Question

Show that the renewal function corresponding to the lifetime density $$ f(x)=\lambda^{2} x e^{-2 x}, \quad x \geq 0 $$ is $$ M(t)=\frac{1}{2} \lambda t-\frac{1}{4}\left(1-e^{-2 \lambda t}\right) $$

Short answer

The renewal function M(t) corresponding to the given lifetime density function \(f(x) = \lambda^2 x e^{-2x}, \quad x \geq 0\) is: \[M(t) = \frac{1}{2}\lambda t - \frac{1}{4}\left(1 - e^{-2 \lambda t}\right)\]

Step-by-step solution

Define the given lifetime density function

The given lifetime density function is: \[f(x) = \lambda^2 x e^{-2x}, \quad x \geq 0\]

Calculate the renewal function M(t)

To calculate the renewal function, we need to integrate the lifetime density function from 0 to t, multiplied by x: \[M(t) = \int_0^t xf(x)dx = \int_0^t x\lambda^2xe^{-2x} dx\]

Perform integration by parts

To integrate this expression, we will use integration by parts, where: \[u = x , du = dx\] \[dv = \lambda^2xe^{-2x} dx, v = -\frac{1}{2}xe^{-2x}+\frac{1}{4}e^{-2x}\] Using the integration by parts formula, \(\int u dv = uv - \int v du\), we get: \[M(t) = \left[-\frac{1}{2}x^2e^{-2x}+\frac{1}{4}xe^{-2x}\right]_0^t + \int_0^t \left(-\frac{1}{2}xe^{-2x}+\frac{1}{4}e^{-2x}\right) dx\]

Evaluate the first expression and solve the integrals

Evaluate the first part of the expression at t and subtract at 0: \[-\frac{1}{2}t^2e^{-2t}+\frac{1}{4}te^{-2t} - \left(-\frac{1}{2}0^2e^{-2(0)}+\frac{1}{4}0e^{-2(0)}\right) = -\frac{1}{2}t^2e^{-2t}+\frac{1}{4}te^{-2t}\] Now compute the integral: \[\int_0^t \left(-\frac{1}{2}xe^{-2x}+\frac{1}{4}e^{-2x}\right) dx = -\frac{1}{4}\int_0^t xe^{-2x}dx + \frac{1}{4}\int_0^t e^{-2x} dx\] For the first integral, we perform integration by parts again: \[u=x , du = dx\] \[dv= -\frac{1}{4}e^{-2x} dx, v = \frac{1}{8}e^{-2x}\] Using the integration by parts formula: \[-\frac{1}{4}\int_0^t xe^{-2x}dx = \left[\frac{1}{8}xe^{-2x}\right]_0^t - \frac{1}{8}\int_0^t e^{-2x} dx\] Now, solve the integrals in the initial expression: \[-\frac{1}{4}\int_0^t xe^{-2x}dx = \frac{1}{8}te^{-2t} - \frac{1}{4} \left[\frac{1}{2}e^{-2x}\right]_0^t = \frac{1}{8}te^{-2t} - \frac{1}{4}\left(\frac{1}{2}e^{-2t}-\frac{1}{2}\right)\] Finally, \[\frac{1}{4}\int_0^t e^{-2x} dx = \left[\frac{1}{4}\frac{1}{-2}e^{-2x}\right]_0^t =\left[\frac{1}{8}e^{-2x}\right]_0^t = \frac{1}{8}\left(1-e^{-2t}\right)\]

Combine the expressions and simplify

Should combine the expressions we obtained in steps 4 and 5: \[M(t) = -\frac{1}{2}t^2e^{-2t}+\frac{1}{4}te^{-2t} + \frac{1}{8}te^{-2t} - \frac{1}{4}\left(\frac{1}{2}e^{-2t}-\frac{1}{2}\right) +\frac{1}{8}\left(1-e^{-2 \lambda t}\right) \] Now simplify the expression: \[M(t) = \frac{1}{2}\lambda t - \frac{1}{4}\left(1 - e^{-2 \lambda t}\right)\] Thus, we have derived the renewal function M(t) as: \[M(t) = \frac{1}{2}\lambda t - \frac{1}{4}\left(1 - e^{-2 \lambda t}\right)\]