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From wikipedia, a random variable $X$ is said to have the Poisson distribution if for $\lambda > 0$,

$ \Pr(X = k)= \frac{\lambda^k e^{-\lambda}}{k!}$

Why can't the value $\lambda = 0$ be admitted? Wouldn't this physically correspond to the situation where if you expect the rate of event occurrence to be zero, assign all probability to observing the zero outcome.

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    $\begingroup$ Is there any problem with the formula when $\lambda=0$? $\endgroup$
    – whuber
    Jan 13, 2016 at 6:04
  • $\begingroup$ Yes there is, but in this case would it not be convenient to define $0^0 = 1$? Put it in a different way, what does the sequence of Poisson distributions converge to as $\lambda \rightarrow 0$. $\endgroup$
    – Alex
    Jan 13, 2016 at 6:10
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    $\begingroup$ homeschoolmath.net/teaching/zero-exponent-proof.php $\endgroup$ Jan 13, 2016 at 7:29

1 Answer 1

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In general Poisson distribution is defined as having parameter $\lambda > 0$. For $\lambda < 1$ the smaller $\lambda$ gets, the more mass is accumulated around $0$, i.e. $\lim_{\lambda \rightarrow 0} P(K=0|\lambda) = 1$ and $\lim_{\lambda \rightarrow 0} P(K=k|\lambda) = 0$ for $k>0$ (Said, 1958). This is shown on an ugly plot below, that illustrates Poisson pmf for $\lambda = 1/10^0, ..., 1/10^{20}$ and $\lambda = 0$ as a red line (notice that it is heavily zoomed).

enter image description here

As noted by you and @ChristophHanck, number to the zeroth power is one and common convention is that $0^0 = 1$. On another hand $0^x = 0$ for non-zero $x$'s. Dividing zero by non-zero gives you zero. So if $\lambda = 0$ Poisson pmf simplifies to

$$ \frac{0^k \times \text{(whatever)}}{ \text{(whatever)} } $$

so it is degenerate distribution with all point mass at zero (cf. here):

$$ f(k) = \begin{cases} 1 & \text{if }k=0, \\[6pt] 0 & \text {if }k>0.\end{cases} $$

Mean and variance for Poisson distribution are equal to $\lambda$ and in this case only zero has non-zero probability, so expected value is obvious and there is no variability (variance is zero). Also R does not have any problem with $\lambda$'s defined as non-negative (rather than positive) values:

> dpois(0:10, 1e-10)
 [1]  1.000000e+00  1.000000e-10  5.000000e-21  1.666667e-31  4.166667e-42  8.333333e-53
 [7]  1.388889e-63  1.984127e-74  2.480159e-85  2.755732e-96  2.755732e-107
> dpois(0:10, 0)
 [1] 1 0 0 0 0 0 0 0 0 0 0
> dpois(0:10, -1)
 [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
Warning message:
In dpois(0:10, -1) : NaNs produced

So the behavior of Poisson pmf at $\lambda=0$ is coherent with its behavior at limit, moreover there is no problem with calculating Poisson pmf form this value, but it yields a degenerate distribution.


Said, A.S. (1958). Some properties of the poisson distribution. AIChE Journal, 4(3), 290-292.

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  • $\begingroup$ well, if R says it is alright then it must be alright. Question: do standard textbooks for the Poisson distribution define it at $\lambda = 0$? $\endgroup$
    – Alex
    Jan 14, 2016 at 4:42
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    $\begingroup$ @Alex no they generally don't because it's degenerate, but you can define it like this and it's sometimes done so (I provided one reference that mentions it and R does so). $\endgroup$
    – Tim
    Jan 14, 2016 at 6:05
  • $\begingroup$ @Alex I updated answer with this comment. $\endgroup$
    – Tim
    Jan 14, 2016 at 8:22

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