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11.1.5 Solved Problems

Problem

Let $\{N(t), t \in [0, \infty) \}$ be a Poisson process with rate $\lambda=0.5$.

  1. Find the probability of no arrivals in $(3,5]$.
  2. Find the probability that there is exactly one arrival in each of the following intervals: $(0,1]$, $(1,2]$, $(2,3]$, and $(3,4]$.

  • Solution
      1. If $Y$ is the number arrivals in $(3,5]$, then $Y \sim Poisson(\mu=0.5 \times 2)$. Therefore, \begin{align*} P(Y=0) &=e^{-1} \\ &=0.37 \end{align*}
      2. Let $Y_1$, $Y_2$, $Y_3$ and $Y_4$ be the numbers of arrivals in the intervals $(0,1]$, $(1,2]$, $(2,3]$, and $(3,4]$. Then $Y_i \sim Poisson(0.5)$ and $Y_i$'s are independent, so \begin{align*} P(Y_1=1,Y_2=1,Y_3=1,Y_4=1) &=P(Y_1=1) \cdot P(Y_2=1) \cdot P(Y_3=1) \cdot P(Y_4=1) \\ &=\bigg[0.5 e^{-0.5}\bigg]^4\\ &\approx 8.5 \times 10^{-3}. \end{align*}

Problem

Let $\{N(t), t \in [0, \infty) \}$ be a Poisson process with rate $\lambda$. Find the probability that there are two arrivals in $(0,2]$ and three arrivals in $(1,4]$.

  • Solution
    • Note that the two intervals $(0,2]$ and $(1,4]$ are not disjoint. Thus, we cannot multiply the probabilities for each interval to obtain the desired probability. In particular, \begin{align*} (0,2] \cap (1,4]=(1,2]. \end{align*} Let $X$, $Y$, and $Z$ be the numbers of arrivals in $(0,1]$, $(1,2]$, and $(2,4]$ respectively. Then $X$, $Y$, and $Z$ are independent, and \begin{align*} X \sim Poisson(\lambda \cdot 1),\\ Y \sim Poisson(\lambda \cdot 1),\\ Z \sim Poisson(\lambda \cdot 2). \end{align*} Let $A$ be the event that there are two arrivals in $(0,2]$ and three arrivals in $(1,4]$. We can use the law of total probability to obtain $P(A)$. In particular, \begin{align*} P(A)&=P(X+Y=2 \textrm{ and }Y+Z=3)\\ &=\sum_{k=0}^{\infty} P\big(X+Y=2 \textrm{ and }Y+Z=3 | Y=k \big)P(Y=k)\\ &=P\big(X=2, Z=3 | Y=0\big)P(Y=0)+P(X=1, Z=2 | Y=1)P(Y=1)+\\ &\hspace{40pt} +P(X=0, Z=1 | Y=2)P(Y=2)\\ &=P\big(X=2, Z=3\big)P(Y=0)+P(X=1, Z=2)P(Y=1)+\\ &\hspace{40pt} P(X=0, Z=1)P(Y=2)\\ &=P(X=2)P(Z=3)P(Y=0)+P(X=1)P(Z=2)P(Y=1)+\\ &\hspace{40pt}P(X=0) P(Z=1)P(Y=2)\\ &=\left(\frac{e^{-\lambda} \lambda^2}{2}\right) \cdot \left(\frac{e^{-2\lambda} (2\lambda)^3}{6}\right) \cdot\left(e^{-\lambda}\right)+ \left(\lambda e^{-\lambda}\right) \cdot \left(\frac{e^{-2\lambda} (2\lambda)^2}{2}\right) \cdot\left(\lambda e^{-\lambda}\right)+\\ &\hspace{40pt} \left(e^{-\lambda}\right) \cdot \left(e^{-2\lambda} (2\lambda)\right) \cdot\left(\frac{e^{-\lambda} \lambda^2}{2}\right). \end{align*}

Problem

Let $\{N(t), t \in [0, \infty) \}$ be a Poisson Process with rate $\lambda$. Find its covariance function \begin{align*} C_N(t_1,t_2)&=\textrm{Cov}\big(N(t_1),N(t_2)\big), \quad \textrm{for }t_1,t_2 \in [0,\infty) \end{align*}

  • Solution
    • Let's assume $t_1 \geq t_2 \geq 0$. Then, by the independent increment property of the Poisson process, the two random variables $N(t_1)-N(t_2)$ and $N(t_2)$ are independent. We can write \begin{align*} C_N(t_1,t_2)&=\textrm{Cov}\big(N(t_1),N(t_2)\big)\\ &=\textrm{Cov}\big( N(t_1)-N(t_2) + N(t_2), N(t_2) \big)\\ &=\textrm{Cov}\big( N(t_1)-N(t_2), N(t_2) \big)+\textrm{Cov}\big(N(t_2), N(t_2) \big)\\ &=\textrm{Cov}\big(N(t_2), N(t_2) \big)\\ &=\textrm{Var}\big(N(t_2)\big)\\ &=\lambda t_2, \quad \textrm{since }N(t_2) \sim Poisson(\lambda t_2). \end{align*} Similarly, if $t_2 \geq t_1 \geq 0$, we conclude \begin{align*} C_N(t_1,t_2)&=\lambda t_1. \end{align*} Therefore, we can write \begin{align*} C_N(t_1,t_2)&=\lambda \min(t_1,t_2), \quad \textrm{for }t_1,t_2 \in [0,\infty). \end{align*}

