Poisson Tail Approximations and Complement Events

Why Poisson Tail Calculations Need Care

A Poisson count can model the number of events observed in a fixed interval when events occur independently at a stable average rate. For a sufficiently large rate, a normal distribution can provide a useful approximation. Tail questions still require care: the count is discrete, while the normal model is continuous, and a small boundary or complement error can noticeably change the answer.

The key decisions are to translate the integer event into the correct half-unit boundary and to calculate the tail that is easiest to handle. A high-probability event may be simpler as one minus a small complementary tail. A rare tail is often clearer to calculate directly. These are two equivalent routes only when their integer boundaries are matched correctly.

Model and approximation: If \(X\sim\operatorname{Poisson}(\lambda)\), then \(E[X]=\lambda\) and \(\operatorname{Var}(X)=\lambda\). For a sufficiently large \(\lambda\), approximate \(X\) by \(Y\sim N(\lambda,\lambda)\), whose standard deviation is \(\sqrt{\lambda}\). This is an approximation, not an exact identity. It may be unreliable when the Poisson rate is too small for a normal approximation to be suitable.

Translate the Event Before Choosing a Tail

For an integer-valued count, include or exclude the endpoint in the event before applying a continuity correction. The correction places the normal boundary halfway between the last included integer and the first excluded integer.

\begin{aligned} P(X\le k)&\approx \Phi\left(\frac{k+0.5-\lambda}{\sqrt{\lambda}}\right),\\ P(X<k)=P(X\le k-1)&\approx \Phi\left(\frac{k-0.5-\lambda}{\sqrt{\lambda}}\right),\\ P(X\ge k)&\approx 1-\Phi\left(\frac{k-0.5-\lambda}{\sqrt{\lambda}}\right),\\ P(X>k)=P(X\ge k+1)&\approx 1-\Phi\left(\frac{k+0.5-\lambda}{\sqrt{\lambda}}\right). \end{aligned}

Here, \(\Phi(z)\) denotes the cumulative distribution function of a standard normal random variable. For example, \(X\ge k\) includes \(k\), so its normal region begins at \(k-0.5\). In contrast, \(X>k\) starts at the next integer, \(k+1\), so its region begins at \(k+0.5\). Keeping the strict or non-strict inequality visible is the most reliable way to avoid a one-unit error.

Why the Half-Unit Boundary Works

The continuity correction treats each integer count \(j\) as occupying a normal interval of width one, from \(j-0.5\) to \(j+0.5\). These intervals meet at their endpoints and cover the number line. The normal approximation to the probability of the count \(j\) is therefore the normal area in that interval. Adding the intervals for all included integer counts gives the tail boundary.

\begin{aligned} P(X\ge k) &\approx \sum_{j=k}^{\infty}P(j-0.5\le Y<j+0.5)\\ &=P(Y\ge k-0.5)\\ &=1-\Phi\left(\frac{k-0.5-\lambda}{\sqrt{\lambda}}\right). \end{aligned}

The same reasoning for \(X\le k\) combines the intervals through \(k+0.5\). The approximation replaces a discrete probability by a continuous area; it does not make the Poisson count continuous or change its possible values.

A Direct-or-Complement Strategy

A complement is especially useful when the requested event has probability near one. Rather than calculate a large area directly, calculate the smaller excluded event and subtract it from one. The identities \(P(X\ge k)=1-P(X\le k-1)\) and \(P(X>k)=1-P(X\le k)\) show why the boundary changes with the inequality.

Write the count event exactly.
Record whether the endpoint is included. For example, distinguish \(X\ge k\) from \(X>k\).

Decide whether the event or its complement is smaller.
For a small upper tail, calculate the upper normal tail directly. For a high-probability upper-tail event, its lower-tail complement may be more convenient.

Apply the matching half-unit boundary.
For \(X\ge k\), use \(k-0.5\); for its complement \(X\le k-1\), the upper boundary is also \(k-0.5\).

Standardize and retain precision.
Subtract \(\lambda\), divide by \(\sqrt{\lambda}\), and keep sufficient precision in \(\Phi\) values until the final rounding.

Worked Tail Calculations

Worked Example: A Small Upper Tail Calculated Directly

A hypothetical monitoring system counts alerts in a 24-hour period. Assume the count \(X\) has a Poisson distribution with parameter \(\lambda=100\), so the mean and variance are both 100. Estimate the probability of at least 112 alerts. Interest and mortality assumptions do not apply to this count model.

Method. This is a relatively small upper tail, so use the direct upper-tail formula. The event is \(X\ge112\), and its continuity-corrected boundary is \(111.5\), not \(112.5\).

The normal approximation is \(Y\sim N(100,100)\), with standard deviation \(\sqrt{100}=10\). Standardizing the boundary gives

z=\frac{111.5-100}{10}=\frac{11.5}{10}=1.15.

Thus

P(X\ge112)\approx P(Y\ge111.5)=1-\Phi(1.15) =1-0.874928\ldots=0.125071\ldots.

The estimate is 0.1251 to four decimal places. As a check, the complementary event is \(X\le111\), whose corrected boundary is also \(111.5\); its approximate probability is \(0.874928\ldots\). The two estimates sum to one before rounding.

