Exponential versus Poisson Distribution::mjtsai1974

25 Apr 2018

Prologue To Exponential versus Poisson Distribution::mjtsai1974

In the world of stochasticity, pass and failure evaluation often been proceeded under the assumption that the intensity of event occurrence is constant over time, which is debatable. When we involve the reinforcement learning issues, or the statistics regression topics, although the discrete case is easily constructed and simulated by bootstrapping algorithm in conjunction with the maximum likelihood estimation, the migration to continuous case would be just to split the test horizon into many, many, uncountable subintervals. Then, that connects the Poisson distribution to the exponential distribution, and the random arrivals probability could be modelled by gamma distribution as a result of the nature of exponential distribution.

Overview The Similarities And Differences

[1] the Poisson distribution

➀we have an assumption that the intensity of event occurrence over time is invariant for Poisson process.
➁suppose it is $\lambda$=$\frac {event\;counts}{time\;length}$
Below exhibits the case where $\lambda$=$\frac {1}{time\;length}$.
This is the case where $\lambda$=$\frac {k}{time\;length}$.
➂it takes up time length of $\frac {k}{\lambda}$ to have $k$ random arrivals, for $\lambda$=$\frac {k}{time\;length}$. So, it holds to say that the adjacent two events of random arrival would just take time $\frac {1}{\lambda}$, which is the ideal inter-arrival times.
➃each distinct inter-arrival times could then be modelled by random variable distributed in exponential probability.
➄the joint distribution of random arrivals would be modelled by gamma distribution.
➅the Poisson process has $n$ random arrivals in time interval $[a,b]$, the locations of these points are independent distributed, and each of them has a uniform distribution.

[2] the exponential distribution

➀suppose $X$ is the rate of event occurrences during time period $t$.
➁suppose $V$ is the space where events take place within, then the success probability over time period $t$ is $\underset{Succss}P$=$\frac {X\cdot t}{V}$, and the failure probability is $\underset{Fail}P$=$1$-$\frac {X\cdot t}{V}$.
➂we divide time period $t$ by $n$, where $n\rightarrow\infty$, then success probability over time period $t$ becomes $\underset{Succss}P$=$\frac {X\cdot t}{V\cdot n}$, and the failure probability becomes $\underset{Fail}P$=$1$-$\frac {X\cdot t}{V\cdot n}$.
After time period $t$, what is the probability for no event occurrence? Equivalently, it is asking the probability for the very first event taking place after time period $t$.
➃This presumes that each subinterval $\frac {t}{n}$ is a failure case of the event, then:

$P(T>t)$

=$\lim_{n\rightarrow\infty}(\underset{Fail}P)^{n}$
=$\lim_{n\rightarrow\infty}(1-\frac {X\cdot t}{V\cdot n})^{n}$
$\approx e^{-\frac {X\cdot t}{V}}$
➄the success probability within time period $t$ could be expressed:
$\underset{Succss}P$=$1$-$e^{-\frac {X\cdot t}{V}}$=$P(T\le t)$
➅take $\lambda$=$\frac {X}{V}$, then, $P(T\le t)$=$1$-$e^{-\lambda\cdot t}$
$f_{T}(t)$=$D_{t}(1-e^{-\lambda\cdot t})$=$\lambda\cdot e^{-\lambda\cdot t}$
, where $\lambda$ is the intensity, the rate of success, or of event occurrence.

Example: Illustration Of The Similarities And Differences

[1] Given the vehicles pass through a highway toll station is $6$ per minute, what is the probability that no cars within $30$ seconds?

➀by exponential distribution, we take $\lambda$=$6$ ($\frac {vehicles}{1\;minute}$), its PDF is $f_{exp}(t)$=$6\cdot e^{-6\cdot t}$, where $t>0$.
No cars within $30$ seconds asks for no car within $0.5$ minute, and we are figuring out the probability that $t>0.5$ will we just have the very first car come in, then:
$F_{exp}(t>0.5)$
=$\int_{0.5}^{\infty}6\cdot e^{-6\cdot t}\operatorname dt$
=$-e^{-6\cdot t}\vert_{0.5}^{\infty}$
=$e^{-3}$
➁by Poisson distribution, we can still use $\lambda$=$6$ ($\frac {vehicles}{1\;minute}$), its PDF is $f_{Pois}(x,t)$=$\frac {(6\cdot t)^{x}}{x!}\cdot e^{-6\cdot t}$, $t$ is now $0.5$.
Therefore, $f_{Pois}(0,0.5)$=$\frac {(6\cdot 0.5)^{0}}{0!}\cdot e^{-6\cdot 0.5}$=$e^{-3}$

We have found that the probability for no vehicles within the dedicated time interval is the same in both exponential and Poisson distribution. This is fully compliant with the claim that the very first inter-arrival times is itself an exponential distribution in Introduction To The Poisson Process Inter-arrival Times.

