Why Approximate a Poisson Count?
A Poisson model is useful for counts of events over a fixed exposure, such as claims recorded by a hypothetical insurer over a year. When its rate is large, summing the exact Poisson probabilities for a tail or interval can be inconvenient. A normal distribution can provide a quick approximation, but its usefulness depends on the size of the Poisson rate and on how the event is expressed.
The key feature is that a Poisson random variable has the same value for its mean and variance. Consequently, its normal approximation has mean equal to the Poisson rate and variance equal to that rate—not standard deviation equal to the rate. This tutorial uses the direct, unadjusted normal approximation at the stated count thresholds. Because a count is integer-valued and a normal variable is continuous, this approximation can be rough, especially at small rates.
The Poisson model assigns probability \(e^{-\lambda}\lambda^n/n!\) to \(N=n\), for \(n=0,1,2,\ldots\). Its rate \(\lambda\) is the expected count over the specified exposure. If a homogeneous Poisson model is appropriate and the exposure is multiplied by a constant, the rate is multiplied by the same constant. The approximation is not useful when \(\lambda=0\): in that case the count is zero with certainty, rather than a nondegenerate normal random variable.
Standardizing Poisson Count Events
To use the approximation, first identify the Poisson rate for the full exposure period. Then use that rate for both the normal mean and variance, and standardize the threshold with \(\sqrt{\lambda}\). For lower tails, upper tails, and intervals, the resulting calculations are:
In these expressions, \(k\), \(a\), and \(b\) are count thresholds, with \(a\leq b\). The interval formula subtracts the approximated lower tail through \(a\) from the approximated lower tail through \(b\), using the direct threshold convention shown. It does not make the Poisson count continuous: it uses a continuous distribution to approximate probabilities for a discrete count. If the requested event is a single exact count, its probability is a point probability, not a normal cumulative probability; the exact Poisson formula remains available for checking it.
Why the Normal Distribution Appears
The approximation is supported by a limiting argument. For a Poisson variable \(N\), the mean and variance are both \(\lambda\). Standardizing gives \(Z=(N-\lambda)/\sqrt{\lambda}\), which has mean zero and variance one. More specifically, its moment-generating function approaches that of a standard normal variable as \(\lambda\) grows.
Set \(Z=(N-\lambda)/\sqrt{\lambda}\), so that the centered count is measured in units of its standard deviation.
Since the Poisson moment-generating function is \(E[e^{sN}]=\exp(\lambda(e^s-1))\), substituting \(s=t/\sqrt{\lambda}\) and centering gives \(E[e^{tZ}]=\exp\{\lambda(e^{t/\sqrt{\lambda}}-1-t/\sqrt{\lambda})\}\).
The Taylor expansion \(e^u=1+u+u^2/2+u^3/6+\cdots\) makes the exponent \(t^2/2+t^3/(6\sqrt{\lambda})+\cdots\). As \(\lambda\) grows, this tends to \(t^2/2\), the exponent of the standard normal moment-generating function.
The standardized Poisson distribution approaches the standard normal distribution for large \(\lambda\). This supports the approximation but does not guarantee high accuracy for every finite rate or every threshold.
The Poisson distribution is generally right-skewed at smaller rates. Its standardized skewness is \(1/\sqrt{\lambda}\), which decreases as \(\lambda\) increases. At a small rate, the probability mass near zero and the right-skewed shape can differ substantially from a symmetric normal curve. No single cutoff makes the approximation uniformly accurate for all events; assess the rate, the threshold, and the consequences of approximation error.
Worked Applications
Worked Example: An Interval of Annual Claim Counts
A hypothetical portfolio has 8,000 independent exposure units. Assume each unit has a Poisson claim count with rate \(0.008\) for the year, and that the total count follows a Poisson model. Approximate the probability that the portfolio has between 56 and 72 claims, inclusive, using the direct normal approximation.
Method. Find the total Poisson rate, use it as the normal mean and variance, and apply the interval formula. This calculation avoids the mistake of treating the interval as though its endpoints were already standardized.
Step 1: Calculate the total rate. The rate for 8,000 exposure units is \( \lambda=8{,}000(0.008)=64 \). Thus \(N\sim\operatorname{Poisson}(64)\), and the approximating variable is \(Y\sim N(64,64)\).
Step 2: Find the standard deviation. The normal variance is 64, so its standard deviation is \(\sqrt{64}=8\).
Step 3: Standardize both endpoints. \(z_{56}=(56-64)/8=-1\), and \(z_{72}=(72-64)/8=1\).
Step 4: Subtract the normal cumulative probabilities. Using \(\Phi(1)=0.8413\) and \(\Phi(-1)=0.1587\), \(P(56\leq N\leq72)\approx0.8413-0.1587=0.6826\), or \(0.6827\) using unrounded standard normal values.
