How to find a partial derivative from such a complicated object

[Giovanni1]

Hi guys.

I have the following equilibrium equation from which I want to extract \(\displaystyle \d{\theta}{\nu}\)


\(\displaystyle z+\theta m(\theta)[\,\frac{\int_{0}^{n^*} \,W(n)g(n)dn+h(n^{*})W(n^{*})G^{*}}{1-(1-h(n^{*}))G^{*}}-U\,]=-c+m(\theta)J'(0)\)

Where \(\displaystyle \nu, z, c, n, r, \delta, \xi\) are parameters, m(.), w(.) and h(.) are functions, and G(.) is a CDF. W(n), J(n) and U are Bellman equations of the following form:

\(\displaystyle (r+\delta+\xi)W(n)=w(n)+[m(\theta)-\delta n]W'(n)+(\delta+\xi)U
\)


\(\displaystyle (r+\delta)J(n)=F(n)+\nu-nw(n)+max\{0;-c+m(\theta)J'(n)\}-\delta nJ'(n) \)\(\displaystyle rU=z+\theta m(\theta)[\,\frac{{\int_{0}^{n^*} \,}W(n)g(n)dn+h(n^{*})W(n^{*})G^{*}}{1-(1-h(n^{*}))G^{*}}-U\,]
\)Moreover, \(\displaystyle W'(n), J'(n)\) are with respect to \(\displaystyle n\in[0,\infty)\)

Thank you

 

[jvicens]

for sharing your equilibrium equation with us. To extract \d{\theta}{\nu} from this equation, we can use the chain rule and differentiate both sides with respect to \nu.

Starting with the left side of the equation:

\d{}{\nu}\left(z+\theta m(\theta)\left[\frac{\int_{0}^{n^*}W(n)g(n)dn+h(n^{*})W(n^{*})G^{*}}{1-(1-h(n^{*}))G^{*}}-U\right]\right)

Applying the chain rule, we get:

\d{}{\nu}\left(z\right)+\d{}{\nu}\left(\theta m(\theta)\right)\left[\frac{\int_{0}^{n^*}W(n)g(n)dn+h(n^{*})W(n^{*})G^{*}}{1-(1-h(n^{*}))G^{*}}-U\right]+\theta m(\theta)\left[\d{}{\nu}\left(\frac{\int_{0}^{n^*}W(n)g(n)dn+h(n^{*})W(n^{*})G^{*}}{1-(1-h(n^{*}))G^{*}}\right)\right]

Using the quotient rule, we can further simplify the last term:

\theta m(\theta)\left[\frac{\d{}{\nu}\left(\int_{0}^{n^*}W(n)g(n)dn+h(n^{*})W(n^{*})G^{*}\right)(1-(1-h(n^{*}))G^{*})-\left(\int_{0}^{n^*}W(n)g(n)dn+h(n^{*})W(n^{*})G^{*}\right)\d{}{\nu}\left((1-(1-h(n^{*}))G^{*}\right)}{(1-(1-h(n^{*}))G^{*})^2}\right]

Now, moving on to the right side of the equation:

\d{}{\nu}\left(-c+m(\theta)J'(0)\right)

Applying the chain rule and using the fact that J'(0) = 0, we get:

\d{}{\nu}\left(-c\right)+\d{}{\nu}\left(m(\theta)\right)J'(0)

Combining the left and right sides, we