A Theorem of Probability

Let (Ω, F , P ) be a probability space and X n := X n (ω), (n = 1, 2, 3, • • •) be its random variables with X n (ω) ≥ 0 for ∀ω ∈ Ω. E[X] := E[X(ω)] := Ω X(ω)dP (ω) denotes the expectation value(or mean) of the random variable X = X(ω). We put {K m } ∞ m=1 to be a natural number sequence with K m < K m+1 and K m → +∞ (as m → +∞). Then we have

then we have

proof The conclusion of the theorem is equivalent to

Firstly we assume our denying of this conclusion, that is, we assume

which will lead to a contradiction later.

(4) is equivalent to P {A} > 0 with

We put the followings:

Then we have trivially

which shows the existence of

and

Next we divide {m j } ∞ j=1 into {m j (k)} ∞ j=1 ’s as follows:

with

where ♯A denotes the number of elements of the set A.

We also put

Then we also have trivially

which leads to the existence of

and

From ( 20), it follows that when N tends to ∞

which leads to ∃l N such that P {D (l) } > 0 (22) because of P {A} > 0 and ♯Λ = ♯N. We put

Because of the assumption of the theorem (1), that is,

there exists a natural number N such that

we have

with sufficiently large N. From (25), • • • ,(31), we have a contradiction:

Therefore we cannot have (4) or (5) which means the conclusion of the theorem:(2) or (3). This completes the proof.

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