Poisson Recurrence, Mode, and Tail Computation

From the Poisson Mass Function to a Recurrence

For a Poisson random variable, probabilities at neighboring counts are closely related. This makes the mass function easier to calculate than a list of separate factorial expressions suggests. The same relationship also reveals where the probabilities reach their largest value and helps organize tail calculations.

Let $X\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$. Its support is the set of nonnegative integers, and its probability mass function is

p_k=P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}, \qquad k=0,1,2,\ldots.

Here $k!$ is the factorial, with $0!=1$. Dividing the probability at count $k+1$ by the probability at count $k$ eliminates the common exponential term.

Theorem: If $X\sim\operatorname{Poisson}(\lambda)$ with $\lambda>0$, then for every integer $k\geq 0$, $$ \frac{P(X=k+1)}{P(X=k)}=\frac{\lambda}{k+1}, \qquad\text{so}\qquad P(X=k+1)=\frac{\lambda}{k+1}P(X=k). $$

To verify the ratio, substitute the mass function at the two counts:

\frac{p_{k+1}}{p_k} = \frac{e^{-\lambda}\lambda^{k+1}/(k+1)!} {e^{-\lambda}\lambda^k/k!} = \frac{\lambda^{k+1}}{\lambda^k} \frac{k!}{(k+1)!} = \frac{\lambda}{k+1}.

The cancellation is valid because $p_k>0$ when $\lambda>0$. Starting from $p_0=e^{-\lambda}$, the recurrence gives every later probability using one multiplication and one division:

p_1=\lambda p_0,\qquad p_2=\frac{\lambda}{2}p_1,\qquad p_3=\frac{\lambda}{3}p_2,\quad\ldots

Worked Example: Generating Probabilities Recursively

Suppose $X\sim\operatorname{Poisson}(2.5)$. Find $P(X=0),\ldots,P(X=4)$, using the recurrence.

Start with $p_0=e^{-2.5}\approx 0.0820850$. Apply $p_{k+1}=2.5p_k/(k+1)$ at each step:

\begin{aligned} p_1&=\frac{2.5}{1}(0.0820850)\approx 0.2052125,\\ p_2&=\frac{2.5}{2}(0.2052125)\approx 0.2565156,\\ p_3&=\frac{2.5}{3}(0.2565156)\approx 0.2137630,\\ p_4&=\frac{2.5}{4}(0.2137630)\approx 0.1336019. \end{aligned}

Thus the probabilities from zero through four are approximately $0.0820850, 0.2052125, 0.2565156, 0.2137630,$ and $0.1336019$. The probabilities increase to $k=2$ and then decrease, consistent with the mode calculation below.

Locating the Mode

A mode of a discrete distribution is a support value at which its probability mass is largest. For the Poisson distribution, compare consecutive probabilities rather than attempting to maximize the factorial formula all at once. The ratio is greater than one while the probabilities rise, equals one when adjacent probabilities tie, and is less than one while they fall.

Theorem: For $X\sim\operatorname{Poisson}(\lambda)$, $\lambda>0$:

If $\lambda$ is not an integer, the unique mode is $\lfloor\lambda\rfloor$, the greatest integer less than $\lambda$.
If $\lambda=m$ is a positive integer, the two modes are $m-1$ and $m$.

Proof. The ratio $p_{k+1}/p_k=\lambda/(k+1)$ is greater than one exactly when $k+1<\lambda$, equals one exactly when $k+1=\lambda$, and is less than one exactly when $k+1>\lambda$. Therefore the sequence of masses increases as long as the next count is below $\lambda$, and decreases once the next count is above $\lambda$.

If $\lambda$ is not an integer and $r=\lfloor\lambda\rfloor$, then $r<\lambda<r+1$. The probabilities increase through $p_r$, because each ratio $p_{k+1}/p_k$ for $k<r$ is greater than one. They decrease after $p_r$, since $p_{k+1}/p_k<1$ for $k\geq r$. Hence $r$ is the unique mode. If $\lambda=m$ is an integer, the ratio $p_m/p_{m-1}=m/m=1$. The masses increase up to $p_{m-1}$, these two adjacent masses are equal, and subsequent masses decrease. Thus $m-1$ and $m$ are both modes. $\square$

Worked Example: Finding the Mode at a Noninteger Rate

Let $X\sim\operatorname{Poisson}(5.6)$. The candidate mode is $\lfloor 5.6\rfloor=5$. Check the ratios on either side:

\frac{p_5}{p_4}=\frac{5.6}{5}=1.12>1, \qquad \frac{p_6}{p_5}=\frac{5.6}{6}\approx 0.9333<1.

The probabilities are still increasing from count 4 to count 5, but are decreasing from count 5 to count 6. The general ratio argument shows they increase at every earlier step and decrease at every later step, so 5 is the unique mode.

Worked Example: The Two Modes at an Integer Rate

Let $X\sim\operatorname{Poisson}(4)$. The ratio between the masses at 4 and 3 is

\frac{p_4}{p_3}=\frac{4}{4}=1,

so $p_4=p_3$. For $k=0,1,2$, the ratios $4/(k+1)$ are greater than one, and for $k\geq 4$, the ratios $4/(k+1)$ are less than one. Therefore the mass rises to count 3, ties at count 4, and falls thereafter. The two modes are 3 and 4.

