Two independent Poisson processes (no successive arrivals etc)

Question

Requests for orange juices and lemonades are according to two independent Poisson processes of rates $\lambda,~\mu$ (per minute), accordingly.

(1) What is the probability in the next one minute there are requests only for lemonades?

(2) Given that two requests were made in the last minute, what is the probability the times of request among them were done in less than 1/2 minute?

(3) What is the probability in the next minute no successive requests for the same type of drink are made?

Attempt Let $\{O_t\}, ~\{L_t\}$ the relative PP for oranges and lemonades, respectively.

(1) we get some ($\geq0$) lemonades and no orange juices in the next minute, so:

$$\sum_{k=0}^{\infty}P(L_1=k,~0_1=0)~P(L_1=k,~O_1=0/L_1+O_1=k)= \sum_{k=0}^{\infty}e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^k}{k!}\bigg(\frac{\mu}{\lambda+\mu}\bigg)^k=e^{-\lambda}.$$

(2) Let $S_1,~S_2$ be the (total) times of the first, second arrival of $\{L_t+O_t\}$, respectively, so under our hypothesis they are uniformly distributed on $[0,1]$ and $$P(|S_2-S_1|<1/2)=1-2\times 1/8=3/4.$$

(3) This one i don't know how to approach.

Edit. An approach to (3). No successive arrivals for $O_1,~L_1$ means that we have $k$ total arrivals for some $k=0,1,2,\ldots$, with probability $e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^k}{k!}$ and given the $k$ total arrivals,

$\bullet$ for $k=2m$ even we have a sequence of arrivals $OLOL...OL$, with probability $\bigg(\frac{\lambda}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)^m$ or a sequence of arrivals $LOLO...LO$, with the same probability,

$\bullet$ for $k=2m+1$ odd we have a sequence of arrivals $OLOL...OLO$, with probability $\bigg(\frac{\lambda}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)^m\bigg(\frac{\lambda}{\lambda+\mu}\bigg)$ or a sequence of arrivals $LOLO...LOL$, with probability $\bigg(\frac{\lambda}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)$.

Working the sums, the desired probability is:

$$e^{-(\lambda+\mu)}\bigg(2\cosh(\sqrt{\lambda \mu})+\frac{1}{\sqrt{\lambda \mu}}\sinh(\sqrt{\lambda \mu})\bigg).$$

Thank you in advance.

Answer 1

Let's call the two variables $X$ and $Y$, to avoid confusions when using the $O$ letter.

The number of events, obeying to a Poisson distribution, over a generic period of time equal to $t$ time units is $$ \bbox[lightyellow] { P\left( {X = k;\;t} \right) = {{\left( {\lambda \,t} \right)^{\,k} e^{\, - \lambda \,t} } \over {k!}} } \tag {a} $$ where $\lambda$ is the (average/expected) rate of occurrence in the unit period.

For two independent variables, with rates $\lambda$ and $\mu$ $$ \bbox[lightyellow] { P\left( {X = k,Y = j;\;t} \right) = {{\left( {\lambda \,t} \right)^{\,k} e^{\, - \lambda \,t} } \over {k!}}{{\left( {\mu \,t} \right)^{\,j} e^{\, - \mu \,t} } \over {j!}} } \tag {b} $$

So we will have

1) What is the probability that in the next unit time there are only $X$ events ?

a) including no $X$ event $$ \bbox[lightyellow] { P\left( {0 \le X,Y = 0;\;t = 1} \right) = 1{{1\,e^{\, - \mu \,1} } \over {0!}} = e^{\, - \mu \,} } \tag {1.a} $$ b) at least one $X$ event $$ \bbox[lightyellow] { P\left( {1 \le X,Y = 0;\; t=1} \right) = \left( {1 - e^{\, - \lambda } } \right)e^{\, - \mu \,} } \tag {1.b} $$

2) Given that two events occurred in the last unit time, what is the probability that they occured with a time difference of less than 1/2 unit time ?

