Unit 11, lesson 71

Poisson and Normal Approximations

When a binomial probability can be approximated by a Poisson law in the rare-event regime or by a normal law with a continuity correction, and how absolute error, relative error and tail accuracy judge the result.

  • Video 71
  • Exact, Approximate and Simulated Answers

One hundred independent parts each fail with probability 0.030.03. The exact chance that none fails is 0.971000.97^{100}, and a Poisson law with mean 33 offers the simpler value e−3e^{-3}. Are the two close enough, and would a normal curve do as well? This lesson describes when a binomial count can be replaced by a Poisson or a normal law, shows how to turn an event about whole numbers into an interval for a continuous variable, and measures each approximation against the exact binomial value for the event that is actually asked about.

Key terms

Rare-event regime

Definition The rare-event regime for a count X∼Bin⁡(n,p)X \sim \operatorname{Bin}(n, p) is the setting of many trials, each with a small success probability, and a moderate mean npnp. The limit theorem behind it reads as follows. Let λ>0\lambda > 0, and for each n≥1n \ge 1 let Xn∼Bin⁡(n,pn)X_n \sim \operatorname{Bin}(n, p_n) with npn→λnp_n \to \lambda as n→∞n \to \infty. Then for every fixed k=0,1,2,…k = 0, 1, 2, \dots

lim⁡n→∞P(Xn=k)=e−λλkk!.\lim_{n \to \infty} P(X_n = k) = \frac{e^{-\lambda}\lambda^{k}}{k!}.

In this regime Bin⁡(n,p)\operatorname{Bin}(n, p) is approximated by Pois⁡(np)\operatorname{Pois}(np), the Poisson law with the same mean. Le Cam’s inequality bounds the error at a fixed nn. If Y∼Pois⁡(np)Y \sim \operatorname{Pois}(np), then ∣P(X∈A)−P(Y∈A)∣≤np2\lvert P(X \in A) - P(Y \in A) \rvert \le np^{2} for every set AA of counts.

In plain words Many independent chances, each of them unlikely, produce a count that is close to Poisson with the same mean. A matching mean alone is not enough. The count Bin⁡(3,1)\operatorname{Bin}(3, 1) has mean 33 but always equals 33, while Pois⁡(3)\operatorname{Pois}(3) gives the value 33 a probability of only about 0.2240.224. Its Le Cam bound is 3⋅12=33 \cdot 1^{2} = 3, which says nothing, since two probabilities never differ by more than 11.

Example For X∼Bin⁡(100,0.03)X \sim \operatorname{Bin}(100, 0.03) the mean is 33. With Y∼Pois⁡(3)Y \sim \operatorname{Pois}(3), the exact probability of no successes and its approximation are

P(X=0)=0.97100≈0.047553,P(Y=0)=e−3≈0.049787.\begin{aligned} P(X = 0) &= 0.97^{100} \\[0.4em] &\approx 0.047553, \\[0.4em] P(Y = 0) &= e^{-3} \\[0.4em] &\approx 0.049787. \end{aligned}

The approximation is too large by about 0.0022350.002235, well inside the Le Cam bound 100⋅0.032=0.09100 \cdot 0.03^{2} = 0.09.

Normal approximation to the binomial

Definition Fix pp with 0<p<10 < p < 1 and let Xn∼Bin⁡(n,p)X_n \sim \operatorname{Bin}(n, p). The count XnX_n is a sum of nn independent Bernoulli indicators with finite positive variance p(1−p)p(1 - p), so the central limit theorem gives, for every real zz,

P(Xn−npnp(1−p)≤z)→Φ(z)as n→∞.\begin{aligned} &P\left( \frac{X_n - np}{\sqrt{np(1 - p)}} \le z \right) \\[0.4em] &\quad \to \Phi(z) \quad \text{as } n \to \infty. \end{aligned}

The normal approximation replaces probabilities of X∼Bin⁡(n,p)X \sim \operatorname{Bin}(n, p) by those of Y∼N(np,np(1−p))Y \sim N\bigl(np, np(1 - p)\bigr), the normal law with the same mean and variance. The theorem is a limit statement and gives no error bound at a fixed nn. Screening rules, such as asking for both npnp and n(1−p)n(1 - p) to be at least 1010, are rough guides rather than theorems, and no single threshold makes every event accurate.

In plain words When the binomial bars spread over many values and are not too lopsided, their outline resembles a normal curve with the same center and spread. The approximation suits sums of many trials with a moderate pp, and it suits the rare-event regime poorly. A normal curve is always symmetric, while a binomial law with 0<p<10 < p < 1 is symmetric only when p=12p = \tfrac12. The curve also has no gaps between integers, and it puts positive probability below 00 and above nn, where a binomial count never falls.

