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In [Calvo, Guillermo A., and Maurice Obstfeld. Optimal time-consistent fiscal policy with finite lifetimes. Econometrica: Journal of the Econometric Society (1988): 411-432], the authors derive $$ \mu(b)\equiv\int_{b}^{\infty}u\left[c(b,t)\right]\left[1-F(t-b)\right]e^{-\beta(t-b)}dt $$ as the utility of an individual where $b$ is the birth time of the individual, $c$ is their consumption, $F$ is the CDF of a random variable corresponding to the length of their life, $u$ is their utility, and $\beta$ is their discount rate.

I would like to instead study the case in which there is a terminal time $T>b$ beyond which we do not care about the individual's utility. How can I formulate this as an integral?

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Calvo and Obstfeld's problem

In [Calvo, Guillermo A., and Maurice Obstfeld. Optimal time-consistent fiscal policy with finite lifetimes. Econometrica: Journal of the Econometric Society (1988): 411-432], they study $$ \mu(b) \equiv \int_{b}^{\infty}u\left[c(b,t)\right]\left[1-F(t-b)\right]e^{-\beta(t-b)}dt $$ where $b$ is the birth time of an individual, $c$ is their consumption, $F(n)$ is the CDF of the random variable $\tilde{n}$ corresponding to the length of their life, $u$ is their utility, and $\beta$ is their discount rate.

Assumptions. $F(0)=0$, $\tilde{n}$ is absolutely continuous (with density function $f$), and $u \geq 0$. All measurability requirements for the integrals to be well-defined should also be satisfied.

The definition of $\mu(b)$ is motivated by noting that \begin{align*} \mathbb{E}\left[\int_{b}^{b+\tilde{n}}u\left[c(b,t)\right]e^{-\beta(t-b)}dt\right] & =\int_{0}^{\infty}\left(\int_{b}^{b+n}u\left[c(b,t)\right]e^{-\beta(t-b)}dt\right)f(n)dn\\ & =\int_{0}^{\infty}\left(\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\boldsymbol{1}_{(t-b,\infty)}(n)dt\right)f(n)dn\\ & =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left(\int_{0}^{\infty}f(n)\boldsymbol{1}_{(t-b,\infty)}(n)dn\right)dt\\ & =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\mathbb{P}(n>t-b)dt\\ & =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[1-\mathbb{P}(n\leq t-b)\right]dt\\ & =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[1-F(t-b)\right]dt \equiv \mu(b) \end{align*} where we have used the fact that $\boldsymbol{1}_{(t-b,\infty)}(n)=1$ if and only if $t\leq b+n$ and the Fubini-Tonelli theorem for nonnegative functions to switch the order of integration (I am assuming that $u \geq 0$).

Your problem

For brevity, let $a\wedge b=\min\{a,b\}$ and $a\vee b=\max\{a,b\}$. From my understanding, you want to study the expectation $$ \mathbb{E}\left[\int_{b}^{T\wedge(b+\tilde{n})}u\left[c(b,t)\right]e^{-\beta(t-b)}dt\right] $$ where there is an expiry time $T > b$ beyond which we do not care about the individual's utility from consumption. Note that unlike the original problem, we only track the individual's utility on the time horizon $[b, T \wedge (b + \tilde{n})]$, corresponding to the time between $b$ (birth) and the smaller of $T$ (expiry) and $b + \tilde{n}$ (death).

Remark. Note that this is not the same thing as what you wrote in your original question (in fact, I was unable to understand the expectation you wrote down; we can chat about this if necessary).

One way to handle this would be to split up the expectation into two parts (one corresponding to the individual living beyond $T$ and one not). Then, under the same assumptions as above, \begin{multline} \mathbb{E}\left[\int_{b}^{T\wedge(b+\tilde{n})}u\left[c(b,t)\right]e^{-\beta(t-b)}dt\right]\\ =\mathbb{E}\left[\boldsymbol{1}_{\{T\leq b+\tilde{n}\}}\int_{b}^{T}u\left[c(b,t)\right]e^{-\beta(t-b)}dt+\boldsymbol{1}_{\{T>b+\tilde{n}\}}\int_{b}^{b+\tilde{n}}u\left[c(b,t)\right]e^{-\beta(t-b)}dt\right]\\ =\boxed{\left(1-F(T-b)\right)\int_{b}^{T}u\left[c(b,t)\right]e^{-\beta(t-b)}dt+\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[F(T-b)-F(t-b)\right]dt} \tag{*} \label{eq:result} \end{multline} where the second expectation is obtained by a Fubini's argument as in the previous paragraph: \begin{align*} \mathbb{E}\left[\boldsymbol{1}_{\{T>b+\tilde{n}\}}\int_{b}^{b+\tilde{n}}u\left[c(b,t)\right]e^{-\beta(t-b)}dt\right] & =\int_{0}^{T-b}\left(\int_{b}^{b+n}u\left[c(b,t)\right]e^{-\beta(t-b)}dt\right)f(n)dn\\ & =\int_{0}^{T-b}\left(\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\boldsymbol{1}_{(t-b,\infty)}(n)dt\right)f(n)dn\\ & =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left(\int_{0}^{T-b}\boldsymbol{1}_{(t-b,\infty)}(n)f(n)dn\right)dt\\ & =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\mathbb{P}(t-b\leq n\leq T-b)\\ & =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[F(T-b)-F(t-b)\right]dt. \end{align*}