• the rate of occurrence of events of any kind ($X$ or $Y$) is of course $\lambda + \mu$;
• the probability that any two events occur in the unit time is $$ \bbox[lightyellow] { P\left( {X + Y = 2;\;t=1} \right) = {{\left( {\left( {\lambda + \mu } \right)\,1} \right)^{\,2} e^{\, - \left( {\lambda + \mu } \right)1} } \over {2!}} = {{\left( {\lambda + \mu } \right)^{\,2} e^{\, - \left( {\lambda + \mu } \right)} } \over 2} } \tag {2.a} $$ and in fact $$ \bbox[lightyellow] { \eqalign{ & P\left( {X + Y = 2;\;t=1} \right) = P\left( {0 = X,Y = 2;\;1} \right) + P\left( {1 = X,Y = 1;\;1} \right) + P\left( {2 = X,Y = 0;\;1} \right) = \cr & = {{\lambda ^{\,0} e^{\, - \lambda \,} } \over {0!}}{{\mu ^{\,2} e^{\, - \mu } } \over {2!}} + {{\lambda ^\, e^{\, - \lambda \,} } \over {1!}}{{\mu e^{\, - \mu } } \over {1!}} + {{\lambda ^{\,2} e^{\, - \lambda \,} } \over {2!}}{{\mu ^{\,0} e^{\, - \mu } } \over {1!}} = \cr & = {{\left( {\lambda + \mu } \right)^{\,2} e^{\, - \left( {\lambda + \mu } \right)} } \over 2} \cr} } \tag {2.b} $$
• the probability that there are no events till time $t$ and that one (the first) event occurs in the interval $t, \, t+dt$ is $$ \bbox[lightyellow] { p(\lambda ,\;t)\,dt = P\left( {X = 0;\;t} \right)\lambda \,dt = e^{\, - \lambda \,t} \lambda \,dt } \tag {2.c} $$ and in fact, we have that the probability of having just one event in time $t$ is $$ \bbox[lightyellow] { P\left( {X = 1;\;t} \right) = \int_{\tau \, = \,0}^{\;t} {p(\lambda ,\;\tau )\,d\tau \;P\left( {X = 0;\;t - \tau } \right)} = \lambda \,t\,e^{\, - \lambda \,t} } \tag {2.d} $$
• the probability that there is just one event in the period $t$ and that it occurs in the interval $\tau, \, \tau+d\tau$ is $$ \bbox[lightyellow] { p_{\,e} (\lambda ,\;\tau ,t)\,d\tau = P\left( {X = 0;\;\tau } \right)\lambda \,d\tau P\left( {X = 0;\;t - \tau } \right) = e^{\, - \lambda \,t} \lambda \,d\tau } \tag {2.e} $$

• the probability that there are only two events $X$, separated by more than $1/2$ unit time will be $$ \bbox[lightyellow] { \eqalign{ & P\left( {X = 2;\;t = 1,\;1/2 < \Delta t \le 1} \right) = \int_{\,\tau \, = \,0}^{\;1/2} {p_{\,e} (\lambda ,\;\tau ,\tau + 1/2)P\left( {X = 1;\;t = 1 - \left( {\tau + 1/2} \right)} \right)\,d\tau } = \cr & = \int_{\,\tau \, = \,0}^{\;1/2} {e^{\, - \lambda \,\left( {\tau + 1/2} \right)} \lambda \;\lambda \,\left( {1/2 - \tau } \right)e^{\, - \lambda \left( {1/2 - \tau } \right)} \,d\tau } = \cr & = \lambda ^{\,2} e^{\, - \lambda } \int_{\,\tau \, = \,0}^{\;1/2} {\,\left( {1/2 - \tau } \right)\,d\tau } = {1 \over 8}\lambda ^{\,2} e^{\, - \lambda } \cr} } \tag {2.f} $$

• the probability that there will be only one $X$, followed by one $Y$ and separated by more than $1/2$ unit time will be $$ \bbox[lightyellow] { \eqalign{ & P\left( {X = 1,Y = 1;\;t = 1,\;1/2 < \Delta t \le 1} \right) = \cr & = \int_{\,\tau \, = \,0}^{\;1/2} {p_{\,e} (\lambda ,\;\tau ,1)P\left( {Y = 0;\;t = \tau + 1/2} \right)P\left( {Y = 1;\;t = 1 - \left( {\tau + 1/2} \right)} \right)\,d\tau } = \cr & = \int_{\,\tau \, = \,0}^{\;1/2} {e^{\, - \lambda \,} \lambda \;e^{\, - \mu \,\left( {\tau + 1/2} \right)} \mu \,\left( {1/2 - \tau } \right)e^{\, - \mu \left( {1/2 - \tau } \right)} \,d\tau } = \cr & = \lambda \,\mu \,e^{\, - \,\left( {\lambda + \mu } \right)\,} \int_{\,\tau \, = \,0}^{\;1/2} {\;\,\left( {1/2 - \tau } \right)\,d\tau } = {1 \over 8}\lambda \,\mu \,e^{\, - \,\left( {\lambda + \mu } \right)\,} \cr} } \tag {2.g} $$

Therefore the probability , given that there are two events, that they be separated by more than $1/2$ unit time will be $$ \bbox[lightyellow] { \eqalign{ & P\left( {\left( {X + Y = 2;\;1/2 < \Delta t \le 1} \right)\;|\;\left( {X + Y = 2;\;t = 1} \right)} \right) = \cr & = \sum\limits_{0\, \le \,k\, \le \,2} {{{P\left( {X = k,Y = 2 - k;\;t = 1,\;1/2 < \Delta t \le 1} \right)} \over {P\left( {X + Y = 2;\;t = 1} \right)}}} = \cr & = {1 \over {{{\left( {\lambda + \mu } \right)^{\,2} e^{\, - \left( {\lambda + \mu } \right)} } \over 2}}}\left( \matrix{ P\left( {X = 0;\;t = 1} \right)P\left( {Y = 2;\;t = 1,\;1/2 < \Delta t \le 1} \right) + \hfill \cr + 2P\left( {X = 1,Y = 1;\;t = 1,\;1/2 < \Delta t \le 1} \right) + \hfill \cr + P\left( {Y = 0;\;t = 1} \right)P\left( {X = 2;\;t = 1,\;1/2 < \Delta t \le 1} \right) \hfill \cr} \right) = \cr & = {1 \over {{{\left( {\lambda + \mu } \right)^{\,2} e^{\, - \left( {\lambda + \mu } \right)} } \over 2}}}\left( {e^{\, - \;\lambda } {1 \over 8}\mu ^{\,2} e^{\, - \;\mu } + 2{1 \over 8}\lambda \,\mu \,e^{\, - \;\left( {\lambda + \mu } \right)} + e^{\, - \;\mu } {1 \over 8}\lambda ^{\,2} e^{\, - \;\lambda } } \right) = \cr & = {1 \over 4} \cr} } \tag {2.h} $$

