02πe−r2/2r dr dθ=−∫0∞e−r2/2d(−r2/2)∫02πdθ=−[e−r2/2]0∞[θ]02π=2π\begin{aligned} \int_{-\infty}^{\infty}e^{-x^2/2}\,dx\int_{-\infty}^{\infty}e^{-y^2/2}\,dy &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)/2}\,dx\,dy\\ &=\int_0^{\infty}\int_0^{2\pi}e^{-(r^2\cos^2\theta+r^2\sin^2\theta)/2}r\,dr\,d\theta\\ &=\int_0^{\infty}\int_0^{2\pi}e^{-r^2/2}r\,dr\,d\theta\\ &=-\int_0^{\infty}e^{-r^2/2}d(-r^2/2)\int_0^{2\pi}d\theta\\ &=-\left[e^{-r^2/2}\right]_0^{\infty}[\theta]_0^{2\pi}=2\pi \end{aligned}

Since ∫−∞∞e−x2/2 dx=∫−∞∞e−y2/2 dy\displaystyle\int_{-\infty}^{\infty}e^{-x^2/2}\,dx=\int_{-\infty}^{\infty}e^{-y^2/2}\,dy, we have ∫−∞∞e−x2/2 dx=2π⇒∫0∞e−x2/2 dx=π2\displaystyle\int_{-\infty}^{\infty}e^{-x^2/2}\,dx=\sqrt{2\pi}\Rightarrow\int_0^{\infty}e^{-x^2/2}\,dx=\sqrt{\frac{\pi}{2}}.