Valuation of Cryptocurrency Without Intrinsic Value: A Promise of Future Payment System and Implications to De-dollarization

Abstract

A random search-based model is proposed to study the valuation of cryptocurrency. We provide a unique approach to valuing cryptocurrency according to people’s acceptance probability of cryptocurrency. Although cryptocurrency is generally considered without intrinsic value, it can have value in an exchange economy. Cryptocurrency has shown huge potential as a future payment system to replace the US dollar’s position in international trade, our model, however, indicates that the value of cryptocurrency is fundamentally unstable, and thus, it has a long way to go before fulfilling its mission.

Appendices

Appendix A: Symbols and Notations

$:

The fiat currency, the US dollar

B:

The cryptocurrency, representing the most popular cryptocurrency Bitcoin

r :

The (risk-free) discount rate, characterizing the time preference of agents.

u :

The utility from consuming one unit of the preferred consumption goods by an agent

x :

The probability of an agent willing to accept randomly met consumption goods

x ² :

The probability of the “double coincidence of wants” for two agents

σ :

The meeting rate of agents in the exchange sector

Π _$ :

The probability of accepting the dollar by a random agent with one unit of consumption goods

Π _B :

The probability of accepting the cryptocurrency by a random agent with one unit of consumption goods, also interpreted as the value of the cryptocurrency

π _$ :

The best choice function by a typical agent with consumption goods when facing the opportunity to sell the consumption goods for the dollar

π _B :

The best choice function by a typical agent with consumption goods when facing the opportunity to sell the consumption goods for the cryptocurrency

ε _$ :

The transaction cost between consumption goods and the dollar balance, paid by the buyer

ε _B :

The transaction cost between consumption goods and the cryptocurrency balance, paid by the buyer

ε ₁ :

The transaction cost for barter, paid by the buyer

α :

The arrival rate of a Poisson process, with which an agent produces one unit of consumption goods randomly drawn from the set of all consumption goods, also representing the economy’s productivity

C _$ :

The total units of the dollar balance

C _B :

The total units of the cryptocurrency balance

N ₀ :

The number of agents in the production sector

N ₁ :

The number of agents with one unit of consumption goods in the exchange sector

N _$ :

The number of agents with one unit of the dollar balance in the exchange sector

N _B :

The number of agents with one unit of the cryptocurrency balance in the exchange sector

µ ₁ :

The fraction of agents in the exchange sector with one unit of consumption goods

µ _$ :

The fraction of agents in the exchange sector with one unit of the dollar balance

µ _B :

The fraction of agents in the exchange sector with one unit of the cryptocurrency balance

V ₀ :

The value of an agent without consumption goods in the production sector

V ₁ :

The value of an agent with one unit of consumption goods in the exchange sector

V _$ :

The value of an agent with one unit of the dollar balance in the exchange sector

V _B :

The value of an agent with one unit of the cryptocurrency balance in the exchange sector

Appendix B: Proofs of Propositions

Proof of Proposition 1

If µ_B is known, according to the definitions of µ_$ in Eq. (3) and µ_B in Eq. (4), we obtain:

$$ \frac{{\mu_{\$ } }}{{\mu_{{\text{B}}} }} = \frac{{C_{\$ } }}{{C_{{\text{B}}} }} $$

(B-1)

Then Eq. (15) is derived as $\mu_{\$ } = \frac{{C_{\$ } }}{{C_{{\text{B}}} }}\mu_{{\text{B}}}$.

According to the definition of µ₁ in Eq. (5): µ₁=1−µ_$−µ_B, then, Eq. (16) is derived as:

$$ \mu_{1} = 1 - \frac{{C_{\$ } }}{{C_{{\text{B}}} }}\mu_{{\text{B}}} - \mu_{{\text{B}}} = 1 - \frac{{C_{{\text{B}}} + C_{\$ } }}{{C_{{\text{B}}} }}\mu_{{\text{B}}} $$

(16)

Next step, to derive the formulae for N₁ and N₀ when we know that N_$ =C_$ and N_B=C_B.

