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Can I apply the Fundamental Theorem of Calculus for $$\int_{-\infty}^{t_1} \int_{-\infty}^{t_2} \frac{\partial \phi\left(\frac{z_2 - \rho z_1}{\sqrt{1 - \rho^2}}\right)}{\partial z_2} dz_2 dz_1$$ in order to get $$\int_{-\infty}^{t_1} \phi\left(\frac{t_2 - \rho z_1}{\sqrt{1 - \rho^2}}\right)dz_1,$$ where $\phi(\cdot)$ is standard normal pdf?

I am not sure whether I can treat $z_1$ as fixed and take $\lim_{a \rightarrow -\infty} \phi\left(\frac{a - \rho z_1}{\sqrt{1 - \rho^2}}\right)$.

Kolibris
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1 Answers1

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Yes. You can write $$\int_{a}^{b}\frac{\partial \phi(z_{1},z_{2})}{\partial z_{2}}d z_{1}=\frac{d}{d z_{2}}\int_{a}^{b}\phi(z_{1},z_{2})dz_{1}$$

Assuming $t_{1},t_{2}$ are constants, we have $$\int_{-\infty}^{t_{2}} \int_{-\infty}^{t_{1}}\frac{\partial \phi(u(z_{1},z_{2}))}{\partial z_{2}}dz_{1}dz_{2}=\int_{-\infty}^{t_{1}}\phi(u(t_{2},z_{1}))dz_{1}-\int_{-\infty}^{t_{1}}\lim _{x \to -\infty}\phi((u(x,z_{1}))dz_{1}$$ where we use various continuity assumptions to move the limit around. Now, $$\lim_{x \to -\infty}\frac{x-\rho z_{1}}{\sqrt{1-\rho^{2}}}=-\infty$$, so as $x \to -\infty$, $u \to -\infty$ and $\phi(u) \to 0$

preferred_anon
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  • This step is clear. My main concern is whether $\lim_{a \rightarrow -\infty} \phi\left(\frac{a - \rho z_1}{\sqrt{1 - \rho^2}}\right) = 0$. I found here: http://www.math.ucsd.edu/~jverstra/20e-lecture16.pdf that I have to calculate $\lim_{(a,b) \rightarrow -\infty}\int_{a}^{t_1} \int_{b}^{t_2} \frac{\partial \phi\left(\frac{z_2 - \rho z_1}{\sqrt{1 - \rho^2}}\right)}{\partial z_1} dz_2 dz_1$ and now I don't have a joint limit. – Kolibris Nov 23 '14 at 15:40
  • Isn't it a property of pdfs that $\lim_{u \to -\infty} \phi(u)=0$? You can certainly treat $z_{1}$ as fixed inside the integral. – preferred_anon Nov 23 '14 at 15:44
  • Yes, it is. But since I take a limit only wrt one variable, isn't it the case that I miss situations when $u$ actually does not go to infinity. For instance, if $u = a - \rho b$ and I take a limit wrt to $(a,b) \rightarrow -\infty$, and $\rho = -0.99$, isn't it the case that the limit of $u$ is not necessarily $-\infty$? – Kolibris Nov 23 '14 at 15:52
  • I've added some more detail. Does that answer your question? – preferred_anon Nov 23 '14 at 16:10