Modes and Shape of the Poisson Distribution

Why the Poisson Mode Matters

The mean of a Poisson count describes its expected level, but it does not tell us which individual count is most probable. That count is the mode. For a discrete distribution, the mode is a value with probability at least as large as the probability at every other possible value. Determining it helps summarize where probability is concentrated and how that concentration changes when the rate changes.

The key tool is the ratio of two consecutive Poisson probabilities. This ratio shows directly whether probabilities are rising or falling as the count increases. It also reveals a detail that can be missed by simply rounding the rate: if the rate is a positive integer, there are two adjacent modes.

Definition: If $X\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$, then $X$ takes the nonnegative integer values $0,1,2,\ldots$, with probability mass function $$ P(X=k)=p_k=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ A mode is a possible value of $X$ at which $p_k$ is greatest. If more than one value attains that greatest probability, each is a mode.

Here $\lambda$ is the Poisson rate for the exposure being modeled. The earlier material on Poisson count models interprets this rate as the expected count over the specified exposure. The mode depends on the value of $\lambda$, including whether it is an integer.

Successive Probabilities Reveal the Mode

For $\lambda>0$, divide the probability of $k+1$ by the probability of $k$. The common exponential factor cancels, as do all but one factor in the factorial:

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

Because every $p_k$ is positive when $\lambda>0$, this ratio tells us how adjacent probabilities compare. If it exceeds 1, then $p_{k+1}>p_k$; if it is less than 1, then $p_{k+1}<p_k$; and if it equals 1, the two probabilities are equal. As $k$ increases, the denominator $k+1$ increases, so the ratios decrease. The probabilities therefore rise and then fall, except that one pair can tie at the maximum.

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

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

For $\lambda=0$, $P(X=0)=1$, so 0 is the unique mode.

Justifying the Mode Rule

The result follows by locating where the successive-probability ratio changes from greater than 1 to less than 1. The following steps include the equality case, which is why an integer rate gives two modes.

Compare adjacent probabilities.
For each nonnegative integer $k$, $p_{k+1}/p_k=\lambda/(k+1)$. Thus $p_{k+1}>p_k$ exactly when $k+1<\lambda$.

Locate the turning point.
When $\lambda$ is not an integer, let $m=\lfloor\lambda\rfloor$, so $m<\lambda<m+1$. Probabilities increase through $p_m$, because the ratios leading up to it exceed 1. The next ratio is $p_{m+1}/p_m=\lambda/(m+1)<1$, so probabilities decrease after $p_m$. Hence $m$ is the unique mode.

Check the integer boundary.
If $\lambda=n$, then $p_n/p_{n-1}=n/n=1$, so $p_{n-1}=p_n$. Ratios before this one exceed 1, and ratios after it are less than 1. The equal probabilities at $n-1$ and $n$ are therefore both maximal.

For $\lambda=0$, the positive-rate ratio calculation is not needed: the Poisson probability mass function assigns probability 1 to zero and probability 0 to every positive count. This special case has just one mode.

Worked Example: Locating a Mode for a Noninteger Rate

A hypothetical roadside-assistance portfolio is modeled as having a Poisson number $X$ of calls in a specified week, with rate $\lambda=3.6$ calls per week. Find the mode and check the probabilities around it. Assume the Poisson model with this rate; no interest or mortality assumption is involved.

Method. Compare consecutive probabilities using $p_{k+1}/p_k=3.6/(k+1)$. Since $3<3.6<4$, the proposed mode is $k=3$. The neighboring ratios verify the increase into 3 and the decrease after 3:

\frac{p_3}{p_2}=\frac{3.6}{3}=1.2>1, \qquad \frac{p_4}{p_3}=\frac{3.6}{4}=0.9<1.

