I am working on a bivariate probit model and I want to calculate the gradient. In order to do that I have to take the devirative w.r.t. $\beta$ in the following expression: $$\int_{-\infty}^{x_{g1}\beta} \int_{-\infty}^{x_{g2}\beta}\phi_2(z_1, z_2, \rho)dz_2dz_1$$ Could somebody give me a hint how to do it? I tried to use differentiation under the integral sign, however, I could not get the required result.
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$\begingroup$ I know little about this other than that probit models are almost always simulated for the precise reason that these integrals are very difficult to calculate. I believe Kenneth Train's book would be a good place to start getting a perspective on this. $\endgroup$ – gregmacfarlane Jan 24 '13 at 21:29
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$\begingroup$ Do you want to calculate the derivatives on the log likelihood? $\endgroup$ – Dimitriy V. Masterov Jan 25 '13 at 2:06
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$\begingroup$ Yes, I want to calculate the derivative of the log-likelihood function. I didn't write the entire expression, since I have a problem only with this part. $\endgroup$ – Kolibris Jan 25 '13 at 10:15
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1$\begingroup$ That expression is a function $f(\beta)$. The numerator of the difference quotient, $df(h)=f(\beta+h)-f(\beta)$, is an integral over two semi-infinite strips whose widths are proportional to $h$. Assuming suitable continuity properties of $\phi_2$ in neighborhoods of these strips, the limiting value of $df(h)/h$ as $h\to 0$ will therefore be a linear combination of integrals over the two corresponding rays: one extending to the left from $(x_{g1}\beta, x_{g2}\beta)$ and the other down from that point. With this geometric image in mind you can obtain the derivative with almost no computation. $\endgroup$ – whuber♦ Apr 28 '14 at 20:18
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Let's take a step back and solve a simpler problem - how do derivatives with respect to a variable in the limits of the integral work, at least when everything is sufficiently nice? Let's take a very basic approach:
$\frac{d}{dz} \int_{a}^{h(z)} g(x) \,dx$
Let $\frac{dG}{dz}=g(z)$.
$=\frac{d}{dz} [G(h(z))-G(a)]=\frac{d}{dz} G(h(z))=h'(z)\,g(h(z))$
The same approach should be sufficient for your problem.
(whuber's approach gets you there faster though)
answered Mar 17 '15 at 14:37
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