Solve ODE for ETI beliefs

[john-p-ryan]

In the case of variable ETI, we get an extra term for the JJZ formula:

$$g(z) = 1 + \theta_z \varepsilon(z) \frac{T'(z)}{1-T'(z)} + \varepsilon(z) \frac{zT''(z)}{(1-T'(z))^2}+ \varepsilon'(z)\frac{zT'(z)}{1-T'(z)}$$

If $T(z)$ and $g(z)$ are exogenous, we can view the endogenous $\varepsilon(z)$ as a first order differential equation from the JJZ formula with variable ETI above. This equation can be written as

$$ \varepsilon'(z)\left[\frac{zT'(z)}{1-T'(z)}\right] + \varepsilon(z)\left[\theta_z \frac{T'(z)}{1-T'(z)} +\frac{zT''(z)}{(1-T'(z))^2}\right]+ (1-g(z)) = 0$$

This is a differential equation of the form

$$ a(x) f'(x) + b(x) f(x) + c(x) = 0 $$

Renormalizing,

$$ f'(x) + P(x) f(x) = Q(x)$$

where $P(x) = \frac{b(x)}{a(x)}$ and $Q(x) = -\frac{c(x)}{a(x)}$. Let $\mu(x) = \exp(\int P(x) dx)$, and note that $\mu'(x) = \mu(x) P(x)$. Next, multiply by $\mu(x)$:

$$ \mu(x) f'(x) + \mu(x) P(x) f(x) = \mu(x) Q(x) $$ $$ \implies \frac{d}{dx}\left[\mu(x) f(x)\right] = \mu(x)Q(x) $$ $$ \implies f(x) = \frac{1}{\mu(x)}\left[\int \mu(x) Q(x) dx + C\right]$$ $$ = \exp\left(-\int \frac{b(x)}{a(x)}dx\right)\left[ \int \exp\left(\int\frac{b(y)}{a(y)}dy\right)dx + C\right] $$

Thus, our ETI formula becomes

$$\varepsilon(z) = \frac{1}{\mu(z)} \left[ \varepsilon_0 + \int_0^{z} Q(z') \mu(z')dz' \right] $$

Where:

$$\mu(z) = \exp \left(\int_0^z \frac{1}{z} + \frac{f'(z)}{f(z)} + \frac{T''(z)}{T'(z)(1-T'(z))}dz\right)$$ $$Q(z) = \frac{(g(z) - 1) (1-T'(z))}{z T'(z)}$$

Thus, in the thought experiment in which we ask "What beliefs on the ETI does Clinton need to justify her policies under Trump's social welfare weights?", we get the following result (assuming Trump's proposal is under $\varepsilon_z = .25$):

The interpretation is that Clinton needs lower beliefs on the ETI to justify her policies assuming Trump's weights. However, the differences are modest. Clinton can justify her policy differences by only believing an average ETI of about $.242$ rather than $.25$.

[codecov-commenter]

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[jdebacker]

@john-p-ryan Thanks for working this out. LGTM. Merging.