Optimal switch from a fossil-fueled to an electric vehicle

Abstract

In this paper we propose and solve a real options model for the optimal adoption of an electric vehicle. A policymaker promotes the abeyance of fossil-fueled vehicles through an incentive, and the representative fossil-fueled vehicle’s owner decides the time at which buying an electric vehicle, while minimizing a certain expected cost. This involves a combination of various types of costs: the stochastic opportunity cost of driving one-unit distance with a traditional fossil-fueled vehicle instead of an electric one, the cost associated to traffic bans, and the net purchase cost. After determining the optimal switching time and the minimal cost function for a general diffusive opportunity cost, we specialize to the case of a mean-reverting process. In such a setting, we provide a model calibration on real data from Italy, and we study the dependency of the optimal switching time with respect to the model’s parameters. Moreover, we study the effect of traffic bans and incentive on the expected optimal switching time. We observe that incentive and traffic bans on fossil-fueled transport can be used as effective tools in the hand of the policymaker to encourage the adoption of electric vehicles and hence to reduce air pollution.

Appendices

A. Facts on the underlying diffusion

Here we collect some properties of the process X. We refer the reader to Ch. II in Borodin and Salminen (2002) for further details. For some reference point $\tilde{x}\in \mathcal {I}$, we introduce the derivative of the scale function of $\left\{ X_{t}^{x}\right\} _{t\ge 0}$ as

$$\begin{aligned} S^{\prime }\left( x\right) := \exp \left\{ \displaystyle -\int _{\tilde{x}}^x \frac{2\mu \left( y\right) }{\sigma ^{2}\left( y\right) }dy\right\} , \quad x\in \mathcal {I}\text {.} \end{aligned}$$

(29)

Moreover, we introduce the speed measure density of $\left\{ X_{t}^{x}\right\} _{t\ge 0}$ as

$$\begin{aligned} m^{\prime }\left( x\right) :=\frac{2}{\sigma ^{2}\left( x\right) S^{\prime }\left( x\right) }, \quad x\in \mathcal {I}\text {.} \end{aligned}$$

(30)

For a given parameter $\rho >0$ (representing in the model the subjective discount factor of the fossil-fueled vehicle owner) we introduce the functions $\psi _{\rho }$ and $\varphi _{\rho }$ as the fundamental solutions to the ordinary differential equation (ODE)

$$\begin{aligned} \left( \mathcal {L}_{X}-\rho \right) u\left( x\right) =0 \text {, \ }x\in \mathcal {I}\text {.} \end{aligned}$$

(31)

The function $\psi _{\rho }$ can be chosen to be strictly increasing, while $\varphi _{\rho }$ strictly decreasing; both $\psi _{\rho }$ and $\varphi _{\rho }$ are strictly positive. The Wronskian between $\psi _{\rho }$ and $\varphi _{\rho }$ (normalized by the scale function density) is the positive constant

$$\begin{aligned} W:=\frac{\psi _{\rho }^{\prime }\left( x\right) \varphi _{\rho }\left( x\right) -\psi _{\rho }\left( x\right) \varphi _{\rho }^{\prime }\left( x\right) }{S^{\prime }\left( x\right) }> 0 \text {, } x\in \mathcal {I}\text {.} \end{aligned}$$

For future use, note that, by the linear independence of $\psi _{\rho }$ and $\varphi _{\rho }$, any solution to (31) can be written as

$$\begin{aligned} u\left( x\right) =A\psi _{\rho }\left( x\right) +B\varphi _{\rho }\left( x\right) \text {,} x \in \mathcal {I}, \end{aligned}$$

for some suitable parameters A and B.

We now recall additional properties of the fundamental solution to (31) $\psi _{\rho }$ and $\varphi _{\rho }$. The fact that $\underline{x}$ and $\overline{x}$ are assumed to be natural (i.e., unattainable) translates into the analytic conditions:

$$\begin{aligned}&\lim _{x \downarrow \underline{x}}\psi (x) = 0,\,\,\,\,\lim _{x \downarrow \underline{x}}\varphi (x) = + \infty ,\,\,\,\,\lim _{x \uparrow \overline{x}}\psi (x) = + \infty ,\,\,\,\,\lim _{x \uparrow \overline{x}}\varphi (x) = 0, \end{aligned}$$

