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Let $(a_n)_n$ be a descending convergent sequences with $\lim_n a_n = a$ and $(b_n)_n$ a convergent series with $\lim_n b_n = b$. Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space with $X$, $Y$ random variables.

Is there a way to show

$$a_n + b_n Y \ge X \ \ [\mathbb{P}]\ \ \text{for all}\ \ n \quad \implies\quad a+bY \ge X \ \ [\mathbb{P}].$$

We know that for every $n$ exists a set $\Omega_n \in \mathcal{F}$ with $\mathbb{P}(\Omega_n)=1$, such that $a_n + b_n Y(\omega) \ge X(\omega)$ for every $\omega \in \Omega_n$. I was thinking of using $\sigma$-continutity of the probability measure, however, I don't know how to show the required inclusions of the sets $\Omega_{n+1} \subset \Omega_n$. Then I'd have this:

$$1 = \mathbb{P}\left(\bigcap_n \left\{a_n + b_n Y \ge X \right\}\right) \\ = \lim_n \ \mathbb{P}\big(a_n + b_n Y \ge X \big) \\ = \mathbb{P}\big(a + b Y \ge X \big).$$

Thanks for reading and ideas.

Friedrich
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You are overcomplicating the isue.

$$\bigcap_{n=1}^\infty\{a_n + b_n Y \geq X\}\subseteq \{a+bY \geq X\}$$

The left set has probability $1$ by your assumption, so the right set as well.

The assumption that the sequences are increasing/decreasing is not necessary.

J. De Ro
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  • Why does this inclusion hold? – Friedrich Jan 04 '20 at 18:39
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    Take an $\omega$ in the left set. Then $a_n + b_n Y(\omega) \geq X(\omega)$ for all $n \geq 1$. Taking the limit, we see that $a + bY(\omega) \geq X(\omega)$, so $\omega$ is in the right set. – J. De Ro Jan 04 '20 at 18:39
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    I don't understand your last question. – J. De Ro Jan 04 '20 at 18:42
  • Why is it eligible to extend the inequality on towards the limit, when I only have that it holds for members of the sequences? – Friedrich Jan 04 '20 at 18:43
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    If $x_n \leq y_n$ for all $n \geq 1$ and the limits exist, then $\lim_n x_n \leq \lim_n y_n$. See for example here: https://math.stackexchange.com/questions/432319/inequality-between-two-sequences-preserved-in-the-limit. – J. De Ro Jan 04 '20 at 18:44
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    Feel free to ask addititional questions if something is unclear! – J. De Ro Jan 04 '20 at 18:47