Why is the Poisson distribution only defined for a positive rate 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.
 A: 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).

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.
