Proving if a function is a valid probability distribution

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To prove that the function P_{k} = 20/5^{k} for k ≥ 2 is a valid probability distribution, it must satisfy two conditions: each P_{k} must be non-negative and the sum of all P_{k} values must equal 1. The first condition is met since P_{k} is positive for k ≥ 2. To verify the second condition, calculate the infinite series sum from k=2 to infinity, which should equal 1. A function is not a valid probability distribution if it fails to meet either of these criteria.
kioria
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Hi,

Given the function:

P_{k} = \frac{20}{5^{k}} for k \geq 2

How would you prove that P is a probability distribution? I would think that you prove that P is bounded by 0 and 1 (i.e., 0 \leq \Sigma P_{k} \geq 1)

And I guess the leading question is how you would prove that a function is not a probability distribution?
 
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You also need that
\sum_{k=2}^\infty\frac{20}{5^k}=1
 
You would prove that a function is NOT a valid probability distribution by showing that at least one of those conditions is not true. That is, that
1) Pk < 0 for some k or
2) Pk > 1 for some k or
3) \sum_{k=2}^\infty\frac{20}{5^k}\ne 1
 
Cheers
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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