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Cumulative Probabilities and Poisson Tails

Use cumulative distribution notation and Poisson point probabilities to calculate integer-range probabilities and lower or upper tails accurately.

Intermediate 10 min read

What You'll Learn

  • Define the cumulative distribution function for a Poisson count.
  • Translate verbal count conditions into inclusive integer bounds.
  • Calculate a probability over a finite integer range by summing point probabilities.
  • Express lower-tail and upper-tail events using cumulative notation.
  • Evaluate an infinite Poisson upper tail by summing terms and controlling the remaining error.
  • Avoid endpoint mistakes when subtracting cumulative probabilities.

Why Cumulative Probabilities Matter

A Poisson model often describes a count over a fixed period or amount of exposure. A question may ask for the probability of exactly four events, but it may instead ask for a range: no more than four, between two and five inclusive, or at least six. Those questions concern several possible count values, so the corresponding point probabilities must be combined carefully.

Cumulative distribution notation gives a concise way to state these events and relate them to sums of point probabilities. The key is to translate the wording into integer endpoints before calculating. Because counts are integers, changing an endpoint by one changes which outcomes are included.

Definition: If \(X\sim\operatorname{Poisson}(\lambda)\), where \(\lambda>0\) is the expected count for the stated exposure, then \(X\) can take the values \(0,1,2,\ldots\), with point probabilities $$ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ The cumulative distribution function is \(F_X(x)=P(X\le x)\). For an integer \(m\ge 0\), $$ F_X(m)=\sum_{k=0}^{m}P(X=k). $$ For \(x<0\), \(F_X(x)=0\); for a noninteger \(x\ge0\), \(F_X(x)=F_X(\lfloor x\rfloor)\).

A Poisson model is appropriate when the count is modeled over a specified exposure and has a fixed mean \(\lambda\) for that exposure. In applications, it is also important to consider whether the underlying process reasonably supports a Poisson model; the calculations below assume that model has already been specified.

From Count Wording to a Sum

For integer bounds \(a\) and \(b\), with \(0\le a\le b\), the event \(a\le X\le b\) includes every integer from \(a\) through \(b\). Its probability is the sum of those point probabilities. Cumulative notation gives the same probability by subtracting the cumulative probability through the integer immediately below \(a\).

$$ P(a\le X\le b)=\sum_{k=a}^{b}P(X=k) =F_X(b)-F_X(a-1). $$

The lower-tail event “no more than \(m\)” means \(X\le m\), so its probability is \(F_X(m)\). The upper-tail event “at least \(r\)” means \(X\ge r\), and can be written as the infinite sum of point probabilities beginning at \(r\). A finite upper range, such as \(r\le X\le s\), is summed only from \(r\) to \(s\).

$$ P(X\le m)=F_X(m)=\sum_{k=0}^{m}P(X=k), \qquad P(X\ge r)=\sum_{k=r}^{\infty}P(X=k). $$

Worked Example: A Probability Over an Inclusive Range

A hypothetical service desk models the number \(X\) of urgent requests received during one shift as Poisson with mean \(\lambda=2.5\) requests per shift. Calculate the probability that the shift receives from two through five requests, inclusive. Give the probability to four decimal places.

Method. The range includes \(X=2,3,4,5\), so calculate the four point probabilities and add them. The Poisson parameter for this one-shift exposure is \(\lambda=2.5\).

First, \(e^{-2.5}\approx0.0820850\). Using \(P(X=k)=e^{-2.5}(2.5)^k/k!\):

  • \(P(X=2)=0.0820850(2.5)^2/2=0.0820850(3.125)=0.2565156\).
  • \(P(X=3)=0.0820850(2.5)^3/6=0.0820850(2.6041667)=0.2137630\).
  • \(P(X=4)=0.0820850(2.5)^4/24=0.0820850(1.6276042)=0.1336019\).
  • \(P(X=5)=0.0820850(2.5)^5/120=0.0820850(0.8138021)=0.0668010\).