Apply the definition of µ_B in Eq. (4) again and then substitute C_$ for N_$ and C_B for N_B:

$$ \mu_{{\text{B}}} = \frac{{N_{{\text{B}}} }}{{N_{1} + N_{\$ } + N_{{\text{B}}} }} = \frac{{C_{{\text{B}}} }}{{N_{1} + C_{\$ } + C_{{\text{B}}} }} $$

(B-2)

Solve for N₁ from Eq. (B-2), we obtain:

$$ N_{1} = \frac{{\left( {1 - \mu_{{\text{B}}} } \right)C_{{\text{B}}} - \mu_{{\text{B}}} C_{\$ } }}{{\mu_{{\text{B}}} }} $$

(17)

Plug N_$ =C_$, N_B=C_B, and $N_{1} = \frac{{\left( {1 - \mu_{{\text{B}}} } \right)C_{{\text{B}}} - \mu_{{\text{B}}} C_{\$ } }}{{\mu_{{\text{B}}} }}$ into Eq. (6): N₀+N₁+ N_$ + N_B =1, we obtain:

$$ N_{0} = 1 - N_{1} - N_{\$ } - N_{{\text{B}}} = 1 - \frac{{\left( {1 - \mu_{{\text{B}}} } \right)C_{{\text{B}}} - \mu_{{\text{B}}} C_{\$ } }}{{\mu_{{\text{B}}} }} - C_{\$ } - C_{{\text{B}}} = \frac{{\mu_{{\text{B}}} - C_{{\text{B}}} }}{{\mu_{{\text{B}}} }} $$

(18)

Proof of Proposition 2

(1) We first prove that only Eq. (19) is independent and the other three equations (Eqs. 20, 21, and 22) are redundant.

Equation (21) can be reduced to µ₁N_$ = µ_$N₁, which exactly comes from the definitions of µ₁ and µ_$, i.e.,

$$ \frac{{\mu_{1} }}{{\mu_{\$ } }} = \frac{{N_{1} }}{{N_{\$ } }} $$

(B-3)

In the same way, Eq. (22) can be reduced to µ₁N_B = µ_BN₁, which exactly comes from the definitions of µ₁ and µ_B, i.e.,

$$ \frac{{\mu_{1} }}{{\mu_{{\text{B}}} }} = \frac{{N_{1} }}{{N_{{\text{B}}} }} $$

(B-4)

Equation (20) can be rearranged as:

$$ \sigma \mu_{1} x^{2} N_{1} + \sigma \mu_{\$ } x \, \Pi_{\$ } N_{1} + \sigma \mu_{{\text{B}}} x\Pi_{{\text{B}}} N_{1} = \alpha \, N_{0} $$

(B-5)

Substitute Eqs. (21) and (22) for the second and the third term of Eq. (B-5):

$$ \sigma \mu_{1} x^{2} N_{1} + \sigma \mu_{1} x \, \Pi_{\$ } N_{\$ } + \sigma \mu_{1} x \, \Pi_{{\text{B}}} N_{{\text{B}}} = \alpha \, N_{0} $$

(B-6)

Equation (B-6) is exactly Eq. (19). Thus, only Eq. (19) is independent.

(2) Then we show how to apply Eq. (19) to pin down µ_B.

Plug Eqs. (16), (17), and (18) into Eq. (19), we obtain:

$$ \alpha \frac{{\mu_{{\text{B}}} - C_{{\text{B}}} }}{{\mu_{{\text{B}}} }} = \sigma \left( {1 - \frac{{C_{{\text{B}}} + C_{\$ } }}{{C_{{\text{B}}} }}\mu_{{\text{B}}} } \right)x^{2} \frac{{\left( {1 - \mu_{{\text{B}}} } \right)C_{{\text{B}}} - \mu_{{\text{B}}} C_{\$ } }}{{\mu_{{\text{B}}} }}\; + \;\sigma \left( {1 - \frac{{C_{{\text{B}}} + C_{\$ } }}{{C_{{\text{B}}} }}\mu_{{\text{B}}} } \right)x \, \Pi_{\$ } C_{\$ } + \sigma \left( {1 - \frac{{C_{{\text{B}}} + C_{\$ } }}{{C_{{\text{B}}} }}\mu_{B} } \right)x \, \Pi_{{\text{B}}} C_{{\text{B}}} $$