For a numerical check, calculate probabilities directly from the probability mass function, retaining the unrounded values until the final reporting step:

\begin{aligned} p_2&=e^{-3.6}\frac{3.6^2}{2!} =e^{-3.6}(6.48)\approx0.1771,\\ p_3&=e^{-3.6}\frac{3.6^3}{3!} =e^{-3.6}(7.776)\approx0.2125,\\ p_4&=e^{-3.6}\frac{3.6^4}{4!} =e^{-3.6}(6.9984)\approx0.1912. \end{aligned}

The neighboring probabilities support the ratio argument: $p_3$ exceeds both $p_2$ and $p_4$. The full sequence rises up to 3 and falls afterward, so the mode is 3 calls. The probabilities above are rounded to four decimal places. The calculation avoids a common mistake: multiplying a previously rounded probability by a ratio and reporting a result as though no rounding had occurred.

When an Integer Rate Produces Two Modes

The integer-rate case is not a rounding convention. It is an exact equality in the probability mass function. If $\lambda=n$, the ratio from $n-1$ to $n$ is exactly 1, while the ratios on either side show an increase before the pair and a decrease after it.

Worked Example: Two Modes at an Integer Rate

A hypothetical monitoring system records a Poisson number $X$ of alert events in a fixed operating period, with rate $\lambda=4$ alerts per period. Find all modes and compare the probabilities at the peak with the probabilities immediately around it. Assume this Poisson model and rate.

Method. The rate is the positive integer 4, so the candidate modes are 3 and 4. The ratios give the complete local comparison:

\frac{p_3}{p_2}=\frac{4}{3}>1, \qquad \frac{p_4}{p_3}=\frac{4}{4}=1, \qquad \frac{p_5}{p_4}=\frac{4}{5}<1.

To verify the equal peak values, calculate each directly from the Poisson probability mass function:

\begin{aligned} p_3&=e^{-4}\frac{4^3}{3!} =e^{-4}\frac{64}{6} =e^{-4}\frac{32}{3} \approx0.1954,\\ p_4&=e^{-4}\frac{4^4}{4!} =e^{-4}\frac{256}{24} =e^{-4}\frac{32}{3} \approx0.1954. \end{aligned}

The probabilities immediately outside the pair are smaller:

p_2=e^{-4}\frac{16}{2}\approx0.1465, \qquad p_5=e^{-4}\frac{1024}{120}\approx0.1563.

Thus the two modes are 3 and 4 alerts. The equal probabilities are exact, not a coincidence created by rounding. All displayed probabilities are rounded to four decimal places. A common error is to report only 4 because it is the integer rate; the ratio $p_4/p_3=1$ shows that 3 must also be included.

How the Rate Changes the Shape

A Poisson distribution’s mode is near its rate, but its shape is not determined by the mode alone. For a Poisson random variable, $E[X]=\lambda$ and $\operatorname{Var}(X)=\lambda$, so its standard deviation is $\sqrt{\lambda}$. These facts help compare both the location and the absolute spread of distributions with different rates.

As $\lambda$ grows, the center moves to larger counts and the standard deviation also grows. The distribution remains defined only on nonnegative integers, and its probabilities generally rise toward the mode and then decline. A larger rate often makes the distribution look less strongly right-skewed relative to its center, but it does not make the distribution exactly symmetric. These are descriptions of shape, not alternative rules for locating the mode; the successive-probability ratio remains the exact method.

Worked Example: Comparing Two Poisson Shapes

Consider two hypothetical service desks, each observed for the same length of time. Desk A’s count $X_A$ is modeled as Poisson with rate $\lambda_A=1.5$, and Desk B’s count $X_B$ as Poisson with rate $\lambda_B=5.5$. Find each mode, compare probabilities around the modes, and compare their standard deviations. Assume the stated Poisson distributions and rates.

Desk A. The rate lies between 1 and 2, so the candidate mode is 1. The ratios on either side are

\frac{p_1}{p_0}=\frac{1.5}{1}=1.5>1, \qquad \frac{p_2}{p_1}=\frac{1.5}{2}=0.75<1.

The peak and its neighboring probabilities are

p_0=e^{-1.5}\approx0.2231,\quad p_1=1.5e^{-1.5}\approx0.3347,\quad p_2=\frac{1.5^2}{2}e^{-1.5}\approx0.2510.

