1
$\begingroup$

Suppose we require the following for a probability measure

1) $0 \leq P(A) \leq 1$

2) $\mathbb P\left(\displaystyle\bigcup_{k=1}^\infty A_k\right) = \displaystyle\sum_{k=1}^\infty \mathbb P(A_k)$

3) $\mathbb P(\Omega) = 1$

for an arbitrary event $A$ and an arbitrary sequencce $\left\{A_k\right\}_{k=1}^\infty$ of pairwise disjoint sets, i.e. $\left\{A_k\right\}_{k=1}^\infty$ satisfies $A_i \cap A_j = \varnothing \; \forall i \neq j$.

Let's now define the function $F(x) := \mathbb P(\mathbb X \leq x)$, which we know exists by assumption. Now, proving it's increasing (non-decreasing) is easy since for every $\varepsilon > 0$ and for every fixed $x_0 \in \mathbb R$

$F(x_0 + \varepsilon) = \mathbb P(\mathbb X \leq x_0 + \varepsilon) = \mathbb P(\mathbb X \leq x_0) + \mathbb P(x_0 \leq \mathbb X \leq \varepsilon) = F(x_0) + \mathbb P(x_0 \leq \mathbb X \leq \varepsilon) \geq F(x_0)$

But right-continuity and the limits $\begin{cases}\displaystyle\lim_{x\to-\infty} F(x) = 0 \\ \displaystyle\lim_{x\to\infty} F(x) = 1\end{cases}$ seem much harder to prove. It's easy to see that the limit $\displaystyle\lim_{\varepsilon \to 0_+} F(x_0 + \varepsilon) = F(x_0)$ is equivalent to $\displaystyle\lim_{\varepsilon \to 0_+} P(x_0 \leq \mathbb X \leq x_0 + \varepsilon) = 0$, but I am not getting any further than that.

I'm assuming I want to use axiom (2) somehow – my original idea was to use (2) together by (1) to show that you can "subtract" infinitely many intervals from $[x_0, x_0+\varepsilon]$ (thus making $\varepsilon$ smaller and smaller), and ultimately reach the limit "at $\varepsilon = 0$" since $ \mathbb P(\varnothing) = 0$ – but I'm not sure I can do that.

After all, assuming an uniform distribution, I can remove infinitely many intervals without reaching the limit by using the fact $\displaystyle\sum_{k=1}^\infty \frac{1}{n^2} = \frac{\pi^2}6$.

$\endgroup$
1
$\begingroup$

You will need the intermediate lemma

$$\mathbb{P} \left ( \bigcap_{n=1}^\infty A_n \right ) = \lim_{n \to \infty} \mathbb{P}(A_n)$$

for a sequence of events $A_n$ such that $A_{n+1} \subset A_n$. This requires both axiom $2$ and one of the others to prove (importantly it does not hold for infinite measures despite the fact that they satisfy axiom 2).

That lemma will get you the right-continuity and the limit at $-\infty$. The limit at $+\infty$ is easier: $\mathbb{P} \left ( \bigcup_{n=1}^\infty \mathbb{P}(\mathbb{X} \leq n) \right ) = \mathbb{P}(\Omega)$ for a finite valued random variable.

$\endgroup$
  • 1
    $\begingroup$ Could you explain how this doesn't contradict with what I just said about removing infinitely many intervals? $\endgroup$ – Markus Klyver Apr 6 '18 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.