(32)

$$\begin{aligned}&\lim _{x \downarrow \underline{x}}\frac{\psi '(x)}{S'(x)} = 0,\,\,\,\,\lim _{x \downarrow \underline{x}}\frac{\varphi '(x)}{S'(x)} = -\infty ,\,\,\,\,\lim _{x \uparrow \overline{x}}\frac{\psi '(x)}{S'(x)} = + \infty ,\,\,\,\,\lim _{x \uparrow \overline{x}}\frac{\varphi '(x)}{S'(x)} = 0. \end{aligned}$$

(33)

Furthermore, for any $\underline{x}<\alpha<\beta <\bar{x}$, one has

$$\begin{aligned} \frac{\psi _{\rho }^{\prime }\left( \beta \right) }{S^{\prime }\left( \beta \right) }-\frac{\psi _{\rho }^{\prime }\left( \alpha \right) }{S^{\prime }\left( \alpha \right) }=\rho \int _{\alpha }^{\beta }\psi _{\rho }\left( y\right) m^{\prime }\left( y\right) dy\text {,} \end{aligned}$$

(34)

and

$$\begin{aligned} \frac{\varphi _{\rho }^{\prime }\left( \beta \right) }{S^{\prime }\left( \beta \right) }-\frac{\varphi _{\rho }^{\prime }\left( \alpha \right) }{S^{\prime }\left( \alpha \right) }= \rho \int _{\alpha }^{\beta }\varphi _{\rho }\left( y\right) m^{\prime }\left( y\right) dy\text {.} \end{aligned}$$

(35)

Finally, it is also worth noticing the probabilistic representation of the fundamental solutions $\psi _{\rho }$ and $\varphi _{\rho }$ in terms of the Laplace transform of hitting times. Letting $\tau _y := \inf \{t \ge 0: X_t =y\}$, $x, y \in \mathcal {I}$, then

$$\begin{aligned} \mathbb {E}_{x}\left[ e^{-\rho \tau _y}\right] =\left\{ \begin{array} [c]{ccc} \frac{\displaystyle \psi _{\rho }\left( x\right) }{\displaystyle \psi _{\rho }\left( y\right) } &{} \text {for} &{} x<y,\\ \frac{\displaystyle \varphi _{\rho }\left( x\right) }{\displaystyle \varphi _{\rho }\left( y\right) } &{} \text {for} &{} x>y. \end{array} \right. \end{aligned}$$

(36)

B. Proof of Theorem 21

Proof

We here prove Theorem 21 by following arguments and techniques as those in Alvarez (2001), among others.

Step 1. We start with the most relevant case in which $\lim _{x \rightarrow \underline{x}} \left( \ell x + \lambda c \right)< \rho \left( I-k\right) <\lim _{x \rightarrow \overline{x}} \left( \ell x + \lambda c \right)$. As already discussed, since $x \mapsto \ell x + \lambda c$ is increasing, we expect that the optimal switching rule is of the form $\tau ^* = \inf \{t \ge 0: X_t^x \ge x^*\}$, for some $x^* \in \mathcal {I}$ to be found. This guess leads to the candidate value function $\widehat{\mathcal {U}}$ given by

$$\begin{aligned} \widehat{\mathcal {U}}\left( x\right) :=\left\{ \begin{array} [c]{lll} \left( I-k-\widehat{V}\left( x^* \right) \right) \mathbb {E}_{x}\left[ e^{-\rho \tau ^{*}}\right] \text {,} &{} \text {for} &{} x<x^{*} \text {,}\\ I-k-\widehat{V}\left( x\right) \text {,} &{} \text {for} &{} x\ge x^{*} \text {.} \end{array} \right. \end{aligned}$$

(37)

Exploiting the probabilistic representation of $\mathbb {E}_{x}\left[ e^{-\rho \tau ^*}\right]$ as in (36), we can write

$$\begin{aligned} \widehat{\mathcal {U}}\left( x\right) =\left\{ \begin{array} [c]{lll} \left( I-k-\widehat{V}\left( x^* \right) \right) \frac{\psi _{\rho }\left( x\right) }{\psi _{\rho }\left( x^{*}\right) }\text {,} &{} \text {for} &{} x<x^{*} \text {,} \\ I-k-\widehat{V}\left( x\right) \text {,} &{} \text {for} &{} x\ge x^{*} \text {.} \end{array} \right. \end{aligned}$$

(38)