Check. The interval is centered at the rate 64, and its endpoints are one standard deviation below and above that rate. The resulting probability is about 0.6827, as expected for the corresponding standard normal interval.
Worked Example: A High-Count Threshold
A hypothetical equipment-monitoring program records events at a Poisson rate of \(0.08\) per device-year. The program covers 1,000 device-years. Approximate the probability of at least 92 events in that exposure period using the direct normal approximation.
Method. Scale the rate to the full exposure and standardize the lower threshold of the upper-tail event. This avoids using the per-device rate as though it were the portfolio rate.
Step 1: Calculate the total rate. \( \lambda=1{,}000(0.08)=80 \). The count model is \(N\sim\operatorname{Poisson}(80)\), and the normal approximation is \(Y\sim N(80,80)\).
Step 2: Calculate the standard deviation. \( \sqrt{80}\approx8.9443 \).
Step 3: Standardize the threshold. \(z=(92-80)/\sqrt{80}=12/8.9443\approx1.3416\).
Step 4: Find the upper tail. The standard normal cumulative probability at \(1.3416\) is approximately \(0.9101\). Therefore, \(P(N\geq92)\approx1-0.9101=0.0899\).
Check. The threshold 92 is above the mean of 80, so the upper-tail probability should be less than one-half. The result, approximately \(0.0899\), has the correct direction and scale.
Worked Example: Testing the Approximation at a Small Rate
A hypothetical low-volume reporting process has a Poisson rate of \(\lambda=1\) event per period. Compare the direct normal approximation to the probability of zero events with the exact Poisson probability, and assess whether the approximation is suitable for this question.
Method. Since \(N\) is a nonnegative integer, \(P(N=0)=P(N\leq0)\). Approximate that lower tail with a normal cumulative probability, then check it against the exact Poisson probability. This comparison avoids assuming that an approximation is accurate merely because its mean and variance match.
Step 1: Set up the normal model. The approximation is \(Y\sim N(1,1)\), with standard deviation \(\sqrt{1}=1\).
Step 2: Approximate the zero-count probability. Standardizing zero gives \(z=(0-1)/1=-1\), so \(P(N=0)=P(N\leq0)\approx\Phi(-1)=0.1587\).
Step 3: Calculate the exact probability. From the Poisson probability formula, \(P(N=0)=e^{-1}1^0/0!=e^{-1}\approx0.3679\).
Step 4: Compare and assess. The absolute difference is \(0.3679-0.1587=0.2092\). The approximation substantially understates the probability of zero events, so it is not suitable for this small-rate event.
Check. The exact probability of at least one event is \(1-0.3679=0.6321\); the normal approximation gives \(1-0.1587=0.8413\). These complementary calculations show the same large discrepancy from the other side.
Common Mistake and Practical Judgment
A frequent error is to divide by the variance instead of by the standard deviation. For the rate 80 in the second example, the incorrect calculation would be \(z=(92-80)/80=0.15\), leading to an upper-tail estimate of \(1-\Phi(0.15)\approx0.4404\). The correct standard deviation is \(\sqrt{80}\), so the correct standardized value is approximately \(1.3416\), and the direct approximation is \(0.0899\). The difference is large enough to change the interpretation of the result.
A second practical check is whether the rate is large enough for the count distribution to look reasonably symmetric near the event of interest. The normal distribution assigns some probability to negative values, while a Poisson count cannot be negative. For the normal approximation \(N(\lambda,\lambda)\), that negative probability is \(\Phi(-\sqrt{\lambda})\); it becomes smaller as \(\lambda\) grows. At small rates, the normal approximation can therefore misrepresent not only the tails but also the concentration of probability near zero.
Check Your Understanding
Use the direct, unadjusted normal approximation unless a question asks for an exact Poisson probability.
- A hypothetical count is \(N\sim\operatorname{Poisson}(49)\). Approximate \(P(N\leq56)\). (Answer: \(0.8413\))
- A hypothetical monitoring system covers 1,600 device-years at a Poisson rate of \(0.04\) events per device-year. Approximate the probability of at least 72 events. (Answer: \(0.1587\))
- A hypothetical portfolio count is Poisson with rate 100. Approximate the probability that the count is between 90 and 110, inclusive. (Answer: \(0.6827\))
- A hypothetical process has count \(N\sim\operatorname{Poisson}(144)\). Approximate \(P(N\leq126)\). (Answer: \(0.0668\))
- For a hypothetical count \(N\sim\operatorname{Poisson}(1)\), compare the normal approximation to \(P(N=0)\) with the exact probability, and state whether the approximation is satisfactory. (Answer: Normal approximation \(0.1587\); exact probability \(e^{-1}=0.3679\); not satisfactory.)