Why is the cdf G considered a strictly increasing function?

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Discussion Overview

The discussion revolves around the properties of cumulative distribution functions (cdf), specifically focusing on why a cdf G is considered a strictly increasing function. Participants explore the implications of definitions involving strict inequalities and equivalences in the context of cdfs.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question why the definition of a strictly increasing function is presented as an equivalence (v1 < v2 <=> G(v1) < G(v2)) rather than as an implication (v1 < v2 => G(v1) < G(v2)).
  • There is a suggestion that the use of equivalence avoids the scenario where "false implies true," indicating a concern about the logical implications of the definitions.
  • One participant asserts that the reverse implication (G(v1) < G(v2) => v1 < v2) is true because G is a cdf, implying that the properties of cdfs support this relationship.
  • Another participant agrees with the conclusion that the definition implies v1 = v2 <=> G(v1) = G(v2), suggesting that this is straightforward to prove.

Areas of Agreement / Disagreement

Participants express varying interpretations of the definitions and implications related to cdfs. There is no consensus on whether the equivalence is necessary or if it introduces confusion, indicating that multiple views remain on this topic.

Contextual Notes

Participants do not fully resolve the implications of the definitions, and there are assumptions about the properties of cdfs that are not explicitly stated or agreed upon.

PAHV
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Let's say we have a cumulative distribution function (cdf) G and random numbers v1 and v2.

The definition of strict increasing function is: v1 < v2 => G(v1) < G(v2).

In a statistics book, the author writes:

"...but with the additional assumption that the cdf G is a strictly increasing function. That is, v1 < v2 <=> G(v1) < G(v2)".

a) He writes the definition with an equivalence (<=>) and not an implication (=>). Could someone explain why? Does in fact the definition also imply that its is an equivalence?

b) The authors definition: "v1 < v2 <=> G(v1) < G(v2)" must imply that:
v1 = v2 <=> G(v1) = G(v2). Correct?

Any help is very appreciated!
 
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PAHV said:
Let's say we have a cumulative distribution function (cdf) G and random numbers v1 and v2.

The definition of strict increasing function is: v1 < v2 => G(v1) < G(v2).

In a statistics book, the author writes:

"...but with the additional assumption that the cdf G is a strictly increasing function. That is, v1 < v2 <=> G(v1) < G(v2)".

a) He writes the definition with an equivalence (<=>) and not an implication (=>). Could someone explain why? Does in fact the definition also imply that its is an equivalence?

b) The authors definition: "v1 < v2 <=> G(v1) < G(v2)" must imply that:
v1 = v2 <=> G(v1) = G(v2). Correct?

Any help is very appreciated!

I think it is because "false implies true" is true, so he wanted to avoid that by using <->. Or maybe it is just an insignificant detail. I think that your 2 conclusion is correct and is easy to prove.
 
PAHV said:
Let's say we have a cumulative distribution function (cdf) G and random numbers v1 and v2.

The definition of strict increasing function is: v1 < v2 => G(v1) < G(v2).

In a statistics book, the author writes:

"...but with the additional assumption that the cdf G is a strictly increasing function. That is, v1 < v2 <=> G(v1) < G(v2)".

a) He writes the definition with an equivalence (<=>) and not an implication (=>). Could someone explain why? Does in fact the definition also imply that its is an equivalence?

b) The authors definition: "v1 < v2 <=> G(v1) < G(v2)" must imply that:
v1 = v2 <=> G(v1) = G(v2). Correct?

Any help is very appreciated!

The reverse implication (<=) is true because G is a cdf.
 
As bpet said, CDF's have this property.
 

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