[2] Still using the same rate that $6$ vehicles pass through a highway toll station per minute, then, what is the probability that all 6 cars crossing the toll station within 30 seconds?

➀from exponential distribution view point, this is asking the success probability within 30 seconds.
$F_{exp}(0.5)$
=$P_{exp}(t\le 0.5)$
=$\int_{0}^{0.5}6\cdot e^{-6\cdot t}\operatorname dt$
=$1$-$e^{-6\cdot 0.5}$=$0.950$
➁for Poisson distribution, this is to calculate the probability of distribution on the number $6$, which is the number of the requested target crossing vehicle counts within $0.5$ minute.
$P_{Pois}(x)$=$\frac {(\lambda\cdot t)^{x}}{x!}\cdot e^{-\lambda\cdot t}$…the Poisson PMF.
Now $t$=$0.5$, $x$=$6$, $P_{Pois}(6)$=$\frac {3^6}{6!}\cdot e^{-3}$=$0.168$

[Notes] Why at this moment, the Poisson and exponential probability come out with different result?

➀as a result of the fact that we treat the pass probability as a whole by integration from each distinct exponential probability from $t$=$0$ to $t$=$0.5$, whereas we only calculate the Poisson probability distributed on the number $6$.
➁if we accumulate the distinct Poisson probability distributed on the numbers from $0$ to $6$, that is $\sum_{x=0}^{6}\frac {(\lambda\cdot t)^{x}}{x!}\cdot e^{-\lambda\cdot t}$, in this example, we get the probability $0.96649146…$, and the bias of $0.01$ could then be found.

By accumulating the distinct Poisson probability distribution from $0$ to $6$ is just to answer different question asking the probability of the number of cars passing by up to 6.

Example: Illustration Of Random Arrivals

[1] Still using the same intensity, the rate that $6$ vehicles pass through a highway toll station per minute, then, what's the probability of the $8$-th vehicle passing through the toll station?

➀random arrivals doesn't reinforce the occurrence of specific event at dedicated time interval, in distinct time interval unit with regard to the given rate, which is $\frac {1}{rate}$, event outcome might just be true or false. Then, the k-th car should pass by after time length of $k\cdot\frac {1}{rate}$, if all goes well, just right at the k-th moment.
➁this involves the distribution of inter-arrival times, which is in exponential distribution itself, and the accumulation of which would leads to the gamma distribution, see Introduction To The Poisson Process Inter-arrival Times.
➂in this question, the most beautiful k-th random arrival takes place at the k-th moment of time interval unit. We need to transform the correct interval unit from the reciprocal value of the given rate.
The time interval unit would be $\frac {1}{\lambda}$. The k-th random arrival takes place at the k-th time interval unit, $\frac {k}{\lambda}$, in this example is $\frac {8}{6}$.
➃we have already proved that $f_{X_{n}}(t)$=$\frac {\lambda\cdot(\lambda\cdot t)^{n-1}\cdot e^{-\lambda\cdot t}}{(n-1)!}$, for $n$=$1$,$2$,..., where $\Gamma(n)$=$(n-1)!$. Then the probability of the 8-th vehicle passing through the toll station at the 8-th time interval unit would be calculated in this way.
$f_{X_{8}}(\frac {8}{6})$
=$\frac {6\cdot(6\cdot \frac {8}{6})^{7}\cdot e^{-6\cdot \frac {8}{6}}}{(7)!}$
=$0.8375191$

[2] What is the probability of the $8$-th, $12$-th, $16$-th car passing by at the $3$-rd time interval unit?

