$X$ is a random variable that can assume only non-negative values (that is its support is non-negative). Further, assume that $a,b$ (with $a>b$) are two positive constants then for what values of $C$ the following probability has non-zero values $$P\left(\frac{bX}{aX+1}>C\right)$$ we can clearly see that if $C\geq 1$ then the above probability is zero. On the other hand if $C\leq 0$ then the above probability is $1$. I want to know the range of $C$ for which the above probability is in the range of $[0,1]$. Thanks in advance for your help.
For what value of $C$ the following probability is non-zero
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probability
probability-theory
probability-distributions
1 Answers
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You just need to rewrite the inequality:
$$ \frac {bX} {aX + 1} > C \iff bX > CaX + C \iff (b - Ca)X > C$$
So when $\displaystyle C \geq \frac {b} {a}$, the probability is $0$ as $$ \Pr\left\{X < \frac {C} {b - Ca} < 0 \right\} = 0$$
Otherwise, when $\displaystyle C < \frac {b} {a}$, the probability becomes $$ \Pr\left\{X > \frac {C} {b - Ca} \right\}$$ and as you said, this becomes $1$ when $C < 0$. So the interesting interval should be $\displaystyle\left(0, \frac {b} {a}\right)$.
Also note that the RHS $$\frac {C} {b - Ca} = \frac {1} {b/C - a} \uparrow +\infty$$ as $\displaystyle C \uparrow \frac {b} {a}$. So the probability may still reach $0$ or $1$ depends on the actual support of $X$ is the whole positive real line or not.