Adding the requested outcomes gives \(0.2565156+0.2137630+0.1336019+0.0668010=0.6706815\). Rounded to four decimal places, the probability is 0.6707.

Check. The cumulative values through 1 and 5 are \(F_X(1)=0.0820850+0.2052125=0.2872975\) and \(F_X(5)=0.9579790\). Their difference is \(0.9579790-0.2872975=0.6706815\), agreeing with the direct sum. The common mistake this method avoids is excluding the lower endpoint when translating an inclusive range.

Why the Cumulative-Difference Formula Works

The cumulative probability \(F_X(b)\) includes the outcomes \(0,1,\ldots,a-1\) as well as \(a,a+1,\ldots,b\). The cumulative probability \(F_X(a-1)\) contains exactly the outcomes below \(a\). Subtracting removes those lower outcomes and leaves only the desired range:

$$ F_X(b)-F_X(a-1) =\left(\sum_{k=0}^{a-1}P(X=k)+\sum_{k=a}^{b}P(X=k)\right) -\sum_{k=0}^{a-1}P(X=k) =\sum_{k=a}^{b}P(X=k). $$

This argument depends on the bounds being integers and the event including both endpoints. If the requested upper endpoint is not included, or the question uses a strict inequality, first identify the integer values that actually satisfy the condition. For example, \(X<4\) means \(X\le3\), not \(X\le4\).

Worked Example: A Lower-Tail Probability and a Range

A hypothetical laboratory models the number \(X\) of instrument alerts during a six-hour monitoring period as Poisson with mean \(\lambda=3\) alerts per period. Find (1) the probability of no more than two alerts and (2) the probability of one or two alerts. Give each result to four decimal places.

Method. “No more than two” is the lower tail \(X\le2\), which is \(F_X(2)\). The range from one to two excludes zero, so its probability is \(F_X(2)-F_X(0)\), or equivalently \(P(X=1)+P(X=2)\).

For this model, \(e^{-3}\approx0.0497871\). The needed point probabilities are

  • \(P(X=0)=0.0497871(3)^0/0!=0.0497871\).
  • \(P(X=1)=0.0497871(3)^1/1!=0.1493612\).
  • \(P(X=2)=0.0497871(3)^2/2!=0.0497871(9/2)=0.2240418\).

Therefore, the lower-tail probability is \(F_X(2)=0.0497871+0.1493612+0.2240418=0.4231901\), or 0.4232 to four decimal places.

For one or two alerts, omit the zero-alert outcome: \(P(1\le X\le2)=0.1493612+0.2240418=0.3734030\), or 0.3734 to four decimal places. As a cumulative check, \(F_X(0)=0.0497871\), and \(F_X(2)-F_X(0)=0.4231901-0.0497871=0.3734030\). Both calculations agree.

Check. The first answer is larger than the second because \(X\le2\) also includes \(X=0\). This is a useful check on the event interpretation.

Calculating an Upper Tail Directly

For a threshold \(r\), “at least \(r\)” includes every value \(r,r+1,\ldots\). Its probability is an infinite sum. When calculating that sum numerically, the point probabilities can be generated successively rather than recalculating factorials. Dividing adjacent Poisson probabilities gives

$$ \frac{P(X=k+1)}{P(X=k)} = \frac{e^{-\lambda}\lambda^{k+1}/(k+1)!} {e^{-\lambda}\lambda^k/k!} =\frac{\lambda}{k+1}. $$

Thus, once one point probability is known, the next is obtained by multiplying by \(\lambda/(k+1)\). For sufficiently large \(k\), these ratios are less than one and decrease as \(k\) increases. That allows the uncomputed remainder of the upper tail to be bounded by a geometric series.

Worked Example: A Poisson Upper Tail by Direct Summation

A hypothetical production cell models the number \(X\) of brief stoppages during a shift as Poisson with mean \(\lambda=2\) stoppages per shift. Calculate \(P(X\ge4)\) to four decimal places by summing the upper-tail probabilities directly, rather than subtracting a lower-tail probability from one.

