Why Poisson probability P(X < λ) is greater than P(X > λ)?

Question

I can intuitively grasp, why Poisson distribution is not symmetric. As I understand, because it is limited by zero, probabilities on the left side from its mean ($λ$) are being "thickened". But why sums of probabilities on both sides from the mean are not equal?

For example, with $λ=1.5$, Poisson probability of $P(X < λ)$ would be $0.55783$. But shouldn't there be equal probabilities of getting more and less than $1.5$ event's occurrences? And does this difference repeat with examples of data from the real world or it is just some kind of inaccuracy of Poisson distribution?

Answer 1

For a continuous distribution, it is the median (not the mean) that satisfies the equation:

$$\mathbb{P}(X < x_\text{median}) = \mathbb{P}(X > x_\text{median}).$$

This does not hold exactly for a discrete distribution, owing to the fact that there is positive probability at the point $x_\text{median}$, but still, it should hold approximately. Now, the Poisson distribution is a positively skewed distribution, and its mean is higher than its median. Thus, the result you are observing is quite unsurprising.

Answer 2

Your assertion is correct for integer $\lambda$ because of the skewness of the Poisson distribution.

But it is not correct for all positive real $\lambda$, and in a handwaving sense, you might say your assertion is only correct for two-thirds of possible $\lambda$.

For example with $\lambda = 1.8$ you get $P(X < 1.8) \approx 0.463$ and $P(X > 1.8) \approx 0.537$, so the reverse of your assertion.

This reversal happens for some values between every pair of integers and the interval in which it occurs gets slightly wider as the integers get larger, initially for $\lambda \in (\log_e(2),1)$ i.e. about $(0.69315,1)$, and as integer $n$ becomes larger then the interval approaches $(n-\frac13,n)$ such as $\lambda \in (6.67,7)$.