Example For X∼Bin⁡(20,0.5)X \sim \operatorname{Bin}(20, 0.5) the mean is 1010 and the variance is 20⋅0.5⋅0.5=520 \cdot 0.5 \cdot 0.5 = 5, so the approximating law is N(10,5)N(10, 5) with standard deviation 5≈2.236\sqrt{5} \approx 2.236. The binomial law is symmetric here, which helps. For Bin⁡(100,0.03)\operatorname{Bin}(100, 0.03) the matching law is N(3,2.91)N(3, 2.91), and it puts about 0.0200.020 of its probability below −0.5-0.5, on counts that cannot occur.

Continuity correction

Definition Let XX take integer values and let YY be a continuous variable used to approximate it, such as a normal variable with the same mean and variance. First write the event as a set of consecutive integers. Then let each integer kk stand for the interval from k−12k - \tfrac12 to k+12k + \tfrac12. For integers a≤ba \le b this gives

P(a≤X≤b)≈P(a−12<Y<b+12).\begin{aligned} &P(a \le X \le b) \\[0.4em] &\quad \approx P\bigl(a - \tfrac12 < Y < b + \tfrac12\bigr). \end{aligned}

In the same way P(X≥a)P(X \ge a) is approximated by P(Y>a−12)P(Y > a - \tfrac12) and P(X≤b)P(X \le b) by P(Y<b+12)P(Y < b + \tfrac12). A strict inequality is rewritten first, so X>13X > 13 means X≥14X \ge 14 and becomes Y>13.5Y > 13.5, while X≥13X \ge 13 becomes Y>12.5Y > 12.5.

In plain words Each count is a bar of width one centered on an integer, and the continuous curve has to cover whole bars. An event whose smallest count is aa starts at the left edge of its bar, a−12a - \tfrac12. Without the correction a single count would get probability 00, since P(Y=k)=0P(Y = k) = 0 for a continuous YY, and an interval of counts would lose half of each end bar. The correction is a rule for choosing the interval. It does not make the binomial law normal, and it cannot rescue a normal model that is unsuitable for the parameters.

Example For X∼Bin⁡(20,0.5)X \sim \operatorname{Bin}(20, 0.5) and the inclusive event 8≤X≤128 \le X \le 12, the exact probability adds five masses,

P(8≤X≤12)=1220∑k=812(20k)=96577131072≈0.736824.\begin{aligned} &P(8 \le X \le 12) \\[0.4em] &\quad = \frac{1}{2^{20}} \sum_{k=8}^{12} \binom{20}{k} \\[0.4em] &\quad = \frac{96577}{131072} \approx 0.736824. \end{aligned}

With Y∼N(10,5)Y \sim N(10, 5) the corrected interval runs from 7.57.5 to 12.512.5, which standardizes to ±2.5/5≈±1.118\pm 2.5/\sqrt{5} \approx \pm 1.118, and P(7.5<Y<12.5)≈0.736448P(7.5 < Y < 12.5) \approx 0.736448, an error of about 0.0003770.000377. The uncorrected interval from 88 to 1212 gives only P(8<Y<12)≈0.628907P(8 < Y < 12) \approx 0.628907, an error of about 0.1079170.107917.

Absolute error and relative error

Definition Let p∗p_* be the exact probability of an event under the model and aa an approximation to it. The symbol p∗p_* is the target probability, not the success probability pp of a single trial. The absolute error is ∣a−p∗∣\lvert a - p_* \rvert, measured on the probability scale. When p∗>0p_* > 0 the relative error is

∣a−p∗∣p∗,\frac{\lvert a - p_* \rvert}{p_*},

often written as a percent. When p∗=0p_* = 0 the relative error is not defined.

In plain words Absolute error says how far off the answer is in probability units, and relative error compares that distance with the size of the answer. An absolute error of 0.010.01 is one percentage point. Reporting two in a million for an event whose probability is one in a million has an absolute error of only 0.0000010.000001 but a relative error of 100100 percent. Which measure matters depends on how the number will be used, and small probabilities usually call for relative error.

Example In the Poisson comparison for Bin⁡(100,0.03)\operatorname{Bin}(100, 0.03) the exact value is p∗=0.97100≈0.047553p_* = 0.97^{100} \approx 0.047553 and the approximation is a=e−3≈0.049787a = e^{-3} \approx 0.049787. The absolute error is about 0.0022350.002235, or 0.220.22 percentage points, and the relative error is

0.0022350.047553≈0.047,\frac{0.002235}{0.047553} \approx 0.047,

about 4.74.7 percent. For the corrected normal interval in the previous card, the absolute error 0.0003770.000377 against p∗≈0.736824p_* \approx 0.736824 is a relative error of about 0.050.05 percent.

Tail accuracy

Definition Tail accuracy is the accuracy of an approximation for a tail probability such as P(X≥t)P(X \ge t) or P(X≤t)P(X \le t) with tt far from the mean. Tail probabilities are small, so tail accuracy is judged by relative error. An error found for one event need not carry over to another event, even with the same parameters, so each approximation is checked for the event actually asked about.