Problem

Let $\{N(t), t \in [0, \infty) \}$ be a Poisson process with rate $\lambda$, and $X_1$ be its first arrival time. Show that given $N(t)=1$, then $X_1$ is uniformly distributed in $(0,t]$. That is, show that \begin{align*} P(X_1 \leq x | N(t)=1)&=\frac{x}{t}, \quad \textrm{for }0 \leq x \leq t. \end{align*}

  • Solution
    • For $0 \leq x \leq t$, we can write \begin{align*} P(X_1 \leq x | N(t)=1)&=\frac{P(X_1 \leq x, N(t)=1)}{P\big(N(t)=1\big)}. \end{align*} We know that \begin{align*} P\big(N(t)=1\big)=\lambda t e^{-\lambda t}, \end{align*} and \begin{align*} P(X_1 \leq x, N(t)=1)&=P\bigg(\textrm{one arrival in $(0,x]$ $\;$ and $\;$ no arrivals in $(x,t]$}\bigg)\\ &=\left[\lambda x e^{-\lambda x}\right]\cdot \left[e^{-\lambda (t-x)}\right]\\ &=\lambda x e^{-\lambda t}. \end{align*} Thus, \begin{align*} P(X_1 \leq x | N(t)=1)&=\frac{x}{t}, \quad \textrm{for }0 \leq x \leq t. \end{align*} Note: The above result can be generalized for $n$ arrivals. That is, given that $N(t)=n$, the $n$ arrival times have the same joint CDF as the order statistics of $n$ independent $Uniform(0,t)$ random variables. This fact is discussed more in detail in the End of Chapter Problems.

Problem

Let $N_1(t)$ and $N_2(t)$ be two independent Poisson processes with rates $\lambda_1=1$ and $\lambda_2=2$, respectively. Let $N(t)$ be the merged process $N(t)=N_1(t)+N_2(t)$.

  1. Find the probability that $N(1)=2$ and $N(2)=5$.
  2. Given that $N(1)=2$, find the probability that $N_1(1)=1$.

  • Solution
    • $N(t)$ is a Poisson process with rate $\lambda=1+2=3$.
      1. We have \begin{align*} P(N(1)=2, N(2)=5)&=P\bigg(\textrm{$\underline{two}$ arrivals in $(0,1]$ and $\underline{three}$ arrivals in $(1,2]$}\bigg)\\ &=\left[ \frac{e^{-3} 3^2}{2!}\right]\cdot \left[\frac{e^{-3} 3^3}{3!}\right]\\ &\approx .05 \end{align*}
      2. $ $ \begin{align*} P(N_1(1)=1 | N(1)=2)&=\frac{P\big(N_1(1)=1, N(1)=2\big)}{P(N(1)=2)}\\ &=\frac{P\big(N_1(1)=1, N_2(1)=1\big)}{P(N(1)=2)}\\ &=\frac{P\big(N_1(1)=1\big) \cdot P\big(N_2(1)=1\big)}{P(N(1)=2)}\\ &=\left[e^{-1} \cdot 2 e^{-2} \right] \big{/} \left[\frac{e^{-3} 3^2}{2!}\right]\\ &=\frac{4}{9}. \end{align*}

Problem

Let $N_1(t)$ and $N_2(t)$ be two independent Poisson processes with rates $\lambda_1=1$ and $\lambda_2=2$, respectively. Find the probability that the second arrival in $N_1(t)$ occurs before the third arrival in $N_2(t)$. Hint: One way to solve this problem is to think of $N_1(t)$ and $N_2(t)$ as two processes obtained from splitting a Poisson process.

  • Solution
    • Let $N(t)$ be a Poisson process with rate $\lambda=1+2=3$. We split $N(t)$ into two processes $N_1(t)$ and $N_2(t)$ in the following way. For each arrival, a coin with $P(H)=\frac{1}{3}$ is tossed. If the coin lands heads up, the arrival is sent to the first process ($N_1(t)$), otherwise it is sent to the second process. The coin tosses are independent of each other and are independent of $N(t)$. Then
      1. $N_1(t)$ is a Poisson process with rate $\lambda p=1$;
      2. $N_2(t)$ is a Poisson process with rate $\lambda (1-p)=2$;
      3. $N_1(t)$ and $N_2(t)$ are independent.


      Thus, $N_1(t)$ and $N_2(t)$ have the same probabilistic properties as the ones stated in the problem. We can now restate the probability that the second arrival in $N_1(t)$ occurs before the third arrival in $N_2(t)$ as the probability of observing at least two heads in four coin tosses, which is \begin{align*} \sum_{k=2}^{4} {4 \choose k} \left(\frac{1}{3} \right)^k \left(\frac{2}{3} \right)^{4-k}. \end{align*}

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