(B-7)

Equation (B-7) can further be reduced:

$$ \alpha \frac{{\mu_{{\text{B}}} - C_{{\text{B}}} }}{{\mu_{{\text{B}}} }} = \sigma x^{2} \left( {1 - \frac{{C_{{\text{B}}} + C_{\$ } }}{{C_{{\text{B}}} }}\mu_{{\text{B}}} } \right)\frac{{\left( {1 - \mu_{{\text{B}}} } \right)C_{{\text{B}}} - \mu_{{\text{B}}} C_{\$ } }}{{\mu_{{\text{B}}} }} + \sigma x\left( {1 - \frac{{C_{{\text{B}}} + C_{\$ } }}{{C_{{\text{B}}} }}\mu_{{\text{B}}} } \right)(\Pi_{\$ } C_{\$ } + \Pi_{{\text{B}}} C_{{\text{B}}} ) $$

(B-8)

Finally, we obtain a quadratic equation for µ_B:

$$ \sigma x(C_{{{\text{B}} }} + C_{\$ } )\left[ {x(C_{{{\text{B}} }} + C_{\$ } } \right) - (\Pi_{\$ } C_{\$ } + \Pi_{{\text{B}}} C_{{\text{B }}} )]\mu_{{\text{B}}}^{2} + \left\{ {\sigma xC_{{\text{B}}} \left[ {(\Pi_{\$ } C_{\$ } + \Pi_{{\text{B}}} C_{{{\text{B}} }} } \right) - 2x(C_{{\text{B }}} + C_{\$ } )] - \alpha C_{{\text{B}}} } \right\}\mu_{{\text{B}}} + C_{{\text{B}}}^{2} \left( {\sigma x^{2} + \alpha } \right) = 0 $$

(B-9)

Define: $m = \sigma x(C_{{{\text{B}} }} + C_{\$ } )\left[ {x\left( {C_{{{\text{B}} }} + C_{\$ } } \right) - (\Pi_{\$ } C_{\$ } + \Pi_{{\text{B}}} C_{{{\text{B}} }} )} \right]$

$$ n = \sigma xC_{{\text{B}}} \left[ {\left( {\Pi_{\$ } C_{\$ } + \Pi_{{\text{B}}} C_{{{\text{B}} }} } \right) - 2x(C_{{{\text{B}} }} + C_{\$ } )} \right] - \alpha C_{{\text{B}}} $$

$$ l = C_{{\text{B}}}^{2} \left( {\sigma x^{2} + \alpha } \right) $$

Equation (B-9) can be rewritten as Eq. (23)

$$ m\mu_{{\text{B}}}^{2} + n\mu_{{\text{B}}} + l = 0 $$

(23)

Proof of Proposition 3

Assume (1) ε₁ =ε_$ =ε_B= ε (2) i_$= i_B=0, then the four Bellman equations from Eqs. (7, 8, 9, and 10) can be rewritten as below:

$$ r\;V_{0} = \alpha \;(V_{1} - V_{0} ) $$

(B-10)

$$ r\;V_{1} = \sigma \mu_{1} x^{2} (u - \varepsilon + V_{0} - V_{1} ) + \sigma \mu_{\$ } x\mathop {\max }\limits_{{\pi_{\$ } }} \pi_{\$ } (V_{\$ } - V_{1} ) + \sigma \mu_{{\text{B}}} x\mathop {\max }\limits_{{\pi_{{\text{B}}} }} \pi_{{\text{B}}} (V_{{\text{B}}} - V_{1} ) $$

(B-11)

$$ r\;V_{\$ } = \sigma \mu_{1} x \, \Pi_{\$ } (u - \varepsilon + V_{0} - V_{\$ } ) $$

(B-12)

$$ r\;V_{{\text{B}}} = \sigma \mu_{1} x \, \Pi_{{\text{B}}} (u - \varepsilon + V_{0} - V_{{\text{B}}} ) $$

(B-13)

We need to prove that: V₁ >V_$= V_B=0, if π_$=Π_$=π_B=Π_B=0.

