Relation between probabilities such as $P(AB>a) > P(Ab>a)$

Let $$X=AB$$,

$$A$$ and $$B$$ are random variables which are NOT independent and I know that $$A>0$$, $$B\geq b >0$$ with $$b$$ is a deterministic constant.

Then for any constant $$a$$, probability $$P(X>a) > P(Ab>a)$$ because $$p=P(X>Ab)=1$$.

Now if I know $$p=P(B>b')$$ with $$b'>b$$,

is there any relation (equality or inequality) among $$P(X>a)$$, $$P(Ab'>a)$$ and $$p$$ ?

Not quite an answer, but too long for a comment.

I think we need to be careful about equality. In the setup you said $$A>0, B\ge b> 0$$. In that case you can only conclude $$X = AB \ge Ab$$, but you cannot conclude $$P(X > Ab)=1$$. In fact, $$P(X>Ab) = P(B>b)$$ which can be any value from $$0$$ to $$1$$.

Also, you said $$P(X > a) > P(Ab > a)$$, but that is again untrue. What you can conclude is $$P(X > a) \ge P(Ab > a)$$, first because it is possible $$B \equiv b$$ in which case $$X=Ab$$ and the two probabilities are equal, and second because it is possible that for certain values of $$a$$ (e.g. $$a < 0$$) we have $$P(Ab> a) =1$$ already and $$P(X > a)$$ cannot possibly be $$>1$$.

Anyway on to your actual question: In general, $$P(Ab' > a)$$ only has to do with the marginal distribution of $$A$$, and $$P(B > b')$$ only has to do with the marginal distribution of $$B$$, but $$P(X > a)$$ has to do with the joint distribution. If $$A$$ and $$B$$ are dependent, you can make the $$3$$ distributions look like a lot of different things, so I don't think you can prove anything in general.

E.g. one natural way to proceed is:

\begin{align} P(X > a) &= P(X > a | B > b') P(B > b') + P(X > a | B \le b') P(B \le b') \\ & \ge P(Ab' > a | B > b') P(B > b') + 0 \end{align}

So this has $$2$$ of the $$3$$ quantities you want. However, since $$A$$ and $$B$$ are dependent, we cannot even say $$P(Ab' > a | B> b') = P(Ab' > a)$$, so I don't see a way to proceed further. I think I can make up examples where $$P(Ab' > a | B>b')$$ is $$>$$ or is $$< P(Ab' > a)$$.

• Thanks for answer. For the first point, as $B \geq b > 0$, isn't it obvious $P(B>b) = 1$ ? For the second point, I was not careful about these notations, but this did not change the essence of the question. At the last inequality deriving $P(X>a)$, could you please explain why the second term equals to 0? – Anna Noie Mar 23 at 8:01
• $B\ge b$ so it is possible $B=b$, and it all depends on $P(B=b)$. Perhaps you're thinking of classic "continuous" r.v. like $B= Uniform(b, 2b)$, or $B=b + Exp(\lambda)$. In those cases $P(B=b) = 0$ (indeed for any specific value $c, P(B=c) = 0$). However your setup did not exclude cases where $B$ is decrete, e.g. $B$ can be a coin-flip between $b$ and $2b$, or where $B$ is a mix between discrete and continuous. And re: the second term, $P(X>a|B \le b')P(B\le b')$ is not equal to $0$ but rather, it is a probability so it is $\ge 0$. – antkam Mar 23 at 15:25