Worked solutions. From the course's own solution sets, licensed CC BY-NC-SA 4.0.
Solutions — Poisson Processes: Merging, Splitting, and Random Incidence¶
Department of Electrical Engineering & Computer Science¶
6.041/6.431: Probabilistic Systems Analysis¶
(Fall 2010)¶
Recitation 15 Solutions¶
October 28, 2010¶
- (a) Let \(X\) be the time until the first bulb failure. Let \(A\) (respectively, \(B\)) be the event that the frist bulb is of type \(A\) (respectively, \(B\)). Since the two bulb types are equally likely, the total expectation theorem yields
$\(\mathbf{E}[X] = \mathbf{E}[X|A]\mathbf{P}(A) + \mathbf{E}[X|B]\mathbf{P}(B) = 1 \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2} = \frac{2}{3}.\)$
(b) Let \(D\) be the event of no bulb failures before time \(t\). Using the total probability theorem, and the exponential distributions for bulbs of the two types, we obtain
$\(\mathbf{P}(D) = \mathbf{P}(D|A)\mathbf{P}(A) + \mathbf{P}(D|B)\mathbf{P}(B) = \frac{1}{2}e^{-t} + \frac{1}{2}e^{-3t}.\)$
(c) We have
$\(\mathbf{P}(A|D) = \frac{\mathbf{P}(A \cap D)}{\mathbf{P}(D)} = \frac{\frac{1}{2}e^{-t}}{\frac{1}{2}e^{-t} + \frac{1}{2}e^{-3t}} = \frac{1}{1 + e^{-2t}}.\)$
(d) The lifetime of the first type-A bulb is \(X_A\), with PDF given by:
$\(f_{X_A}(x) = \begin{cases} e^{-x} & x \geq 0 \\ 0 & \text{elsewhere} \end{cases}\)$
Let \(Y\) be the total lifetime of two type-B bulbs. Because the lifetime of each type-B bulb is exponential with \(\lambda = 3\), the sum \(Y\) has an Erlang distribution of order 2 with \(\lambda = 3\). Its PDF is:
$\(f_Y(y) = \begin{cases} 9ye^{-3y} & y \geq 0 \\ 0 & \text{elsewhere} \end{cases}\)$
$\(\begin{aligned} \mathbf{P}(G) &= \mathbf{P}(Y \geq X_A) \\ &= \int_{-\infty}^{\infty} f_Y(y) \int_{-\infty}^y f_{X_A}(x)dxdy \\ &= \int_0^{\infty} 9ye^{-3y} \int_0^y e^{-x}dxdy = 9 \int_0^{\infty} ye^{-3y} \left[ -e^{-x} \right]_{x=0}^{x=y} dy \\ &= 9 \int_0^{\infty} ye^{-3y}(1 - e^{-y})dy = 9 \int_0^{\infty} ye^{-3y} - ye^{-4y}dy \\ &= 9 \left[ -\frac{1}{3}ye^{-3y} - \frac{1}{9}e^{-3y} + \frac{1}{4}ye^{-4y} + \frac{1}{16}e^{-4y} \right]_{y=0}^{y=\infty} \\ &= 9\left( \frac{1}{9} - \frac{1}{16} \right) = \frac{7}{16} \end{aligned}\)$
A simpler solution involving no integrals is as follows:
The bulb failure times of interest (1st type-A, 2nd type-B) may be thought of as the arrival times of two independent Poisson processes of rate \(\lambda_A = 1\) and \(\lambda_B = 3\). We may imagine that these two processes were split from a joint Poisson process of rate \(\lambda_A + \lambda_B\), where the splitting probabilities for each arrival are \(P(A) = \frac{\lambda_A}{\lambda_A + \lambda_B} = 1/4\) to process A and \(P(B) = \frac{\lambda_B}{\lambda_A + \lambda_B} = 3/4\) to process B. Now we may just focus on whether arrivals to the joint process go to process A or to process B. Each arrival to the joint process corresponds to an independent trial. There are two possible outcomes: the arrival is handed to process A with probability \(P(A)\) or the arrival is handed to process B with probability \(P(B)\). Then our event of interest occurs when either the first arrival goes to A, or the first arrival goes to B followed by the second going to A. So the corresponding probability is
$\(P(A \text{ or } BA) = P(A) + P(BA) = P(A) + P(B)P(A) = 7/16\)$
(e) Let \(V\) be the total period of illumination provided by type-B bulbs while the process is in operation. Let \(N\) be the number of light bulbs, out of the first 12, that are of type-B. Let \(X_i\) be the period of illumination from the \(i\)th type-B bulb. We then have \(V = Y_1 + \dots Y_N\). Note that \(N\) is a binomial random variable, with parameters \(n = 12\) and \(p = 1/2\), so that
