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When I study the textbook Statistical Inference by Casella and Berger, I have often seen expressions in the form of P(an equation) = 1. However, some other textbooks or lecture notes will instead say "The equation". For example, in the definition of Jensen's inequality:enter image description here But you may find other books saying g(X) = a+ bX only.

Another example: in the definition of complete statistics: enter image description here Other books may say g(T) = 0 only.

I wonder whether these two kinds of expression are equivalent? And why?

Nick Cox
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littletennis
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  • Can you list what (formally published) "other books" say "$g(X) = a + bX$ only " or "$g(T) = 0$ only"? Technically speaking, saying that is not mathematically correct. – Zhanxiong Dec 05 '23 at 13:49
  • These "equations" are shorthand for describing events. After all, probability is defined as a function of events, not equations. If this distinction is unclear, research the "axiom of specification" in set theory. You could start at https://math.stackexchange.com/questions/9542. – whuber Dec 05 '23 at 14:22
  • @Zhanxiong For example, in Statistical Inference by Casella and Berger P287 for the Proof of Basu's thm. See the last sentence at the bottom, it use P(P(S(X)=S|T(X)=T) = P(S(X)=S)) = 1 for all t to get the result P(S(X)=S|T(X)=T) = P(S(X)=S) – littletennis Dec 06 '23 at 14:30

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The concept you are looking for is almost surely: suppose in the probability space $(\Omega,\boldsymbol{\mathfrak A},\Pr),$ we are investigating to see whether a property $P$ holds.

Now what we actually want to check is whether devoid of a null set $N\in\boldsymbol{\mathfrak A}, $ i.e. $\Pr[N]=0,$ all $\omega\in\Omega$ satisfies the property $P, $ that is, whether $$ \Pr[\omega\in\Omega\setminus N:P]=1.$$

Coming to the equality case of Jensen's Inequality, its derivation is based on the fact that for a convex function $g$ defined on an open internal $I, $ we have for every $m\in[(D_\ell g) (x_0), (D_r g) (x_0) ], $ for all $x_0\in I, $ $$g(x)\geq m(x-x_0)+g(x_0),~~x\in I. $$ So, the equality would imply $g$ must be a straight line.

Whenever you are integrating w.r.t. a measure (in this case, probability measure), it only depends on how the measurable function is defined on a non-null set. Since $g$ is defined on $I, $ it is continuous on $I$ and hence $\boldsymbol{\mathfrak B}_\mathbb R/\boldsymbol{\mathfrak B}_\mathbb R$-measurable. If it is defined as $a+bX$ on $\mathbb R\setminus N$ where $N$ is a null set, then you will attain the equality.

User1865345
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