First, plug π_$=Π_$=π_B=Π_B=0 into Eqs. (B-10), (B-11), (B-12), and (B-13), we obtain:

$$ r\;V_{0} = \alpha \;(V_{1} - V_{0} ) $$

(B-10)

$$ r\;V_{1} = \sigma \mu_{1} x^{2} (u - \varepsilon + V_{0} - V_{1} ) $$

(B-14)

$$ r\;V_{\$ } = 0 $$

(B-15)

$$ r\;V_{B} = 0 $$

(B-16)

Equations (B-15) and (B-16) directly show that V_$= V_B=0.

Combining Eqs. (B-10) and (B-14), we can solve for V₁ as below:

$$ V_{1} = \frac{{\left( {r + \alpha } \right)\sigma \mu_{1} x^{2} \left( {u - \varepsilon } \right)}}{{r^{2} + \alpha r + \sigma \mu_{1} x^{2} r}} $$

(B-17)

As long as the utility from consuming the consumption goods is higher than the transaction cost, i.e., u-ε >0, we have V₁ >0= V_$= V_B. Thus, our value functions are well consistent with the optimal behaviors of agents defined in Definition 1.

Proof of Proposition 4

Assume (1) ε₁ =ε_$ =ε_B= ε (2) i_$= i_B=0, re-use the four Bellman equations from Eqs. (B-10) (B-11), (B-12), and (B-13) with π_$=Π_$=1, 0< π_B= Π_B<1, and V_B =V₁, we obtain:

$$ r\;V_{0} = \alpha \;(V_{1} - V_{0} ) $$

(B-10)

$$ r\;V_{1} = \sigma \mu_{1} x^{2} (u - \varepsilon + V_{0} - V_{1} ) + \sigma \mu_{\$ } x(V_{\$ } - V_{1} ) $$

(B-18)

$$ r\;V_{\$ } = \sigma \mu_{1} x(u - \varepsilon + V_{0} - V_{\$ } ) $$

(B-19)

$$ r\;V_{{\text{B}}} = \sigma \mu_{1} x \, \Pi_{{\text{B}}} (u - \varepsilon + V_{0} - V_{{\text{B}}} ) $$

(B-13)

According to Eq. (B-19), we obtain:

$$ V_{\$ } = \frac{{\sigma \mu_{1} x\left( {u - \varepsilon + V_{0} } \right)}}{{r + \sigma \mu_{1} x}} $$

(B-20)

According to Eq. (B-13), we obtain:

$$ V_{{\text{B}}} = \frac{{\sigma \mu_{1} x\Pi_{{\text{B}}} \left( {u - \varepsilon + V_{0} } \right)}}{{r + \sigma \mu_{1} x\Pi_{{\text{B}}} }} $$

(B-21)

Comparing Eq. (20) with (21), we require that: V_$ >V_B, i.e.,

$$ \frac{{\sigma \mu_{1} x\left( {u - \varepsilon + V_{0} } \right)}}{{r + \sigma \mu_{1} x}} > \frac{{\sigma \mu_{1} x\Pi_{{\text{B}}} \left( {u - \varepsilon + V_{0} } \right)}}{{r + \sigma \mu_{1} x\Pi_{{\text{B}}} }} $$

The above condition can be simplified in the following steps:

$$ \frac{1}{{r + \sigma \mu_{1} x}} > \frac{{\Pi_{{\text{B}}} }}{{r + \sigma \mu_{1} x\Pi_{{\text{B}}} }} $$

$$ r + \sigma \mu_{1} x\Pi_{{\text{B}}} > \Pi_{{\text{B}}} \left( {r + \sigma \mu_{1} x} \right) $$

$$ 1 > \Pi_{{\text{B}}} $$

Thus, if 0< Π_B<1, V_B < V_$.