Notice that $\widehat{\mathcal {U}}$ is already continuous at $x^*$ by construction. In order to determine a candidate for the threshold $x^*$, we impose that $\widehat{\mathcal {U}}$ is $C^1$ at $x=x^*$; i.e., $\widehat{\mathcal {U}}'\left( x^*-\right) =\widehat{\mathcal {U}}'\left( x^*+ \right)$, which in turn leads to

$$\begin{aligned} - \left( I-k-\widehat{V}\left( x^* \right) \right) \psi _{\rho }^{\prime }\left( x^{*}\right) + \left( I-k-\widehat{V}\right) ^{\prime }\left( x^{*}\right) \psi _{\rho }\left( x^{*}\right) = 0. \end{aligned}$$

Dividing the latter by $S'\left( x \right)$ (cf. (29)) we find

$$\begin{aligned} \frac{\left( I-k-\widehat{V}\right) ^{\prime }\left( x^{*}\right) \psi _{\rho }\left( x^{*}\right) }{S^{\prime }\left( x^{*}\right) } - \frac{\left( I-k-\widehat{V}\left( x^* \right) \right) \psi _{\rho }^{\prime }\left( x^{*}\right) }{S^{\prime } \left( x^* \right) } =0\text {.} \end{aligned}$$

(39)

Letting $A: \mathcal {I}\rightarrow \mathbb {R}$ be such that

$$\begin{aligned} A\left( x \right) := \frac{ \psi _{\rho }\left( x \right) \left( I-k-\widehat{V}\right) ^{\prime }\left( x \right) - \left( I-k-\widehat{V}\left( x \right) \right) \psi _{\rho }^{\prime }\left( x \right) }{S^{\prime } \left( x \right) } \end{aligned}$$

(40)

we have that (39) is equivalent to $A\left( x^* \right) =0$.

Using the fact that $S^{\prime }$ solves $\left( \mathcal {L}_X S' \right) \left( x \right) =0$, and the fact that $\psi _{\rho }$ solves (31), some algebra shows that

$$\begin{aligned} A^{\prime }(x)= \psi _{\rho } \left( x \right) m'\left( x\right) \left( \mathcal {L}_X - \rho \right) \left( I-k - \widehat{V}\right) \left( x \right) \text {.} \end{aligned}$$

Thanks to (13), we have that $\lim _{x \rightarrow \underline{x}} A\left( x \right) =0$; hence, by the fundamental theorem of calculus, for any $x \in \mathcal {I}$, we have

$$\begin{aligned} A\left( x \right) = \int _{\underline{x}}^x \psi _{\rho } \left( y \right) m'\left( y\right) \left( \mathcal {L}_X - \rho \right) \left( I-k - \widehat{V}\right) \left( y \right) dy \text {.} \end{aligned}$$

(41)

Since $\left( \mathcal {L}_X - \rho \right) \widehat{V}\left( x \right) =- \left( \ell x + \lambda c \right)$, we can write from (41)

$$\begin{aligned} A\left( x \right) = \int _{\underline{x}}^x \psi _{\rho } \left( y \right) m'\left( y\right) \left( - \rho \left( I-k \right) + \ell y + \lambda c \right) dy \text {,} x \in \mathcal {I}\text {.} \end{aligned}$$

(42)

Because it must be $A\left( x^* \right) =0$, then we obtain the equation for $x^*$

$$\begin{aligned} \int _{\underline{x}}^{x^*} \psi _{\rho } \left( y \right) m'\left( y\right) \left( -\rho \left( I-k \right) + \ell y + \lambda c \right) dy =0 \text {.} \end{aligned}$$

(43)

Step 2. We now show that there exists a unique $x^*$ solving (43) such that $x^* > \hat{x}$ with

$$\begin{aligned} \hat{x} := \frac{1}{\ell } \left( \rho \left( I-k\right) - \lambda c \right) \text {.} \end{aligned}$$

With reference to (42), observe that $A\left( \hat{x} \right) <0$ because $y \mapsto \left( - \rho \left( I-k\right) \right.$ $\left. + \ell x + \lambda c \right)$ is increasing and null in $\hat{x}$. Using again (42) one finds

$$\begin{aligned} A^{\prime } \left( x \right) = \psi _{\rho } \left( x \right) m^{\prime } \left( x \right) \left( - \rho \left( I-k\right) + \ell x + \lambda c \right) \end{aligned}$$

and we observe that $A^{\prime } \left( x \right) >0$ $\forall x > \hat{x}$ and $A^{\prime } \left( x \right) < 0$ $\forall x < \hat{x}$.

