I have noticed that when a Poisson distribution has an integer value of $\lambda$, the following holds: $$ \mathbb{P}[X = \lambda] = \mathbb{P}[X = \lambda - 1] $$
I have been able to prove this rather simply using an algebraic method. However, my question is, is there an intuitive reason for why this is the case? For example, if an average of 10 phone calls are received an hour, why is it that 10 phone calls are equally likely as 9 phone calls?
The best way I can see is to consider the Poisson distribution as a binomial limit and try to see intuitively why the recursive formula
$$\mathbb{P}[X = k] = \frac{\lambda}{k}\mathbb{P}[X = k - 1]$$
holds. Then, if $\lambda = k$, the numerator and denominator cancel out.
Say that you recieve an average of $\lambda$ phone calls in your working day. Consider your day as divided into a large number $n$ time slots, and that the probability of recieving a call in a given time slot has independent probability $\lambda/n$. Then in the limit as $n \to \infty$ the number of calls $Y_n$ per day has the Poisson($\lambda$) distribution.
Let $p_k = \mathbb{P}[Y_n = k]$. We now want to understand why
$$ p_k \approx \frac{\lambda}{k}p_{k-1}\,.$$
Warm-up: How to obtain $p_1$ if we know $p_0$? One of the $n$ time slots should have a call instead of being empty, and there are $n$ different ways in which this can happen. Thus, with obvious notation
$$ p_1 = n\frac{p_{\mathrm{call}}}{p_\text{empty}}p_0 = n\frac{\frac{\lambda}{n}}{1-\frac{\lambda}{n}}p_0 \approx n \frac{\lambda}{n}p_0 = \lambda p_0\,.$$
The denominator $p_\text{empty}$ is close to one and will be disregarded from now on.
General case: To obtain $p_k$ from $p_{k-1}$ in general, there are two things going on. First, any given configuration of $k$ calls over your $n$ time slots is a factor $\lambda/n$ less likely to happen than any given configuration of $k-1$ calls:
$$\mathbb{P}(\text{slots } a_1, a_2, \ldots a_k \text{ filled}) \approx \frac{\lambda}{n}\mathbb{P}(\text{slots } b_1, b_2, \ldots b_{k-1} \text{ filled})\,.$$
Second, there are about a factor $n/k$ more ways for $k$ slots to be filled than for $k-1$ ways to be filled: $$\binom{n}{k} \approx \frac{n}{k}\binom{n}{k-1}\,.$$ (It is not hard to argue this intuitively: For each placement of $k-1$ calls there are approximately $n$ empty slots in which to place an extra call, since $n$ is large. However, a given configuration of $k$ calls can be obtained from a ($k-1$)-configuration in $k$ different ways, since each of the $k$ calls could've been the last one added.)
Of course,
$$p_k = \binom{n}{k}\mathbb{P}(\text{slots } a_1, a_2, \ldots a_k \text{ filled}) \quad\text{and}\quad p_{k-1}= \binom{n}{k-1}\mathbb{P}(\text{slots } b_1, b_2, \ldots b_{k-1} \text{ filled})\,.$$ Putting these things together we obtain
$$p_k \approx \frac{n}{k}\frac{\lambda}{n}p_{k-1} = \frac{\lambda}{k}p_{k-1}\,,$$ which is what we wanted.
Intuitive example: It now seems to me that one can get a pretty good intuition of why $p_\lambda = p_{\lambda-1}$ holds by toying with examples and using the ideas above.
Take the example given (slightly changed): If an average of 8 phone calls are received an hour, why is it that 8 phone calls are equally likely as 7 phone calls? Divide the hour into 100 slots. The probability for some 8th time slot to turn into a phone call is 8/100. And there are approximately a factor of 100/8 more ways to recieve 8 phone calls than 7. The factors cancel. As you make the time slots smaller, the approximation becomes more exact.
