Question

Consider a renewal process having interarrival distribution \(F\) such that $$ F(x)=\frac{1}{2} e^{-x}+\frac{1}{2} e^{-x / 2}, \quad x>0 $$ That is, interarrivals are equally likely to be exponential with mean 1 or exponential with mean 2 . (a) Without any calculations, guess the equilibrium distribution \(F_{e}\) (b) Verify your guess in part (a).

Short answer

Our guess for the equilibrium distribution, \(F_e(x)\), is a combination of two exponential distributions with the same means 1 and 2: \[ F_e(x) = a e^{-x} + b e^{-x/2}, x > 0 \] After verifying our guess using the renewal equation and solving for coefficients, we find the equilibrium distribution to be: \[ F_e(x) = \frac{1}{2} e^{-x} + \frac{1}{2} e^{-x/2}, \quad x > 0 \]

Step-by-step solution

Understanding Equilibrium Distribution

Equilibrium distribution represents the long-term behavior of the renewal process when the initial transient behavior has disappeared. In essence, it characterizes the steady-state probability distribution over the states.

Guess the Equilibrium Distribution

Given the interarrival distribution F(x): \[ F(x) = \frac{1}{2} e^{-x} + \frac{1}{2} e^{-x/2}, x > 0 \] We can see that interarrivals have equal probabilities of having exponential distribution with mean 1 or mean 2. Therefore, we can make an educated guess that the equilibrium distribution F_e should have the same property, i.e., the form of F_e should also be a combination of two exponential distributions with the same means: \[ F_e(x) = a e^{-x} + b e^{-x/2}, x > 0 \] Here, we still need to verify our guess and find the correct coefficients `a` and `b`.

Verify the Guess

We know that for a renewal process, the equilibrium distribution can be found using the renewal equation: \[ F_e(x) = F(x) + \int_{0}^{x} F_e(x-u) \,dF(u) \] Let's plug our guess for F_e(x) and F(x) into this equation: \[ a e^{-x} + b e^{-x/2} = \frac{1}{2}e^{-x} + \frac{1}{2}e^{-x/2} + \int_{0}^{x} (a e^{-(x-u)} + b e^{-(x-u)/2}) \, d\left(\frac{1}{2} e^{-u} + \frac{1}{2} e^{-u/2}\right) \] Differentiating both sides with respect to `x`, we get: \[ -a e^{-x} - \frac{1}{2}b e^{-x/2} = (a e^{-x} + b e^{-x/2}) \left(-\frac{1}{2} e^{-x} - \frac{1}{4} e^{-x/2}\right) \] Now, we can compare the coefficients of \(e^{-x}\) and \(e^{-x/2}\) terms on both sides of the equation: For \(e^{-x}\) terms: \[ -a = -\frac{1}{2}a - \frac{1}{4}b \] For \(e^{-x/2}\) terms: \[ -\frac{1}{2}b = -\frac{1}{2}a - \frac{1}{4}b \] Solving the simultaneous equations, we find the values for `a` and `b`: \[ a = \frac{1}{2}, \quad b = \frac{1}{2} \] Now, substituting the values of `a` and `b`, we have the equilibrium distribution: \[ F_e(x) = \frac{1}{2} e^{-x} + \frac{1}{2} e^{-x/2}, \quad x > 0 \] Our guess for the equilibrium distribution was correct, and we have verified it mathematically.