Question

Suppose that \(g(x, 0)=0\) and that $$ D_{w}[g(x, w)]=-\lambda g(x, w)+\lambda g(x-1, w) $$ for \(x=1,2,3, \ldots .\) If \(g(0, w)=e^{-\lambda w}\), show by mathematical induction that $$ g(x, w)=\frac{(\lambda w)^{x} e^{-\lambda w}}{x !}, \quad x=1,2,3, \ldots $$

Short answer

The equation \(g(x, w)=\frac{(\lambda w)^{x} e^{-\lambda w}}{x !}\), \(x=1,2,3, \ldots \), is proved by mathematical induction.

Step-by-step solution

Base Case

For \(x=1\), we compute \(D_{w}[g(1, w)] = -\lambda g(1, w) + \lambda g(0, w)\). Given the boundary condition \(g(0, w) = e^{-\lambda w}\), we substitute and it simplifies to: \(D_{w}[g(1, w)] = -\lambda g(1, w) + \lambda e^{-\lambda w} \). Solving this differential equation, considering the equation for \(g(x, w)\), we find \(g(1, w) = (\lambda w) e^{-\lambda w}\), which is equal to \(\frac{(\lambda w)^1 e^{-\lambda w}}{1!}\), satisfying the base case.

Assume the Assertion is True for \(k\)

We assume the assertion is true up to \(x = k\), that is: \(g(k, w) = \frac{(\lambda w)^{k} e^{-\lambda w}}{k !}\).

Prove Assertion is True for \(k+1\)

Now we prove the assertion is true for \(x = k+1\), that is, we want to show \(g(k+1, w) = \frac{(\lambda w)^{k+1} e^{-\lambda w}}{(k+1) !}\). By differentiating both sides of the assumed statement for \(x=k\), we get: \(D_{w}[g(k+1, w)] = -\lambda g(k+1, w) + \lambda g(k, w)\). Now substitute the assumed statement \(g(k, w) = \frac{(\lambda w)^{k} e^{-\lambda w}}{k !}\) into the equation. Solving the resulting equation for \(g(k+1, w)\), we get \(g(k+1, w) = \frac{(\lambda w)^{k+1} e^{-\lambda w}}{(k+1) !}\), showing that the hypothesis holds for \(x = k+1\).

Key concepts

Differential Equation

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They play a significant role in understanding dynamic systems and changes over time. In our exercise, the differential equation is expressed as:

• \(D_{w}[g(x, w)]=-\lambda g(x, w)+\lambda g(x-1, w)\)

This equation indicates how the function \(g(x, w)\) changes with respect to a variable \(w\). Here, \(\lambda\) acts as a rate constant, impacting how quickly the change occurs. The structure of this equation suggests a coupling between consecutive terms \(g(x,w)\) and \(g(x-1,w)\), typical of recurrence relations. By solving this differential equation, you find specific forms for \(g(x,w)\) that satisfy both the equation and the given boundary conditions. Often, solving such equations involves techniques like separation of variables or integrating factors.

Boundary Condition

Boundary conditions are essential to finding unique solutions to differential equations. They provide additional constraints which ensure the solution is tailored to a specific problem scenario. In the provided exercise, we have a key boundary condition:

• \(g(0, w)=e^{-\lambda w}\)
• \(g(x, 0)=0\) for \(x \geq 1\)

These conditions specify the behavior of the function \(g(x, w)\) at particular points, guiding the solution to meet certain criteria. The condition \(g(0, w)=e^{-\lambda w}\) gives a starting point which, when combined with the differential equation, helps build the solution recursively for higher values of \(x\). Understanding boundary conditions is critical for working accurately with differential equations in various mathematical and physical applications.

Probability Generating Function

A probability generating function is a useful mathematical tool in probability theory. Essentially, it is a function designed to encode the probabilities of a random variable's outcomes into a power series. In our exercise, the solution involves finding a probability generating function for the function \(g(x, w)\):

• \(g(x, w)=\frac{(\lambda w)^{x} e^{-\lambda w}}{x!}\)

This is reminiscent of generating functions used to describe distributions, such as the Poisson distribution. The through-lines in these functions include expressions like \((\lambda w)^x/x!\), which characterize the probabilities of distinct outcomes. By expressing \(g(x, w)\) as a probability generating function, it simplifies the handling of probabilities of events. With mathematical induction, this probability generating function confirms the pattern holds for all natural numbers \(x\). Understanding these functions ensures an easier manipulation and derivation of important statistical distributions in various settings.