Question 7.3: A communication channel is degraded beyond use in a random m......
A communication channel is degraded beyond use in a random manner. A smart student figured out that the duration Y of the intervals between consecutive periods of degradation has the PDF
\,\displaystyle f_Y\left (y\right )=\frac{0.2^4 y^3 e^{-0.2 y}}{3 !} \quad y \geq 0What is the s-transform of the PDF of Y?
First, we must realize that Y is an Erlang random variable. To determine the order of the random variable, we put its PDF in the general form of the Erlang PDF as follows:
\,\displaystyle f_Y\left (y\right )=\frac{\lambda^k y^{k-1} e^{-\lambda y}}{\left (k-1\right ) !} \quad y \geq 0
From the power of y we see that k-1=3, which means that k=4 and so Y is a fourth-order Erlang random variable. Let X be the underlying exponential distribution. Since \lambda=0.2, the PDF of X and its s-transform are given as follows:
\,\displaystyle \begin{aligned} f_X\left (x\right ) & =0.2 e^{-0.2 x} \quad x \geq 0 \\ M_X\left (s\right ) & =\frac{0.2}{s+0.2} \end{aligned}Thus, Y=X_1+X_2+X_3+X_4 , where the X_k are independent and identically distributed with the above PDF. Since the s-transform of the PDF of the sum of independent random variables is equal to the product of their s-transforms, we have that
\,\displaystyle M_Y\left (s\right )=\left[M_X\left (s\right )\right]^4=\left[\frac{0.2}{s+0.2}\right]^4