Worked Example: A Lower Tail

A hypothetical laboratory records the number of particles detected in a fixed sampling interval. Suppose \(X\sim\operatorname{Poisson}(225)\). Estimate the probability that the count is at most 207. Interest and mortality assumptions are not relevant.

Method. The event is \(X\le207\), so the included counts extend through 207. The corrected upper boundary is \(207.5\). The normal approximation is \(Y\sim N(225,225)\), with standard deviation \(\sqrt{225}=15\).

z=\frac{207.5-225}{15}=\frac{-17.5}{15}=-1.166666\ldots

The standard normal probability is

P(X\le207)\approx\Phi(-1.166666\ldots) \approx0.1216725\ldots.

Therefore the estimate is 0.1217 to four decimal places. The result is a lower-tail probability, so it should be below one-half: the boundary 207.5 is below the mean 225. This provides a quick direction check.

Worked Example: A High-Probability Event Using Its Complement

A hypothetical depot tracks the number of small equipment requests received per shift. Assume \(X\sim\operatorname{Poisson}(64)\), and estimate the probability of receiving at least 54 requests. Interest and mortality assumptions do not apply.

Method. Since 54 is below the mean 64, the requested upper-tail event has probability greater than one-half. Calculate its complement, \(X\le53\), which is the smaller lower tail.

The normal approximation is \(Y\sim N(64,64)\), with standard deviation \(\sqrt{64}=8\). The complement includes counts through 53, so its corrected boundary is \(53.5\):

z=\frac{53.5-64}{8}=\frac{-10.5}{8}=-1.3125.

Then

\begin{aligned} P(X\ge54) &=1-P(X\le53)\\ &\approx1-\Phi(-1.3125)\\ &\approx1-0.0947\\ &=0.9053. \end{aligned}

The estimate is 0.9053 to four decimal places. The result is consistent with the event beginning well below the mean. The complement boundary \(53.5\) is also the lower boundary for the event \(X\ge54\), which checks that the integer events align.

Worked Example: A Small Upper Tail With a Large Rate

A hypothetical sensor records \(X\sim\operatorname{Poisson}(144)\) signals in a day. Estimate \(P(X\ge159)\). Interest and mortality assumptions are not applicable.

Method. This is an upper-tail event above the mean, so calculate the direct upper normal tail. Its corrected boundary is \(158.5\). The normal approximation is \(Y\sim N(144,144)\), with standard deviation \(\sqrt{144}=12\).

z=\frac{158.5-144}{12}=\frac{14.5}{12}=1.208333\ldots

Using the unrounded standardized value,

P(X\ge159)\approx1-\Phi(1.208333\ldots) \approx0.1134596\ldots.

The estimate is 0.1135 to four decimal places. As a check, the result is a small probability because 159 is above the mean. Retaining precision through the normal probability calculation matters: rounding an intermediate table value too early can change the final fourth decimal.

Common Boundary and Complement Errors

A frequent mistake is to use the integer threshold itself as the normal boundary. For instance, for \(X\ge159\), using \(159.5\) produces the calculation \(1-\Phi((159.5-144)/12)\). That boundary lies halfway between 159 and 160, so it approximates \(P(X\ge160)\), not \(P(X\ge159)\). The correct boundary is \(158.5\), giving \(1-\Phi((158.5-144)/12)\).

Another error is complementing the wrong event. The complement of \(X\ge k\) is \(X\le k-1\), not \(X\le k\). The complement of \(X>k\), however, is \(X\le k\). Writing the event in integer form before changing it to a normal probability prevents this off-by-one mistake.

Do not use the same boundary for \(X\ge k\) and \(X>k\). Their corrected boundaries are \(k-0.5\) and \(k+0.5\), respectively.
Do not round each table probability before subtracting. Keep enough digits for \(1-\Phi(z)\) or a difference of CDF values, then round the final result to the requested precision.
Do not interpret the approximation as exact. The formulas rely on a normal approximation to a Poisson count; the Poisson model itself remains discrete.

Key takeaway: First determine the exact integer event, then place the half-unit boundary, and only then choose a direct tail or its complement. For \(X\ge k\), the boundary is \(k-0.5\); for \(X>k\), it is \(k+0.5\). Equivalent complement calculations must use the same dividing boundary.

Check Your Understanding

For each problem, use a continuity-corrected normal approximation and round the final probability to four decimal places. Assume the stated Poisson model; interest and mortality assumptions are not applicable.

A hypothetical call center receives \(X\sim\operatorname{Poisson}(81)\) calls in a specified interval. Estimate \(P(X\ge91)\). (Answer: 0.1456)
A hypothetical test station records \(X\sim\operatorname{Poisson}(196)\) defects in a batch. Estimate \(P(X\le180)\). (Answer: 0.1341)
A hypothetical storage unit receives \(X\sim\operatorname{Poisson}(100)\) requests in an hour. Estimate \(P(X>108)\), taking care with the strict inequality. (Answer: 0.1977)
A hypothetical fleet desk processes \(X\sim\operatorname{Poisson}(121)\) service notices in a shift. Estimate \(P(X\ge111)\) using the complementary event. (Answer: 0.8301)

What You'll Learn