So, we have two equations...ok?
One equation tells us the relationship
between the two Lagrange multipliers
In particular, it tells us "Z"
is a function of Lambda 1
Ok?
And the second equation here
two equations, two unknowns
will allow us to solve for Lambda 1 itself
So, "Z" here is in the form of an infinite sum.
But if you remember how your geometric
series work, you can actually write this
out because this, the first couple terms
just to drive your intuition, look like
one plus 'e' to the negative lambda one
plus 'e' to the negative 2 lambda one
and so forth.
And so we know the geometric series
of that form by a very nice argument
that we won't do here.
Sum to one over 1 minus 'e' to the
negative lambda 1.
So we know already how to reduce this
infinite sum to a very elegant form.
The next thing I'll point out,
because I'm an ex-physicist...
I'm always, I'm always gonna...
going to eventually slip up on this
We like to call 'z' the Partition Function
Ok? 'z' here is the sum over these terms
here. And so one thing you notice about
'z' about the partition function is if you
take the derivative of 'z' with respect
'i', look what happens. The derivative
of 'z' - sorry, we just got to lambda 1 -
derivative of 'z' with respect to lambda 1
what happens is the following:
Well, you get a negative sign and you
get the sum. And now what you've done is
pulled down a factor of 'i'. So, now all
of a sudden, you have the same infinite
sum you have in the other equation
appearing. This sum here you can't do
immediately, by the geometric series trick
because instead of it being 1 plus 'p'
plus 'p' squared plus 'p' to the
fourth...plus 'p' cubed and so forth,
it now has these funny terms in the front
here hanging off. But notice that if you
the derivative of the partition function,
you pull down that factor of 'i'.
So, in fact, we can now rewrite this
second constrained equation here as
1 over 'z', minus, 'd' 'Z' 'd' lambda 1
equals 4.
So, first thing you have to do is take
a derivative of 'z' with respect to
lambda 1 and then we have to divide
by 'z'. And so we can do that quite
easily. We have a factor 'e' to the
negative lambda 1 on the top, 1 minus 'e'
to the negative lambda 1 squared - that's
the derivative of 'z' with respect
to lambda 1...minus sign. And then we have
a factor of 1 over 'z', and all that does
is pull one of these out. So 'z' is 1 over
1 minus 'e' to the negative lambda 1. Ok?
So if we divide by that...we just lose
one of these. And so now, we've turned
this equation here, which we would have
trouble solving, because it's an infinite
sum, and you know, putting an infinite sum
into MatLab tends to take a long time to
solve. But instead, we have a analytic
form, a functional form that we can put in
right away. Ok? So now, all we have to do
is solve this equation here. Find the
value of lambda 1 that sets this term here
equal to 4. Once we find that value of
lambda 1, we can then plug into 'z'
here, ok? And once we know both 'z' and
lambda 1, we can recover not just the
functional form of the probability of
waiting for 'x', but actually we'll be
able to compute the waiting time as a
function of 'x', ok? Er, the probability
of a waiting time for a particular value,
ok? So, we're left with this equation here
and instead of solving that exactly,
because we have some logarithms, I can
tell you the answer that lambda 1 is equal
to roughly 0.22. And from there, we can
also compute the value of 'z', and we can
now write out the proportionality,
roughly speaking, ok? So this distribution
here, with appropriate normalization,
will be the exponential distribution, it's
an exponential distribution, linear in
the waiting time 'x'. And what I've shown
to you, is this distribution has the
following properties, ok? It satisfies
this linear constraint. It's normalized.
Ok? And it has the maximum entropy of all
the distributions with those two previous
properties.