➀$f_{X_{8}}(\frac {3}{6})$
=$\frac {6\cdot(6\cdot \frac {3}{6})^{7}\cdot e^{-6\cdot \frac {3}{6}}}{(7)!}$
=$0.12962418871490290869$
➁$f_{X_{12}}(\frac {3}{6})$
=$\frac {6\cdot(6\cdot \frac {3}{6})^{11}\cdot e^{-6\cdot \frac {3}{6}}}{(11)!}$
=$0.00132570193003877975$
➂$f_{X_{16}}(\frac {3}{6})$
=$\frac {6\cdot(6\cdot \frac {3}{6})^{15}\cdot e^{-6\cdot \frac {3}{6}}}{(15)!}$
=$0.00000327783444240358$

We can easily tell, the arrivals in the time interval unit of the order less than it, the probability would just decrease almost to $0$ as the order of arrival increases.

[3] What is the probability of the $5$-th, $4$-th car passing by at the $5$-th, $4$-th time interval unit?

➀$f_{X_{5}}(\frac {5}{6})$
=$\frac {6\cdot(6\cdot \frac {5}{6})^{4}\cdot e^{-6\cdot \frac {5}{6}}}{(4)!}$
=$1.05280421860710423385$
➁$f_{X_{4}}(\frac {4}{6})$
=$\frac {6\cdot(6\cdot \frac {4}{6})^{3}\cdot e^{-6\cdot \frac {4}{6}}}{(3)!}$
=$1.1722008888789875388$

There exists no probability larger than $1.0$, above answer from gamma PDF just guarantees that $5$-th, $4$-th car passing by at the $5$-th, $4$-th time interval unit would definitely happen.

[4] All we have done is by assuming that each i-th event arrival occurrs at the i-th time interval unit, this is not the normal case!!!

➀we can have a more random question about the probability of $10$-th car passing by after 10 minutes.
$f_{X_{10}}(10)$
=$\frac {6\cdot(6\cdot 10)^{9}\cdot e^{-6\cdot 10}}{(9)!}$
=$0.00000000000000145908$
This is quiet small probability for the $10$-th car passing by after 10 minutes, given that 6 cars would just across every one minute is the constant rate.
➁we then regularize the question by the probability of $60$-th car passing by after 10 minutes.
$f_{X_{60}}(10)$
=$\frac {6\cdot(6\cdot 10)^{59}\cdot e^{-6\cdot 10}}{(59)!}$
=$0.30859046994207513678$
This is a much higher probability than the probability for $10$-th car passing by after 10 minutes.
➁we then ask the question by the probability of $59$-th car passing by after 10 minutes.
$f_{X_{59}}(10)$
=$\frac {6\cdot(6\cdot 10)^{58}\cdot e^{-6\cdot 10}}{(58)!}$
=$0.30344729544304055116$
It’s a little smaller.
➂we then ask the question by the probability of $25$-th car passing by after 10 minutes.
$f_{X_{25}}(10)$
=$\frac {6\cdot(6\cdot 10)^{24}\cdot e^{-6\cdot 10}}{(24)!}$
=$0.00000040124226676254$
It’s much smaller.

As we can tell that the most optimized probability of random arrival by means of the gamma distribution would be the order of arrival is in 100 percent proportional to the time length it takes, given that the rate is a constant over time!!
➃soon we verify it with the case the $25$-th arrival at $25$-th time interval unit, that is
$f_{X_{25}}(\frac {25}{6})$
=$\frac {6\cdot(6\cdot \frac {25}{6})^{24}\cdot e^{-6\cdot \frac {25}{6}}}{(24)!}$
=$0.4771377088083926763$
Next, compare it with the $15$-th arrival at $25$-th time interval unit, that is
$f_{X_{15}}(\frac {25}{6})$
=$\frac {6\cdot(6\cdot \frac {25}{6})^{14}\cdot e^{-6\cdot \frac {25}{6}}}{(14)!}$
=$0.03560745811847304799$
Here we concrete above conclusion.

[5] We are here with below probability as a summary

for the final finding.
$f_{X_{8}}(\frac {8}{6})$=$0.8375191$
$f_{X_{25}}(\frac {25}{6})$=$0.4771377088083926763$
$f_{X_{60}}(10)$=$0.30859046994207513678$, where $10$=$\frac {60}{6}$

As the event arrives at the time interval unit of appropriate proportion, which is also more closed to $1$, like $\frac {8}{6}$, it will just have higher probability by using the gamma distribution for a prediction.