In plain words A normal or Poisson approximation can match the middle of a binomial law closely and still be off by a large factor far out in a tail. Both probabilities are small there, so the absolute error looks harmless while the relative error can be large.

Example For X∼Bin⁡(20,0.5)X \sim \operatorname{Bin}(20, 0.5) the corrected normal approximation was off by about 0.050.05 percent on 8≤X≤128 \le X \le 12. For the upper tail X≥18X \ge 18, the exact value and the corrected approximation with Y∼N(10,5)Y \sim N(10, 5) are

P(X≥18)=211220≈0.000201,P(Y>17.5)≈0.000398.\begin{aligned} P(X \ge 18) &= \frac{211}{2^{20}} \\[0.4em] &\approx 0.000201, \\[0.4em] P(Y > 17.5) &\approx 0.000398. \end{aligned}

The absolute error is only about 0.0001970.000197, yet the relative error is about 9898 percent, so the approximation nearly doubles the answer. The Poisson approximation shows the same pattern. For Bin⁡(100,0.03)\operatorname{Bin}(100, 0.03) its relative error is about 4.74.7 percent at zero successes but about 2626 percent for P(X≥10)P(X \ge 10), where the exact value is about 0.0008740.000874 and the Poisson value about 0.0011020.001102.

Check yourself

  1. A count XX is approximated by a normal variable YY with a continuity correction. Write the corrected event for 4≤X≤74 \le X \le 7, for X>6X > 6 and for X<4X < 4.
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The inclusive event 4≤X≤74 \le X \le 7 contains the counts 44, 55, 66 and 77, whose bars together run from 3.53.5 to 7.57.5, so it becomes 3.5<Y<7.53.5 < Y < 7.5. The event X>6X > 6 does not contain 66. Its first included count is 77, so it becomes Y>6.5Y > 6.5, not Y>5.5Y > 5.5. The event X<4X < 4 means X≤3X \le 3, whose last bar ends at 3.53.5, so it becomes Y<3.5Y < 3.5. Whether the normal approximation is accurate for these events still depends on the parameters of XX.

  1. Each of 8080 independent trials succeeds with probability 0.020.02. Find the exact probability of at least one success, its Poisson approximation, and the absolute and relative errors.
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The count is X∼Bin⁡(80,0.02)X \sim \operatorname{Bin}(80, 0.02), and at least one success is the complement of none. The exact value is

P(X≥1)=1−0.9880≈0.801351.\begin{aligned} P(X \ge 1) &= 1 - 0.98^{80} \\[0.4em] &\approx 0.801351. \end{aligned}

The Poisson law with mean 80⋅0.02=1.680 \cdot 0.02 = 1.6 gives 1−e−1.6≈0.7981031 - e^{-1.6} \approx 0.798103. The absolute error is about 0.0032480.003248, and the relative error is about 0.003248/0.801351≈0.00410.003248/0.801351 \approx 0.0041, or 0.410.41 percent. For the probability of no successes, 0.9880≈0.1986490.98^{80} \approx 0.198649 against e−1.6≈0.201897e^{-1.6} \approx 0.201897, the absolute error is the same, but the relative error is about 1.61.6 percent because the target is smaller.

  1. For X∼Bin⁡(20,0.5)X \sim \operatorname{Bin}(20, 0.5), the corrected normal approximation to P(X≥13)P(X \ge 13) is about 0.1317760.131776 against the exact 0.1315880.131588. Someone concludes that the same approximation is just as reliable for P(X≤3)P(X \le 3). Compute both values for P(X≤3)P(X \le 3) and judge the claim.
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The exact value adds the masses at 00, 11, 22 and 33,

P(X≤3)=1+20+190+1140220=13511048576≈0.001288.\begin{aligned} &P(X \le 3) \\[0.4em] &\quad = \frac{1 + 20 + 190 + 1140}{2^{20}} \\[0.4em] &\quad = \frac{1351}{1048576} \approx 0.001288. \end{aligned}

With Y∼N(10,5)Y \sim N(10, 5) the corrected event is Y<3.5Y < 3.5, and P(Y<3.5)=Φ(−6.5/5)≈Φ(−2.907)≈0.001825P(Y < 3.5) = \Phi(-6.5/\sqrt{5}) \approx \Phi(-2.907) \approx 0.001825. The absolute error is about 0.0005370.000537, small on the probability scale, but the relative error is about 0.000537/0.001288≈420.000537/0.001288 \approx 42 percent. For P(X≥13)P(X \ge 13) the relative error was only about 0.140.14 percent. The claim is wrong. A close match on a moderate tail does not certify a far tail, so each event needs its own check.

How this lesson connects