until the first jump occur so the process N t is equal to zero between zero and Xi one.

And at Xi one it jumps,

it is equal to one at the point s one,

and it is also equal to one until the next renewal time S two,

and the difference between S one and S two is indicated by Xi two.

So it is equal to one here,

and afterwards after S two,

it is equal to two until the next time arrival S three and so on.

So, the jump time process with jumps only on

one and the jumps are indicated by random variables S one and S two.

In this respect these random variables are known as renewal times and Xi one,

Xi two are known as interarrival times.

There is no doubt Renewal process

can be applied to various areas for instance to marketing.

If you imagine that you have a shop which opens at time moment zero,

and you wait some time until the first the first customer is coming,

this time is modeled by Xi one.

Then you wait some more time Xi two until the second customers come and so on.

And in this case N t clearly indicates the amount of customers

which came in your shop till time t. Look at this picture,

it's really easy to understand the following.

small proposition that basically demands that S n is larger than t,

coincides with the event that N t is smaller than

n. This is a very simple statement which directly follows from our picture.

So, if we take some moment t,

say t is here,

so what is written in left hand of this equality is that the renewal time number

n occur after time t. For instance S two is after t,

and what is written in

the right hand side is that before time t less than n renewal times occur.

If you look at the picture,

N t is equal exactly to one.

So, it is written here that before time t less than one event occur.

And if you think about these two events,

so events that renewal time number n after t or before time t less than n events,

you will definitely think that due to simple logic these events are completely the same.

In this lecture where you can see there is a renewal theory very precisely.

The question which I would like to address now,

is how to find the mathematical expectation of N t from the distribution of Xi,

it is from F. This question arises many applications.

For instance, here in marketing,

you may ask how many customers do you have till time

t. And you can just ask this from mathematical point of view Xi one,

Xi two and so on is something,

what you can observe.

Actually you know when your first customer came,

the second and so on and from this data,

you can estimate the function F. This is

a very typical task for mathematical statistics because Xi one,

Xi two and so on is a sequence of independent and basically distributed random variables,

and you can either assume that F is from

some parametric class and estimated parametrically,

or you can try to estimate nonparametrically.

So, in any case, this task is well started and you know how to do this.

The question is, if you know the estimator of F,

how to find the mathematical expectation of N t?

This question is rather intricate and we should study it in details.

Nevertheless, let me mention that S n is actually equal to the sum of Xi one and so on,

Xi n. So, this is a sum of independent,

identical distributed ran variables,

and as for this question,

they're actually very closely related to starting such sums.

And as you know from the course of probability theory,

the sum of two independent random variables

is computed via the so-called convolution operation.

And the next subsection I would to recall them in aspects of the convolution,

operation and then we'll apply

this operation to our setup.