Computing Upper-Tail Probabilities

An upper tail collects probability at and above a threshold. Since the distribution is discrete, inclusive and strict inequalities correspond to different integer cutoffs. If $r$ is an integer, then $X\geq r$ includes the mass at $r$, while $X>r$ starts at $r+1$.

Definition: For an integer $r\geq 1$, the Poisson upper tail at $r$ can be computed by the complement identity $$ P(X\geq r)=1-P(X\leq r-1) =1-\sum_{k=0}^{r-1}p_k. $$ For $r\geq 0$, a direct expression is $$ P(X\geq r)=\sum_{k=r}^{\infty}p_k. $$

The complement identity follows because the two disjoint events $\{X\geq r\}$ and $\{X\leq r-1\}$ partition the nonnegative-integer support. Their probabilities therefore add to one. The direct sum is the definition of the probability of a countable union of disjoint point events.

The recurrence can be used in either calculation. For a complement, generate $p_0$ through $p_{r-1}$ and subtract their sum from one. For a direct tail sum, first compute $p_r$, then generate $p_{r+1},p_{r+2},\ldots$ recursively. That latter expression is an infinite series; a finite partial sum alone is not the exact tail unless the remaining probability is also accounted for.

Worked Example: An Upper Tail by Its Complement

Suppose $X\sim\operatorname{Poisson}(4)$. Find $P(X\geq 3)$. The complement consists of counts 0, 1, and 2:

P(X\geq 3)=1-(p_0+p_1+p_2).

Generate the needed probabilities from $p_0=e^{-4}\approx0.01831564$:

\begin{aligned} p_1&=4p_0\approx0.07326256,\\ p_2&=\frac{4}{2}p_1\approx0.14652511. \end{aligned}

Thus

P(X\geq3) \approx 1-(0.01831564+0.07326256+0.14652511) =1-0.23810331 =0.76189669.

The answer exceeds one-half, which is plausible because the mode is at 3 and 4 and the event includes both modes and all larger counts.

Choosing a Calculation and Checking the Cutoff

A complement is often convenient when it requires only a short sum of lower-count probabilities. A direct upper-tail sum can be more natural when the tail contains only a few terms of appreciable size, or when the probabilities are being accumulated from the threshold upward. In either method, use the event’s exact cutoff before calculating.

For example, “at least 5” means $X\geq5$, so its complement is $X\leq4$. “More than 5” means $X\geq6$, so its complement is $X\leq5$. The change of one count is essential; no approximation is involved.

Worked Example: An Upper Tail with a Five-Count Complement

Let $X\sim\operatorname{Poisson}(2.5)$. Find $P(X\geq5)$. Its complement is $X\leq4$, so use the probabilities generated earlier:

\begin{aligned} P(X\geq5) &=1-(p_0+p_1+p_2+p_3+p_4)\\ &\approx1-(0.0820850+0.2052125+0.2565156+0.2137630+0.1336019)\\ &=1-0.8911780\\ &\approx0.1088220. \end{aligned}

The result is a relatively small upper tail, as expected for a threshold of 5 when the rate is 2.5. If the question instead asked for $P(X>5)$, the complement would include counts through 5, not through 4.

Common Mistakes and Why the Recurrence Matters

Using $k$ instead of $k+1$ in the denominator. The correct step from $p_k$ to $p_{k+1}$ is $p_{k+1}=\lambda p_k/(k+1)$, because $(k+1)!=(k+1)k!$. For example, $p_1=\lambda p_0$, not $p_0$.
Reporting only one mode at an integer rate. When $\lambda=m$, the ratio from $m-1$ to $m$ equals one. Both counts have maximal mass.
Dropping or adding the threshold count in a tail. “At least $r$” includes $r$; “greater than $r$” does not. Write the event using $\geq$ before forming its complement.
Treating a truncated direct sum as exact. The direct tail has infinitely many terms. A finite sum from $r$ to some larger count omits the remaining upper tail unless an error bound or remainder calculation is supplied.

The recurrence provides a practical alternative to repeated factorial calculations, while the ratio gives the shape of the mass function without computing every probability. Together with the complement identity, these facts make Poisson mode and tail calculations transparent and easy to check. As an edge case, if $\lambda=0$, the distribution is degenerate at zero: $P(X=0)=1$, so zero is its sole mode and $P(X\geq r)=0$ for every integer $r\geq1$.

Check Your Understanding

Use the recurrence and carefully translate each event into integer counts.

For $X\sim\operatorname{Poisson}(3.7)$, what is the mode, and which two consecutive ratios verify the rise and fall around it?
If $X\sim\operatorname{Poisson}(6)$, how many modes are there, and what are they?
For $X\sim\operatorname{Poisson}(1.8)$, write $P(X\geq4)$ as a complement involving a finite sum. Do not evaluate it.
Explain why the recurrence for $p_{k+1}$ has $k+1$, rather than $k$, in its denominator.
What event is the complement of $X>7$, and how does it differ from the complement of $X\geq7$?