Moreover, given $x>\hat{x}+ \delta$, for some $\delta >0$, the integral mean-value theorem and (41) give for some $\xi \in \left( \hat{x}+ \delta , x\right)$

$$\begin{aligned}&A\left( x \right) = \int _{\underline{x}}^{x} \psi _{\rho } \left( y \right) m'\left( y\right) \left( - \rho \left( I-k\right) + \ell y + \lambda c \right) dy \\&= \int _{\underline{x}}^{\hat{x}+\delta } \psi _{\rho } \left( y \right) m'\left( y\right) \left( - \rho \left( I-k \right) + \ell y + \lambda c \right) dy \\&\quad + \int _{\hat{x}+\delta }^{x} \psi _{\rho } \left( y \right) m'\left( y\right) \left( - \rho \left( I-k \right) + \ell y +\lambda c \right) dy \\&=\int _{\underline{x}}^{\hat{x}+\delta }\psi _{\rho }\left( y\right) m^{\prime }\left( y\right) \left( - \rho \left( I-k\right) + \ell y -\lambda c\right) dy\\&\quad +\left( \frac{ -\rho \left( I-k\right) + \ell \xi + \lambda c}{\rho }\right) \int _{\hat{x}+\delta }^{x}\rho \psi _{\rho }\left( y\right) m^{\prime }\left( y\right) dy \\&=\int _{\underline{x}}^{\hat{x}+\delta }\psi _{\rho }\left( y\right) m^{\prime }\left( y\right) \left( - \rho \left( I-k\right) +\ell y+ \lambda c\right) dy\\&\quad + \left( \frac{ - \rho \left( I-k\right) + \ell \xi + \lambda c}{\rho } \right) \left( \frac{\psi _{\rho }^{\prime }\left( x\right) }{S^{\prime }\left( x\right) }-\frac{\psi _{\rho }^{\prime }\left( \hat{x}+\delta \right) }{S^{\prime }\left( \hat{x} + \delta \right) }\right) \text {.} \end{aligned}$$

Since $- \rho \left( I-k\right) + \ell \left( \hat{x} + \delta \right) + \lambda c >0$ and $\lim _{x \uparrow \overline{x}} \frac{\psi _{\rho }^{\prime }\left( x\right) }{S^{\prime } \left( x \right) } = + \infty ,$ we have that $\lim _{x \uparrow \overline{x}} A\left( x \right) = +\infty .$ This fact, together with $A\left( \hat{x}\right) <0$ and $A^{\prime }\left( x \right) >0$ $\forall x > \hat{x}$, leads to the existence of a unique $x^* >\hat{x}$ such that $A\left( x^* \right) =0$; that is, satisfying (42).

Step 3. We now prove that the $C^1$-function $\widehat{\mathcal {U}}$ of (14) is such that

$$\begin{aligned} (a)\,\,\, \left( \mathcal {L}_{X}-\rho \right) \widehat{\mathcal {U}}\left( x\right) =0 \text {on} x<x^{*} \qquad \text {and} \qquad (b) \,\,\, \widehat{\mathcal {U}}\left( x\right) =I-k-\widehat{V}\left( x\right) \text {on} x\ge x^{*}, \end{aligned}$$

as well as

(c):

$\widehat{\mathcal {U}}\left( x \right) \le I-k - \widehat{V}\left( x \right)$ $\forall x < x^*$,

(d):

$\left( \mathcal {L}_{X}-\rho \right) \widehat{\mathcal {U}}\left( x\right) \ge 0$ $\forall x >x^*$.

Since (a) and (b) above are verified by construction, it thus remains to prove (c) and (d).

Proof of (c). Given (14) it is enough to show

$$\begin{aligned} \frac{I-k-\widehat{V}\left( x^*\right) }{\psi _{\rho }\left( x^{*}\right) }\le \frac{I-k-\widehat{V}\left( x\right) }{\psi _{\rho }\left( x\right) }\text {,} \forall x < x^*\text {.} \end{aligned}$$

(44)

First of all, we notice that $x^*$ is such that

$$\begin{aligned} \left. \frac{d}{dy} \left( \frac{I-k-\widehat{V}\left( y\right) }{\psi _{\rho }\left( y\right) }\right) \right| _{y=x^{*} }= 0 \end{aligned}$$

