Question

Let Z be the rate at which customers are served in a queue. Assume that Z has the p.d.f.

\(\begin{aligned}f\left( z \right) &= 2e{}^{ - 2z},z > 0\\ &= 0,otherwise\end{aligned}\)

Find the p.d.f. of the average waiting time T = 1/Z.

Short answer

The PDF of T is \(g\left( T \right) = \left\{ \begin{array}{l}\frac{{2{e^{\frac{{ - 2}}{t}}}}}{{{t^2}}},t > 0\\0,otherwise\end{array} \right.\)

Step-by-step solution

Given information

The random variable Z has the following distribution

\(\begin{aligned}f\left( z \right) &= 2e{}^{ - 2z},z > 0\\ &= 0,otherwise\end{aligned}\)

Obtain the CDF of Z

\(\begin{aligned}{F_Z}\left( z \right) &= P\left( {Z \le z} \right)\\ &= \int\limits_0^z {2{e^{ - 2z}}} dx\\ &= 1 - {e^{ - 2z}},z > 0\end{aligned}\)

Create a new variable, T, and use the CDF approach

\({\rm{T = }}\frac{{\rm{1}}}{{\rm{Z}}}\)

A CDF approach is a method of random variable transformation wherein the pdf of the new variable is fetched from the CDF of the new variable, which is in terms of the CDF of the old variable.

The CDF approach steps,

• In a CDF approach, we substitute the Y variable in the CDF formula.
• We then substitute Y in terms of X.
• We reduce this form until we bring the CDF in terms of X.
• Since we have already calculated the CDF of X, we replace the form of variable y in the CDF formula for X.
• In the final step, we get the CDF of the Y variable as an expression of the CDF of X with y variables.

By using the CDF approach.

\(\begin{aligned}{F_{T = }}\left( t \right) &= P\left( {T \le t} \right)\\ &= P\left( {\frac{1}{Z} \le t} \right)\\ &= P\left( {Z \ge \frac{1}{t}} \right)\\ &= 1 - {F_Z}\left( {\frac{1}{t}} \right)\\ &= {e^{\frac{{ - 2}}{t}}}\end{aligned}\)

Therefore, the CDF of T is \({e^{\frac{{ - 2}}{t}}}\)

Convert the CDF into PDF

The pdf is obtained from CDF by differentiating it with respect to the variable.

\(\begin{aligned}{f_t} &= \frac{d}{{dx}}\left( {{F_T}\left( t \right)} \right)\\ &= \frac{d}{{dx}}\left( {{e^{\frac{{ - 2}}{t}}}} \right)\\ &= \left\{ \begin{aligned}\frac{{2{e^{\frac{{ - 2}}{t}}}}}{{{t^2}}},t > 0\\0,otherwise\end{aligned} \right.\end{aligned}\)

Therefore, the PDF of T is \( = \left\{ \begin{array}{l}\frac{{2{e^{\frac{{ - 2}}{t}}}}}{{{t^2}}},t > 0\\0,otherwise\end{array} \right.\)