Relationship to original problem

Note that the second problem is a generalization of the first. You can retrieve the original problem by formally taking $T \rightarrow \infty$: \begin{multline*} =\left(1-F(\infty)\right)\int_{b}^{T}u\left[c(b,t)\right]e^{-\beta(t-b)}dt+\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[F(\infty)-F(t-b)\right]dt\\ =\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[1-F(t-b)\right]dt \equiv \mu(b). \end{multline*} This formal limit can be made rigorous using the dominated convergence theorem:

Proposition. Under the assumptions stipulated above, if $t \mapsto u(c(b,t))$ is of polynomial growth in $t$ and $\beta > 0$, then \eqref{eq:result} converges to $\mu(b)$ as $T \rightarrow \infty$.

Proof. This follows immediately from the fact that $$ \left|\int_{b}^{\infty}u(c(b,t))e^{-\beta(t-b)}\right|\leq\int_{b}^{\infty}\left|u(c(b,t))\right|e^{-\beta(t-b)}dt\leq C\int_{b}^{\infty}\left(1+t^{d}\right)e^{-\beta(t-b)}dt $$ is finite.

Memoryless instantaneous probability of death

Suppose $$ F(t)=\lambda\int_{0}^{s}e^{-\lambda s}ds=1-e^{-\lambda t}. $$ is your mortality CDF (i.e., the survival function is $S(t)=1-F(t)=1-(1-e^{-\lambda t})=e^{-\lambda t}$). In this case, direct substitution gives you the following expression: \begin{multline*} \left(1-F(T-b)\right)\int_{b}^{T}u\left[c(b,t)\right]e^{-\beta(t-b)}dt+\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[F(T-b)-F(t-b)\right]dt\\ =e^{-\lambda(T-b)}\int_{b}^{T}u\left[c(b,t)\right]e^{-\beta(t-b)}dt+\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\beta(t-b)}\left[e^{-\lambda(t-b)}-e^{-\lambda(T-b)}\right]dt \\ = e^{\alpha b}\left(\int_{b}^{\infty}u\left[c(b,t)\right]e^{-\alpha t}dt-e^{-\lambda T}\int_{T}^{\infty}u\left[c(b,t)\right]e^{-\beta t}dt\right) \end{multline*} where $\alpha = \beta + \lambda$. The last line reveals that the problem is equivalent to the difference of two integrals without a terminal time (one with discount factor $\beta$, and one with $\alpha$).

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  • $\begingroup$ Thanks a lot for this proof ! It is really helpful! $\endgroup$ – optimal control Aug 7 '18 at 19:29
  • $\begingroup$ No worries. Unfortunate that I missed the bounty period T_T $\endgroup$ – parsiad Aug 7 '18 at 19:35
  • $\begingroup$ How is that? I am so sorry if it is my fault $\endgroup$ – optimal control Aug 7 '18 at 20:08
  • $\begingroup$ No worries; I was mostly kidding. Out of curiosity, was my interpretation of your question about $T$ correct? If so, can I edit your original question to reflect that? It would be nice to clean it up for future reference. $\endgroup$ – parsiad Aug 7 '18 at 20:18
  • $\begingroup$ Yes, I was also thinking as you did. I just did not achieve to make the calculations. Thanks for that. Just a question: Is it possible to have a such probability $\int_{b}^{T}u\left(c\left(b,t\right)\right)\left[1-F\left(t-b\right)\right]e^{-\beta\left(t-b\right)}dt$ inside a utility function? If we say that there is a memoryless, exponential survival probability. To be honest, I need something like that for the stuff that I work. I do not know if it is possible to justify this in a plausible way. $\endgroup$ – optimal control Aug 14 '18 at 23:46

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