What a Short-Interval Error Term Says
In a short interval of length \(h\), a probability approximation is a statement about how its error behaves as \(h\) decreases to zero. The symbol \(o(h)\) does not mean “small” without qualification: it means that the error, divided by \(h\), tends to zero. This scale matters in the infinitesimal description of a Poisson process. A first-order term can be correct even when the approximation is not accurate to second order, and an error of order \(h^2\) is generally not small enough to call \(o(h^2)\).
Let \((N_t)_{t\ge0}\) be a Poisson process of rate \(\lambda>0\) on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(S=\{0,1,2,\ldots\}\) and continuous time \(t\ge0\). For a fixed time \(t\), write \(Y_h=N_{t+h}-N_t\) for the count in the interval \((t,t+h]\). Its distribution is Poisson with parameter \(\lambda h\). The infinitesimal probabilities of interest are those of \(Y_h=0\), \(Y_h=1\), and \(Y_h\ge2\), as \(h\downarrow0\).
Thus \(r(h)=o(h)\) means that the error is negligible relative to \(h\). By contrast, \(r(h)=O(h)\) only bounds the error by a constant multiple of \(h\); it need not be negligible compared with \(h\). For example, \(h\) is \(O(h)\), but \(h\) is not \(o(h)\), because \(h/h=1\). Also, \(h^2=o(h)\), since \(h^2/h=h\to0\), while \(h^2\) is not \(o(h^2)\), since \(h^2/h^2=1\).
Proof. By definition, \(f(h)=ah+o(h)\) means that \(f(h)-ah=r(h)\) for a function \(r\) with \(r(h)/h\to0\). Dividing by \(h>0\) gives \(f(h)/h-a\to0\), which is equivalent to \(f(h)/h\to a\). Conversely, if \(f(h)/h\to a\), then \((f(h)-ah)/h\to0\), so \(f(h)-ah=o(h)\). If both \(f(h)=ah+o(h)\) and \(f(h)=bh+o(h)\), then \(f(h)/h\) tends to both \(a\) and \(b\). A real-valued function cannot have two different limits, so \(a=b\). \(\square\)
Poisson Probabilities at Short Times
The infinitesimal form for a positive-rate Poisson process is \( \mathbb{P}(Y_h=0)=1-\lambda h+o(h)\), \( \mathbb{P}(Y_h=1)=\lambda h+o(h)\), and \( \mathbb{P}(Y_h\ge2)=o(h)\). These statements specify first-order accuracy. They do not, by themselves, specify the coefficient of an \(h^2\) term. The exact Poisson probabilities make it possible to check both the stated orders and stronger proposed orders.
Proof. Set \(x=\lambda h\), so \(x\downarrow0\). The Poisson probability formula gives \(\mathbb{P}(Y_h=0)=e^{-x}\), \(\mathbb{P}(Y_h=1)=xe^{-x}\), and \(\mathbb{P}(Y_h\ge2)=1-(1+x)e^{-x}\). Taylor's theorem for the exponential gives \(e^{-x}=1-x+x^2/2+O(x^3)\). Substituting \(x=\lambda h\) yields the displayed expansion for zero events. Multiplying that expansion by \(x\) gives \(xe^{-x}=x-x^2+O(x^3)\), which yields the one-event expansion. Finally, \((1+x)e^{-x}=(1+x)(1-x+x^2/2+O(x^3))=1-x^2/2+O(x^3)\). Taking the complement gives \(1-(1+x)e^{-x}=x^2/2+O(x^3)\), as claimed. Since \(h^2/h=h\to0\), both \(h^2\) and \(O(h^3)\) are \(o(h)\). Thus the multiple-event probability is \(o(h)\), and the zero- and one-event expansions differ from their first-order terms by \(o(h)\). \(\square\)
The argument uses the exact Poisson count probabilities, not an informal interpretation of “very short.” The \(O(h^3)\) terms in the result have constants depending on the fixed rate \(\lambda\). If the rate were allowed to vary with \(h\), this statement would require a separate uniform bound.
Worked Example: Checking All Three Orders at Rate Two
Take \(\lambda=2\), and let \(Y_h\) count events in an interval of length \(h>0\). The exact probabilities are \(\mathbb{P}(Y_h=0)=e^{-2h}\), \(\mathbb{P}(Y_h=1)=2he^{-2h}\), and \(\mathbb{P}(Y_h\ge2)=1-(1+2h)e^{-2h}\). Taylor expansion gives \(e^{-2h}=1-2h+2h^2+O(h^3)\). Therefore \(\mathbb{P}(Y_h=0)=1-2h+2h^2+O(h^3)\). The error relative to \(1-2h\) is \(2h^2+O(h^3)\); dividing by \(h\) gives \(2h+O(h^2)\to0\). Thus the zero-event approximation has error \(o(h)\).
Multiplying the same expansion by \(2h\) gives \(\mathbb{P}(Y_h=1)=2h-4h^2+O(h^3)\). The error relative to \(2h\) is \(-4h^2+O(h^3)\); after division by \(h\), this tends to zero. For the multiple-event probability, substitute the zero- and one-event probabilities: \(\mathbb{P}(Y_h\ge2)=1-e^{-2h}-2he^{-2h}=2h^2+O(h^3)\). Dividing by \(h\) gives \(2h+O(h^2)\to0\), so this probability is \(o(h)\).
Check: The three exact probabilities sum to \(1\) because \(e^{-2h}+2he^{-2h}+[1-(1+2h)e^{-2h}]=1\). Their first-order terms also sum to \(1\): the zero-event term contributes \(1-2h\), the one-event term contributes \(2h\), and the multiple-event term has no first-order contribution.