$\(\mathbf{E}[N] = 6, \quad \operatorname{var}(N) = 12 \cdot \frac{1}{2} \cdot \frac{1}{2} = 3.\)$
Furthermore, \(\mathbf{E}[X_i] = 1/3\) and \(\operatorname{var}(X_i) = 1/9\). Using the formulas for the mean and variance of the sum of a random number of random variables, we obtain
$\(\mathbf{E}[V] = \mathbf{E}[N]\mathbf{E}[X_i] = 2,\)$
and
$\(\operatorname{var}(V) = \operatorname{var}(X_i)\mathbf{E}[N] + (\mathbf{E}[X_i])^2\operatorname{var}(N) = \frac{1}{9} \cdot 6 + \frac{1}{9} \cdot 3 = 1\)$
(f) Using the notation in parts (a)-(c), and the result of part (c), we have
$\(\begin{aligned} \mathbf{E}[T|D] &= t + \mathbf{E}[T - t|D \cap A]\mathbf{P}(A|D) + \mathbf{E}[T - t|D \cap B]\mathbf{P}(B|D) \\ &= t + 1 \cdot \frac{1}{1 + e^{-2t}} + \frac{1}{3} \left( 1 - \frac{1}{1 + e^{-2t}} \right) \\ &= t + \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{1 + e^{-2t}}. \end{aligned}\)$
- (a) The total arrival process corresponds to the merging of two independent Poisson processes, and is therefore Poisson with rate \(\lambda = \lambda_A + \lambda_B = 7\). Thus, the number \(N\) of jobs that arrive in a given three-minute interval is a Poisson random variable, with \(\mathbf{E}[N] = 3\lambda = 21\), \(\operatorname{var}(N) = 21\), and PMF
$\(p_N(n) = \frac{(21)^n e^{-21}}{n!}, \quad n = 0, 1, 2, \dots.\)$
(b) Each of these 10 jobs has probability \(\lambda_A/(\lambda_A + \lambda_B) = 3/7\) of being type A, independently of the others. Thus, the binomial PMF applies and the desired probability is equal to
$\(\binom{10}{3} \left( \frac{3}{7} \right)^3 \left( \frac{4}{7} \right)^7\)$
(c) Each future arrival is of type A with probability \(\lambda_A/(\lambda_A + \lambda_B) = 3/7\) of being type A, independently of the others. Thus, the number \(K\) of arrivals until the first type A arrival is geometric with parameter \(3/7\). The number of type B arrivals before the first type A arrival is equal to \(K - 1\), and its PMF is similar to a geometric, except that it is shifted by one unit to the left. In particular,
$\(p_K(k) = \left( \frac{3}{7} \right)\left( \frac{4}{7} \right)^k, \quad k = 0, 1, 2, \dots.\)$
- The event \(\{X < Y < Z\}\) can be expressed as \(\{X < \min\{Y, Z\}\} \cap \{Y < Z\}\). Let \(Y\) and \(Z\) be the 1st arrival times of two independent Poisson processes with rates \(\mu\) and \(\nu\). By merging the two processes, it should be clear that \(Y < Z\) if and only if the first arrival of the merged process comes from the original process with rate \(\mu\), and thus
$\(\mathbf{P}(Y < Z) = \frac{\mu}{\mu + \nu}.\)$
Let \(X\) be the 1st arrival time of a third independent Poisson process with rate \(\lambda\). Now \(\{X < \min\{Y, Z\}\}\) if and only if the first arrival of the Poisson process obtained by merging the two processes with rates \(\lambda\) and \(\mu + \nu\) comes from the original process with rate \(\lambda\), and thus
$\(\mathbf{P}(X < \min\{Y, Z\}) = \frac{\lambda}{\lambda + \mu + \nu}.\)$
Note that the event \(\{X < \min\{Y, Z\}\}\) is independent of the event \(\{Y < Z\}\), as the time of the first arrival of the merged process with rate \(\mu + \nu\) is independent of whether that first arrival comes from the process with rate \(\mu\) or the process with rate \(\nu\). Hence,
$\(\begin{aligned} \mathbf{P}(X < Y < Z) &= \mathbf{P}(X < \min\{Y, Z\}) \cdot \mathbf{P}(Y < Z) \\ &= \frac{\lambda\mu}{(\lambda + \mu + \nu)(\mu + \nu)}. \end{aligned}\)$
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