I’m not sure whether most people will be completely lost (possibly because I;ve explained it badly), or whether they’ll think what I’ve just said is completely obvious. Let me give an example to help illustrate.

**Poisson Distribution as a partial infinite sum**

Start with the following identity:

And let's bring the exponential other to the other side.

$$ \sum_{i=0}^{\infty} \frac{ x^i}{i!} e^{-x} = 1$$

If we state a few obvious facts; firstly, this is an infinite sum (which I claims above were related to probability distributions - so good so far). Secondly, the identity is true by the definition of $e^x$, all we need to do to prove the identity is show the convergence of the infinite sum, i.e. that $e{x}$ is well defined. Finally, each individual summand is greater than or equal to 0.

With that established, if we define a function:

$$ F(x) = \sum_{i=0}^{\infty} \frac{ x^i}{i!} e^{-x}$$

That is, a function which specifies as its parameter the number of partial sumummads we should add together. We can see from the above identity that:

- The partial sum is strictly less than 1
- The sum converges to 1 as $k \rightarrow \infty$.

But wait, the formula for $F(x)$ above is actually just the formula for the CDF of a Poisson random variable! That’s interesting right? We started with an identity involving an infinite sum, we then normalised it so that the sum was equal to 1, then we defined a new function equal to the partial summation from this normalised series, and voila, we ended up with the CDF of a well-known probability distribution.

Can we repeat this again? (I’ll give you a hint, we can)

**Exponential Distribution as a partial infinite integral**

Let’s examine an integral this time. We’ll use the following identity:

$$\int_{0}^{ \infty} e^{- \lambda x} dx = \lambda$$

An integral is basically just a type of infinite series, so let’s apply the same process, first we normalise:

$$ \frac{1}{\lambda} \int_{0}^{ \infty} e^{- \lambda x} dx = 1$$

Then define a function equal to the partial integral:

$$ F(y) = \frac{1}{\lambda} \int_{0}^{ y} e^{- \lambda x} dx $$

And we've ended up with the CDF of an Exponential distribution!

**Euler Integral of the first kind**

This construction even works when we use more complicated integrals. The Euler integral of the first kind is defined as:

$$\frac{\int_{0}^{1}t^{{x-1}}(1-t)^{{y-1}}dt}{B(x,y)} = 1$$

And once again, we can construct a probability distribution:

$$B(x;a,b) = \frac{\int_{0}^{x}t^{{a-1}}(1-t)^{{b-1}}dt}{B(a,b)}$$

Which is of course the definition of a Beta Distribution, this definition bears some similarity to the definition of an exponential distribution in that our normalisation constant is actually defined by the very integral which we are applying it to.

**Conclusion**

So can we do anything useful with this information? Well not particularly. but I found it quite insightful in terms of how these crazy formulas were discovered in the first place, and we could potentially use the above process to derive our own distributions – all we need is an interesting integral or infinite sum and by normalising and taking a partial sum/integral we've defined a new way of partitioning the unit interval.

Hopefully you found that interesting, let me know if you have any thoughts by leaving a comment in the comment box below!