Testing Proposed Approximations
To test a proposed first-order approximation, subtract it from the exact probability and divide the difference by \(h\). The limit must be zero for an \(o(h)\) error. If the limit is nonzero, the coefficient of \(h\) is wrong. If the limit is zero, the first-order claim is valid, but this does not establish a more precise order such as \(o(h^2)\).
Worked Example: A Wrong Rate Coefficient and a Flexible Remainder
Take a Poisson process with rate \(\lambda=3\). For \(h>0\), the exact zero-event probability is \(e^{-3h}=1-3h+\tfrac92h^2+O(h^3)\). Consider the proposed approximation \(1-2h+o(h)\). Its difference from the exact probability, divided by \(h\), is \(\big(e^{-3h}-(1-2h)\big)/h = -1+\tfrac92h+O(h^2)\), which tends to \(-1\), not \(0\). The proposed coefficient \(2\) is incorrect; the first-order coefficient must be \(3\).
Now consider the proposed one-event approximation \(3h+h^{3/2}\). Since \(\mathbb{P}(Y_h=1)=3he^{-3h}=3h-9h^2+O(h^3)\), the difference between the exact probability and the proposed expression is \(-h^{3/2}-9h^2+O(h^3)\). After division by \(h\), this is \(-\sqrt{h}-9h+O(h^2)\to0\). So the approximation is valid to order \(o(h)\), even though its remainder includes \(h^{3/2}\), which is not \(O(h^2)\).
Check: The first test has a nonzero limiting error ratio, so it fails the required order. The second has a zero limiting error ratio, so it passes. The distinction follows directly from the definition and does not depend on how small either error looks at one selected value of \(h\).
Worked Example: Why the Multiple-Event Probability Is Not o(h²)
Again take \(\lambda=3\). The exact multiple-event probability is \(\mathbb{P}(Y_h\ge2)=1-(1+3h)e^{-3h}\). Using \(e^{-3h}=1-3h+\tfrac92h^2+O(h^3)\), multiplication gives \((1+3h)e^{-3h}=1-\tfrac92h^2+O(h^3)\). Hence \(\mathbb{P}(Y_h\ge2)=\tfrac92h^2+O(h^3)\).
Dividing by \(h\) gives \(\tfrac92h+O(h^2)\to0\), which verifies \(o(h)\). But dividing by \(h^2\) gives \(\tfrac92+O(h)\to\tfrac92\), not zero. Therefore the probability is not \(o(h^2)\), even though it is \(O(h^2)\). The stronger, correctly calibrated second-order statement is \(\mathbb{P}(Y_h\ge2)=\tfrac92h^2+o(h^2)\).
Check: The coefficient is \(\lambda^2/2=3^2/2=9/2\). Substitution into the limiting ratio gives \(\mathbb{P}(Y_h\ge2)/h^2\to9/2\), agreeing with the expansion and ruling out \(o(h^2)\).
Useful Order Rules and Common Pitfalls
If \(r(h)=O(h^2)\), then \(r(h)=o(h)\): indeed, for sufficiently small \(h>0\), \(|r(h)|\le Ch^2\), so \(|r(h)/h|\le Ch\to0\). The converse does not hold in general. The function \(h^{3/2}\) is \(o(h)\), because \(h^{3/2}/h=\sqrt h\to0\), but it is not \(O(h^2)\), because \(h^{3/2}/h^2=1/\sqrt h\) is unbounded near zero.
Probability normalization gives a useful consistency check. For each \(h>0\), the three events \(Y_h=0\), \(Y_h=1\), and \(Y_h\ge2\) are disjoint and cover all possibilities. Their probabilities must sum to one. If two proposed approximations have first-order errors \(o(h)\), the third probability is determined by subtraction, provided the approximations refer to this same count and interval. This is an algebraic check on the approximations; it is not a replacement for verifying their errors against the model.
Common mistake. A frequent error is to read “the probability of two or more events is \(o(h)\)” as “it is \(o(h^2)\).” For a positive-rate Poisson process the leading term is \(\lambda^2h^2/2\), so dividing by \(h\) gives a quantity tending to zero, while dividing by \(h^2\) gives \(\lambda^2/2>0\). Another mistake is to accept \(1-ch+o(h)\) for the zero-event probability without checking \(c\). The quotient test shows that \(c\) must equal \(\lambda\).
The positive-rate assumption matters for statements about the leading multiple-event term. If \(\lambda=0\), then \(Y_h=0\) with probability one for every \(h\); the one-event and multiple-event probabilities are exactly zero. The positive-rate expansion with a nonzero coefficient \(\lambda^2/2\) is intended for fixed \(\lambda>0\). Always state which rate regime is under discussion before interpreting a leading term.
Check Your Understanding
For each item, use a limit of the error divided by the stated power of \(h\) to justify your answer.
- For fixed \(\lambda>0\), does \(1-\lambda h+7h^2\) approximate the Poisson zero-event probability with error \(o(h)\)?
- Is \(3h+o(h)\) a valid first-order approximation to the one-event probability when \(\lambda=3\)? What coefficient would make it invalid?
- For a Poisson process of rate \(2\), is the multiple-event probability \(o(h^2)\), \(O(h^2)\), or both?
- Give a function that is \(o(h)\) but is not \(O(h^2)\), and verify both claims from the definitions.
- Why does checking a numerical error at one positive value of \(h\) not establish that an error is \(o(h)\)?