While the requested probability will therefore be $3/4$,
which is what you already found and eq. (2.e) confirms that times are uniformly distributed.

3) Given that two events occurred in the last unit time, what is the probability in the next unit time no successive event of the same type occur ?

Putting $P$ as the probability of two events occurring in the unit time AND no same type of events in a successive unit time, will give $$ \bbox[lightyellow] { \eqalign{ & P = \cr & P\left( {X = 0,Y = 2;\;t = 1} \right)P\left( {Y = 0;\;t = 1} \right) + \cr & + P\left( {X = 1,Y = 1;\;t = 1} \right)P\left( {X = 0,Y = 0;\;t = 1} \right) + \cr & + P\left( {Y = 0,X = 2;\;t = 1} \right)P\left( {X = 0;\;t = 1} \right) = \cr & = {{\mu ^{\,2} } \over 2}e^{\, - \left( {\lambda + \mu } \right)} e^{\, - \mu } + \lambda \,\mu \,e^{\, - \left( {\lambda + \mu } \right)} e^{\, - \left( {\lambda + \mu } \right)} + {{\lambda ^{\,2} } \over 2}e^{\, - \left( {\lambda + \mu } \right)} e^{\, - \lambda } = \cr & = {1 \over 2}e^{\, - \left( {\lambda + \mu } \right)} \left( {\mu ^{\,2} e^{\, - \mu } + 2\lambda \,\mu \,e^{\, - \left( {\lambda + \mu } \right)} + \lambda ^{\,2} e^{\, - \lambda } } \right) \cr} } \tag {3.a} $$

So the requested probability is $$ \bbox[lightyellow] { \eqalign{ & {P \over {P\left( {X + Y = 2;\;t = 1} \right)}} = \cr & = {{{1 \over 2}e^{\, - \left( {\lambda + \mu } \right)} \left( {\mu ^{\,2} e^{\, - \mu } + 2\lambda \,\mu \,e^{\, - \left( {\lambda + \mu } \right)} + \lambda ^{\,2} e^{\, - \lambda } } \right)} \over {{1 \over 2}\left( {\lambda + \mu } \right)^{\,2} e^{\, - \left( {\lambda + \mu } \right)} }} = \cr & = {{\mu ^{\,2} e^{\, - \mu } + 2\lambda \,\mu \,e^{\, - \left( {\lambda + \mu } \right)} + \lambda ^{\,2} e^{\, - \lambda } } \over {\left( {\lambda + \mu } \right)^{\,2} }} \cr} } \tag {3.b} $$

Answer 2

Here is a solution to part (1) that exploits the relationship between Poisson and exponential disributions. Some of the rest might also benefit from this point of view.

Let $L$ be the waiting time for the first lemonade order. Then $L \sim \mathsf{EXP}(rate = \mu = 2).$ That is, $F_L(t) = 1 - e^{-\mu t},$ for $t > 0.$ Similarly, let $J$ be the wait for the first orange juice order with $F_J(t) = 1 - e^{-\lambda t},$ for $t > 0.$ Then $$P(L \le 1, J > 1) = P(L \le 1)P(J > 1) = F_L(1)(1-F_J(1)) \\= (1-e^{-\mu t})(e^{-\lambda t}) = e^{-\lambda t} - e^{-(\mu - \lambda)t}.$$

If $\mu = 2$ and $\lambda = 3$, then $P(L \le 1, J > 1) = 0.043.$

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In the simulation below of a million realizations, rexp generates random exponential observations, pexp is the exponential CDF, exp is $e$ to a power, and the mean of a logical vector (with T and F values) is its proportion of T's. The simulated answer is accurate to three places.

m = 10^6; a = rexp(m, 2);  j = rexp(m, 3)
mean(a < 1 & j > 1);  exp(-3)-exp(-5);  pexp(1,2)*(1-pexp(1,3))
## 0.042865     # aprx (1) by simulation
## 0.04304912   # exact (1)
## 0.04304912   # exact (1)

Note: If I understand what (2) is asking--and I'm not at all sure of that--then a rough simulated answer is about 0.82 (for my choice of parameters).