Therefore Desk A’s unique mode is 1. Its standard deviation is $\sqrt{1.5}\approx1.225$.

Desk B. The rate lies between 5 and 6, so the candidate mode is 5. The ratios into and out of that count are

\frac{p_5}{p_4}=\frac{5.5}{5}=1.1>1, \qquad \frac{p_6}{p_5}=\frac{5.5}{6}\approx0.9167<1.

Direct calculation at the mode and its neighbors gives

\begin{aligned} p_4&=e^{-5.5}\frac{5.5^4}{4!}\approx0.1558,\\ p_5&=e^{-5.5}\frac{5.5^5}{5!}\approx0.1714,\\ p_6&=e^{-5.5}\frac{5.5^6}{6!}\approx0.1571. \end{aligned}

Desk B’s unique mode is 5, and its standard deviation is $\sqrt{5.5}\approx2.345$. Thus its most probable count is higher, and its absolute spread is larger. The probabilities and standard deviations are rounded to four decimal places and three decimal places, respectively. The comparison avoids the mistake of treating a higher mean as proof that every individual count is more probable; the mode and spread describe different features of the distribution.

Common Mistakes and Exam Checks

A reliable solution separates the exact shape argument from any decimal probability calculations. First identify whether the rate is an integer. Then use ratios to establish where the probabilities rise, tie, and fall. Only calculate individual probabilities if the question requests them or if they provide a useful check.

Giving only the rate as the mode. The mode is an integer. For a noninteger rate such as 3.6, the mode is 3, not 3.6.
Missing the second mode. For a positive integer rate $n$, both $n-1$ and $n$ are modes because their probabilities are equal.
Using the wrong ratio direction. $p_{k+1}/p_k=\lambda/(k+1)$. A ratio greater than 1 means the probability increases from $k$ to $k+1$.
Rounding inside a recurrence. If using the recurrence to calculate probabilities, carry adequate precision through the intermediate steps. Otherwise, use the direct probability mass function or display the unrounded expression before rounding the final result.
Ignoring the zero-rate case. When $\lambda=0$, the count is 0 with probability 1; the positive-rate mode rule does not apply.

Key takeaway: For a positive Poisson rate, compare $p_{k+1}/p_k=\lambda/(k+1)$. A noninteger rate has the single mode $\lfloor\lambda\rfloor$; a positive integer rate $n$ has the two modes $n-1$ and $n$. The rate locates the distribution, while its variance $\lambda$ describes its absolute spread.

Check Your Understanding

For each question, state the mode or modes and use a ratio or probability comparison to justify the result. Probabilities requested should be rounded to four decimal places.

A hypothetical warehouse is modeled as receiving a Poisson number of damaged-item reports per shift with rate $\lambda=4.7$. Find the most probable report count. (Answer: the unique mode is 4.)
A hypothetical sensor produces a Poisson count with rate $\lambda=6$ per operating interval. Identify both modes and state the ratio that shows their probabilities are equal. (Answer: modes 5 and 6; $p_6/p_5=6/6=1$.)
A Poisson count has rate $\lambda=0.8$. Calculate $p_0$ and $p_1$, then identify the mode. (Answer: $p_0=e^{-0.8}\approx0.4493$, $p_1=0.8e^{-0.8}\approx0.3595$; the unique mode is 0.)
A hypothetical inspection count is Poisson with rate $\lambda=3.2$. Compare $p_3$ with $p_2$ and $p_4$ using consecutive-probability ratios, and give the mode. (Answer: $p_3/p_2=3.2/3\approx1.0667$, $p_4/p_3=3.2/4=0.8$; the unique mode is 3.)
Two hypothetical counts have Poisson rates 2.5 and 7.5 for equal exposure periods. For each distribution, give its mode and standard deviation, and indicate which has greater absolute spread. (Answer: modes 2 and 7; standard deviations $\sqrt{2.5}\approx1.581$ and $\sqrt{7.5}\approx2.739$; the rate-7.5 distribution has greater absolute spread.)

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