Why does the infinite sum of a Poisson distribution add up to 1?

Question

I understand that it has to add up to 1 since it is a distribution; I'm needing the logical explanation for it. The long-winded explanation I've seen online...I need something that's not just copied and pasted from somewhere.

Answer 1

The reason is simply that by definition, e^x = [k=0, ∞]∑(x^k/k!). So for the Poisson distribution:

[k=0, ∞]∑(e^(-λ)λ^k/k!) = e^(-λ) [k=0, ∞]∑(λ^k/k!) = e^(-λ)e^λ = 1

That's really all there is to it.

Answer 2

The probability density
function of the Poisson distribution is given by:

[L^x]*[e^(-L)]
p(X = x) = ----------------
x!

Where I have used capital L to represent the parameter of the
distribution. Traditionally, the Greek letter Lambda is used for this
parameter. I will keep calling it L from now on, though.

Finding E(x) = mean of the Poisson is actually fairly simple. We go
back to the definition of E(x) to see that we need to find the
following infinite sum.

inf
---
\ [L^x]*[e^(-L)]
E(X) = / x * ----------------
--- x!
x=0

We will apply a standard procedure for summing up this series or a
very often-used trick, if you like. (My professors like to say that "a
twice-used trick is a procedure.") We will factor out whatever we can
from under the sigma sign, trying to arrive at something like

(Some Expression) * Infinite Sum of p(x)

And since we already know that p(x) is the density function of the
Poisson, we know that it must sum up to 1. You will see the above
procedure being applied again and again whenever you find expected
values.

Let's do it. First we notice that when x = 0, the entire first term
vanishes, so we can rewrite the expression as:

inf
---
\ [L^x]*[e^(-L)]
E(X) = / ----------------
--- (x-1)!
x=1

Where I have made the additional simplification of cancelling the first
term in the factorial with the x we had in the numerator. Now, the
above looks *almost* the same as p(x), except that we have (x - 1)!
instead of x!. No problem. Let us make a change-of-variable and choose
y = x - 1. We'll see how everything turns out after we do that.

inf if y = x - 1, then x = y + 1
--- |
\ [L^(y + 1)]*[e^(-L)]
E(X) = / -----------------------
--- y!
y=0
|
When x = 1, y = 0

Almost there - except that now we have an extra L that multiplies
every term in the series. So, all we have to do is factor it out.

inf
---
\ [L (y)]*[e^(-L)]
E(X) = L* / ------------------
--- y!
y=0

And there you have it. The resulting summation is nothing more than
adding up all the values of the density function of the Poisson, and
we know that by the definition of the density function, it must all
add up to 1. This gives us the final very neat result:

E(X) = L

The variance is found by very similar means, except that this time we
employ another interesting little procedure (I would call it a trick,
but since it is also used for finding the variance of a Binomial
distribution, it's a procedure). Recall that

Var(X) = E(X^2) - [E(X)]^2

Since we already know E(X), all we have to do is now find E(X^2) in
order to obtain the variance. Unfortunately, this is a very difficult
task, so we instead do something else.

Notice that E[X(X-1)] = E(X^2 - X) = E(X^2) - E(X). This alternate
expression is easier to add up, because of that factorial in the
denominator of the Poisson density. Finding it will give you an
expression containing E(X^2) which can then be used for finding the
variance. In other symbols,

E[X(X - 1] + E(X) - [E(X)]^2 = E(X^2) - E(X) + E(X) - [E(X)]^2
= E(X^2) - [E(X)]^2
= Var(X)

So the only ingredient missing is E(X(X - 1)). This I'll let you find
yourself. You need to add up the following infinite sum:

inf
---
\ [L^x]*[e^(-L)]
E[X(X - 1)] = / x(x - 1) * ---------------
--- x!
x=0

Do it the same way I did it for the E(X). Play around with the
expressions, do a substitution, and try to arrive at an expression of
the form of (Something)*[Sum of p(x)]. I hope this gives you no major
difficulties, once you have fully understood how to do it for E(X).

I can think of an entirely different standard method for finding the
mean and variance of the Poisson distribution. It involves a
theoretical device called moment generating functions, abbreviated
mgf's. Since this method is very straightforward, provided that we
have already derived the mgf of the Poisson, I'll explain it too.

The mgf of ANY distribution is given by the formula:

mgf(t) = E[e^(tX)]

where E( ) denotes the expected value function related to that
distribution and x is a random variable. In particular, the mgf of the
Poisson can be found by evaluating the above expected value (which
involves a lot of fun and summing up again an infinite series). In the
end, one can arrive at the following formula:

mgf(t) = exp[(e^t - 1)L]

where exp[ ] is another way of writing e^( ) and L is again the
parameter related to the Poisson distribution, same L as before.

Now, the beauty of the mgf is that it greatly simplifies calculations
for obtaining means and variances. It is called _moment_ generating
function, because in a sense, it packs of all the moments of the
distribution into one neat expression. Moments of a random variable
are E(X), E(X^2), E(X^3), etc. Notice how the mean is the first
moment, E(X), and the variance is the second moment minus the first
moment squared, i.e.

Var(X) = E(X^2) - [E(X)]^2

There is a standard result that says that if we find the kth
derivative of the mgf and evaluate it at t = 0, we will obtain the kth
moment. This theorem can be obtained by expressing the mgf as an
infinite sum (use the definition), then differentiating term-by-term,
and finally evaluating at t = 0. Using this little theorem and the mgf
of the Poisson we find that:

d mgf(t) | d exp[(e^t - 1)L] | |
-------- | = ----------------- | = exp[(e^t - 1)L]*Le^t | = L
dt |t=0 dt |t=0 |t=0

It is reassuring to see that this agrees with our previous result.

Now we need to find the second derivative in order to get the second
moment. We can then use the second moment to find the variance.

d^2 mgf(t) | d exp[(e^t - 1)L]*Le^t |
------------ | = ------------------------ |
dt^2 |t=0 dt |t=0

|
= exp[(e^t - 1)L] * (Le^t)^2 + exp[(e^t - 1)L] * Le^t |
|t=0

= L^2 + L

And just for an added touch of suspense, I'll let you figure out from
here what the variance of a Poisson random variable is.

Answer 3

Because otherwise it wouldn't be a distribution.