Method. Begin with \(P(X=4)\), then use \(P(X=k+1)=P(X=k)\,2/(k+1)\). Since \(e^{-2}\approx0.1353353\), the point probabilities from four through ten are:

  • \(P(X=4)=0.1353353(2^4/4!)=0.0902235\).
  • \(P(X=5)=P(X=4)(2/5)=0.0360894\).
  • \(P(X=6)=P(X=5)(2/6)=0.0120298\).
  • \(P(X=7)=P(X=6)(2/7)=0.0034371\).
  • \(P(X=8)=P(X=7)(2/8)=0.0008593\).
  • \(P(X=9)=P(X=8)(2/9)=0.0001909\).
  • \(P(X=10)=P(X=9)(2/10)=0.0000382\).

The sum from four through ten is \(0.0902235+0.0360894+0.0120298+0.0034371+0.0008593+0.0001909+0.0000382=0.1428682\).

To check the remaining infinite tail, \(P(X=11)=P(X=10)(2/11)\approx0.0000069\). For every subsequent step, the ratio is at most \(2/12=1/6\). Therefore the sum from 11 onward is no greater than \(0.0000069/(1-1/6)\approx0.0000083\). The full tail is between \(0.1428682\) and \(0.1428766\); both bounds round to 0.1429 to four decimal places.

Check. The calculated value is positive and less than \(P(X\ge3)\), as it must be because \(X\ge4\) is a subset of \(X\ge3\). The common mistake avoided here is to stop at a finite term without checking that the omitted tail cannot affect the requested rounding.

Common Endpoint Mistakes

Cumulative notation is compact, but it does not correct an incorrectly translated event. For example, for the inclusive range \(2\le X\le5\), the incorrect expression \(F_X(5)-F_X(2)\) removes the probability at \(X=2\) as well as the values below 2. In the first example, this wrong calculation gives \(0.9579790-0.5438131=0.4141659\), rather than the correct \(0.6706815\). The correct lower cumulative endpoint is one less than the requested lower bound:

$$ P(2\le X\le5)=F_X(5)-F_X(1), \quad\text{not}\quad F_X(5)-F_X(2). $$

Before evaluating any formula, write the event using integer values. Then use the following distinctions:

WordingEventProbability notation
No more than \(m\)\(X\le m\)\(F_X(m)\)
Fewer than \(m\)\(X\le m-1\)\(F_X(m-1)\)
At least \(r\)\(X\ge r\)\(\sum_{k=r}^{\infty}P(X=k)\)
More than \(r\)\(X\ge r+1\)\(\sum_{k=r+1}^{\infty}P(X=k)\)
From \(a\) through \(b\), inclusive\(a\le X\le b\)\(F_X(b)-F_X(a-1)\)
Key takeaway: A Poisson cumulative probability adds point probabilities from zero through its stated integer endpoint. For a finite inclusive range, subtract the cumulative probability through the integer just below the lower endpoint. For an upper tail, include the threshold itself and all larger counts.

Check Your Understanding

For each problem, identify the included integer outcomes before calculating.

  1. A hypothetical call center models incoming escalation requests per hour as \(X\sim\operatorname{Poisson}(1.5)\). Find the probability of no more than one request. (Answer: \(0.5578\))
  2. A hypothetical field team models equipment faults per day as \(X\sim\operatorname{Poisson}(2)\). Find the probability of two through four faults, inclusive. (Answer: \(0.5413\))
  3. A hypothetical greenhouse models pest detections in a weekly inspection as \(X\sim\operatorname{Poisson}(3.5)\). Find \(P(X\le2)\). (Answer: \(0.3208\))
  4. A hypothetical data center models short network interruptions per shift as \(X\sim\operatorname{Poisson}(2.4)\). Find the probability of three through five interruptions, inclusive. (Answer: \(0.3946\))
  5. A hypothetical repair depot models urgent parts requests per day as \(X\sim\operatorname{Poisson}(1.2)\). Find the probability of three through five requests, inclusive. (Answer: \(0.1190\))

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