Measure Theory and Dominated Convergence Theorem

Please see the attached file for the fully formatted problems.

I have provided a solution to the attached problem. I do not understand or like the solution - I was hoping you could provide an alternate solution or expand upon the solution I have provided in more detail.

Exercise (moment-generating function).

? Let X be a nonnegative random variable, and assume that

is finite for every (real numbers). Assume further that E [XetX] < &#8734; for every . The purpose of this exercise is to show that and, in particular,
? Recall the definition of the derivative:

? The limit above is taken over a continuous variable s, but we can choose a sequence of numbers converging to t and compute:

,

where now we are taking a limit of the expectations of the sequence of random variables:
.

If this limit turns out to be the same, regardless of how we choose the sequence that converges to t, then this limit is also the same as
and is
? The Mean Value Theorem from calculus states that if f(t) is a differentiable function, then for any two numbers s and t, there is a number &#952; between s and t such that:
.

If we fix and define , then this becomes
, (*)
where &#952;(&#969;) is a number depending on &#969; (i.e., a random variable lying
between t and s).

1) Use the Dominated Convergence Theorem (defined in Supplement below) and equation (*) to show that:

.

This establishes that the desired formula .

2) Suppose that the random variable X can take both positive and negative values and and for every . Show that once again:
.

(Hint: Consider a random variable X that can take both + and - values. For such a random variable, we define the positive and negative parts of X by:

, .

Use this notation to write X = X+ - X-.)

Solution:

1) By (*), The last inequality is by X &#8805; 0 and the fact that &#952; is between t and sn, and hence smaller than 2t for n sufficiently large. So, by the Dominated Convergence Theorem:

.

2) Since , for every ,

for every . Similarly, we have for every . So, like Solution 1), we have:

for n sufficiently large.
So, by the Dominated Convergence Theorem:

.

Supplement:

Definition 1: Let (&#937;, ,P) be a probability space. If a set satisfies P(A) = 1, we say that the event A occurs almost surely.

Definition 2: Let f1, f2, f3, ... be a sequence of real-valued, Borel-measurable functions defined on . Let f be another real-valued, Borel-measurable function defined on . We say that f1, f2, f3, ... converges to f almost everywhere and write:

almost everywhere
if the set for which the sequence of numbers f1(x), f2(x), f3(x), ... does not have a limit f(x) is a set with Lebesgue measure zero.

Dominated Convergence Theorem:

Let X1, X2, ... be a sequence of random variables converging almost surely to a random variable X. If there is another random variable Y such that and almost surely for every n, then:

Let f1, f2, ... be a sequence of Borel-measurable functions on converging almost everywhere to a function f. If there is another function g such that and almost everywhere for every n, then:

Definitions from Shreve's Continuous Time Models