Proof of Corollary 1

The existence of a dollar-dominated equilibrium in Proposition 1 also determines the unique value of Π_B. Now we solve for this value as below.

Since V_B =V₁, substitute V₁ for V_B in Eq. (B-21), we obtain:

$$ V_{1} = \frac{{\sigma \mu_{1} x\Pi_{{\text{B}}} \left( {u - \varepsilon + V_{0} } \right)}}{{r + \sigma \mu_{1} x\Pi_{{\text{B}}} }} = \frac{{\sigma \mu_{1} x\Pi_{{\text{B}}} }}{{r + \sigma \mu_{1} x\Pi_{{\text{B}}} }}\left( {u - \varepsilon + V_{0} } \right) $$

(B-22)

From Eq. (B-18), we obtain:

$$ \left( {r + \sigma \mu_{1} x^{2} + \sigma \mu_{\$ } x} \right)\;V_{1} = \sigma \mu_{1} x^{2} (u - \varepsilon + V_{0} ) + \sigma \mu_{\$ } x \, V_{\$ } $$

(B-23)

Substituting Eq. (B-20) for V_$ in Eq. (B-23), we obtain:

$$ \left( {r + \sigma \mu_{1} x^{2} + \sigma \mu_{\$ } x} \right)\;V_{1} = \sigma \mu_{1} x^{2} (u - \varepsilon + V_{0} ) + \sigma \mu_{\$ } x\frac{{\sigma \mu_{1} x\left( {u - \varepsilon + V_{0} } \right)}}{{r + \sigma \mu_{1} x}} $$

(B-24)

Rearranging Eq. (B-24), we obtain:

$$ V_{1} = \frac{{\sigma \mu_{1} x^{2} \left( {r + \sigma \mu_{1} x + \sigma \mu_{\$ } } \right)}}{{\left( {r + \sigma \mu_{1} x^{2} + \sigma \mu_{\$ } x} \right)\left( {r + \sigma \mu_{1} x} \right)}}\left( {u - \varepsilon + V_{0} } \right) $$

(B-25)

Combining Eq. (B-22) with (B-24), we obtain:

$$\frac{{\sigma \mu_{1} x\Pi_{{\text{B}}} }}{{r + \sigma \mu_{1} x\Pi_{{\text{B}}} }}\left( {u - \varepsilon + V_{0} } \right) = \frac{{\sigma \mu_{1} x^{2} \left( {r + \sigma \mu_{1} x + \sigma \mu_{\$ } } \right)}}{{\left( {r + \sigma \mu_{1} x^{2} + \sigma \mu_{\$ } x} \right)\left( {r + \sigma \mu_{1} x} \right)}}\left( {u - \varepsilon + V_{0} } \right)$$

(B-26)

The coefficients of both sides of Eq. (B-26) are equal, then,

$$ \frac{{\sigma \mu_{1} x\Pi_{{\text{B}}} }}{{r + \sigma \mu_{1} x\Pi_{{\text{B}}} }} = \frac{{\sigma \mu_{1} x^{2} \left( {r + \sigma \mu_{1} x + \sigma \mu_{\$ } } \right)}}{{\left( {r + \sigma \mu_{1} x^{2} + \sigma \mu_{\$ } x} \right)\left( {r + \sigma \mu_{1} x} \right)}} $$

(B-27)

Solving for Π_B from Eq. (B-27), we obtain:

$$ \Pi_{{\text{B}}} = \frac{{x\left( {r + \sigma \mu_{1} x + \sigma \mu_{\$ } } \right)}}{{r + \sigma \mu_{1} x + \sigma \mu_{\$ } x}} $$

(B-28)

Thus, Π_B>0. We still need to prove that: Π_B<1, which requires that:

$$ \frac{{x\left( {r + \sigma \mu_{1} x + \sigma \mu_{\$ } } \right)}}{{r + \sigma \mu_{1} x + \sigma \mu_{\$ } x}} < 1 $$

To solve the above condition, we need x<1. This is true according to the definition of x.