(45)

because

$$\begin{aligned} \left. \frac{d}{dy} \left( \frac{I-k-\widehat{V} }{\psi _{\rho }\left( y\right) }\right) \right| _{y=x^{*} }&=\left. \frac{\left( I-k-\widehat{V}\right) ^{\prime }\left( y\right) \psi _{\rho }\left( y\right) -\left( I-k-\widehat{V}\right) \left( y\right) \psi _{\rho }^{\prime }\left( y\right) }{\left( \psi _{\rho }\left( y\right) \right) ^{2}}\right| _{y=x^{*}} \\&=\left. A\left( y\right) \cdot \frac{W\cdot S^{\prime }\left( y \right) }{\left( \psi _{\rho }\left( y \right) \right) ^{2}} \right| _{y=x^*}=0\text {,} \end{aligned}$$

due to (45). Moreover, by (41),

$$\begin{aligned} \left. \frac{d^2}{dy^2} \left( \frac{I-k-\widehat{V}}{\psi _{\rho }}\right) \left( y\right) \right| _{y=x^{*}}=\left. \left[ A \left( y\right) \cdot \left( \frac{W\cdot S^{\prime }\left( y\right) }{\left( \psi _{\rho }\left( y\right) \right) ^{2}}\right) ^{\prime }+A^{\prime }\left( y\right) \cdot \frac{W\cdot S^{\prime }\left( y\right) }{\left( \psi _{\rho }\left( y\right) \right) ^{2}}\right] \right| _{y=x^{*}} \text {.} \end{aligned}$$

Now, since $A \left( x^* \right) =0$, using again (41) and the fact that $x^* >\hat{x}$ we have

$$\begin{aligned}&\left. \frac{d^2}{dy^2} \left( \frac{I-k-\widehat{V}}{\psi _{\rho }}\right) \left( y\right) \right| _{y=x^{*}} = A^{\prime }\left( x^{*}\right) \cdot \frac{W\cdot S^{\prime }\left( x^{*}\right) }{\left( \psi _{\rho }\left( x^{*}\right) \right) ^{2}} \\ =&\frac{W\cdot S^{\prime }\left( x^{*}\right) }{\left( \psi _{\rho }\left( x^{*}\right) \right) ^{2}}\cdot \left. \frac{d}{dy}\left[ \int _{\underline{x}}^{y}\psi _{\rho }\left( z\right) m^{\prime }\left( z\right) \left( - \rho \left( I-k\right) +\ell z + \lambda c \right) dz\right] \right| _{y=x^{*}} \\&= \psi _{\rho }\left( x^{*}\right) m^{\prime }\left( x^{*}\right) \left( - \rho \left( I-k\right) + \ell x^{*} + \lambda c\right) >0 \text {.} \end{aligned}$$

This proves that the function $\frac{I-k-\widehat{V}\left( x\right) }{\psi _{\rho }\left( x\right) }$ attains a minimum at $x=x^*$ and thus gives (44).

Proof of (d). For any $x> x^*$ we have

$$\begin{aligned} \left( \mathcal {L}_{X}-\rho \right) \widehat{\mathcal {U}}\left( x\right) =\left( \mathcal {L}_{X}-\rho \right) \left( I-k-\widehat{V}\right) \left( x\right) =-\rho \left( I-k\right) +\ell x+\lambda c>0\text {,} \end{aligned}$$

where in the second equality we use $\left( \mathcal {L}_{X}-\rho \right) \widehat{V}\left( x\right) = - \left( \ell x+\lambda c\right)$ and for the last inequality we use the fact that $x^* > \hat{x}$.

Step 4. The final verification of the optimality of $\widehat{\mathcal {U}}$ and of $\tau ^* =\inf \{ t \ge 0: X_t^x \ge x^*\}$ follows by a standard application of Itô’s formula (up to a localization argument) and the use of inequalities (a)–(d) above. We refer to Peskir and Shiryaev (2006) for proofs in related settings.

Step 5. The proof of the fact that $\widehat{\mathcal {U}}=0$ and $\widehat{\mathcal {U}}=I-k - \widehat{V}$ in the other two cases easily follows by noticing that $V\left( x \right) = \inf _{\tau \ge 0} \mathbb {E}_x \left[ \int _0^{\tau } e^{-\rho t} \left( \ell X_t + \lambda c - \rho \left( I-k \right) \right) dt \right] + \left( I-k \right)$ and (8).