1
$\begingroup$

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 $p$ ?

$\endgroup$
0
$\begingroup$

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 $< P(Ab' > a)$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Anna Noie Mar 23 at 8:01
  • $\begingroup$ $B\ge b$ 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$. $\endgroup$ – antkam Mar 23 at 15:25

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.