According to the scheme which I introduced previously,

let me provide these three steps.

The first step, we should calculate

the Laplace transform of P from P. This is a very simple step because

LP of AS is equal to one-half L of the function exponent minus X at

the point S plus Laplace transform of function exponent to the power of minus two X of S.

I show to ranges that Laplace transform of such function is equal to S plus one,.

And for this function,

it is equal to one divided by S plus two.

In other words, the Laplace transform of the function P is equal to

three S plus four divided by two S plus one S plus two.

Okay, the first step is done.

The second step, we should use of the formula which we

showed previously to calculate the Laplace transform of the function U,

of the mathematical expectation of NT from

the Laplace transform of P. This formula is the following formula.

So, it is Laplace transform of P at the point S divided by S

multiplied by one minus LP of S. Here,

we just use the formula for LP of S obtained on the previous step.

And finally get that this Laplace transform of the function

U is equal to three S plus four

divided by S squared

two S plus three.

You can check these calculations.

It is just an very easy exercise.

The most important is the step number three,

where we should guess which U was used for this Laplace of U.

So, we should somehow involve this Laplace transform.

But as I said before,

the general formula for this inversion operation is very difficult.

And therefore, we should just apply what we know about Laplace transform.

We know Laplace transforms from many functions.

And this knowledge is many situations

sufficient to find the function U with given Laplace transform.

More precisely, I mean is the following.

So, what we have here is just Laplace transform as a function U.

And this Laplace transform can be also represented in the following form.

So, it can be some constant divided by S squared plus another

constant divided by S plus another constant C divided by two S plus three.

So, these constants can be explicitly written.

If we just compare this expression with this sum.

To do this, we should multiply the first fraction by two S plus three,

the second fraction by two S squared plus three S and the third fraction two S squared.

And afterwards, we should

just write an equality that this fraction should be equal to this one.

What you will finally get is that three S plus four should be equal to,

let me think about,

it is two B plus C multiplied by S squared plus two

A plus three B multiplied by S and plus three A.

Therefore, three A should be equal to four.

Two A plus three B should be equal to three.

And tow B plus three C should be equal to zero.

We have three equations with three unknown variables A,

B and C. And solving this system,

we finally arrive at the following answer,

that A is equal to four divided by three,

constant B is equal to one divided by

nine and constant C is equal to minus two divided by nine.

What we have now,

we have the composition of the Laplace transform of U into the sum of three fractions.

And basically, we know the functions such that

these fractions are exactly Laplace transforms all of these functions.

And therefore, we can basically write the answer.

So, it turns out that the function U to point T is equal to four divided by three,

multiplied by a function such as this Laplace transform is equal to S squared.

This function is known is equal to T.

Now, we should write plus then take B one divided by

nine and multiply by a function with Laplace

transform equal to one divided by S. This function is just a constant one.

And finally, we should write here minus two divided by nine,

but also here we have one-half minus one divided by nine and multiply with

a function such that it's Laplace transform is equal to

one divided by S plus one and a half.

This function also known as exponent in the power minus one and a half T.

So, we get the final answer,

mathematical expectation of NT is equal to the following formula.

So, the problem which was formulated in the beginning is solved.

Basically, you can use this methodology for any distribution of random variables,

Xi one, Xi two and so on.

But this take into account in the last step of the scheme is rather difficult.

And here, we managed to make this an inverse operation but in some situations,

it is going be much more complicated.