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 (unintelligible)

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 expert at this...I'm always, I'm always

gonna...going to, uh,

eventually (unintelligible) from 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.