# Find constants $A, B$ for cumulative density functions (probability) [closed]

I'm stuck with this question and can't seem to find $A$ & $B$. A continuous random variable $X$, which can only take positive values, has cumulative distribution function of the form $$F(x) = \frac{A+Bx}{9+8x}$$ for $x\ge 0$ where $A$ and $B$ are constants which you will need to evaluate. Calculate to $4$ decimal places $P(X>2)$.

Help would be sincerely appreciated. Thank you so much!

## closed as off-topic by Jack D'Aurizio, Hakim, Adam Hughes, user147263, AlexRDec 4 '14 at 20:11

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• A big hint: what do you know about the behavior of $F$ at the ends of its interval (here, at $x=0$ and as $x\to\infty$)? – Steven Stadnicki Dec 4 '14 at 18:27
• The wording of the question ix suboptimal. "Can only take positive values" is consistent with $F(17)=0$. – André Nicolas Dec 4 '14 at 18:57

For every distribution function $F$ it is true that
1. $\displaystyle\lim_{x\to +\infty}F(x)=1$ and
2. $\displaystyle\lim_{x\to -\infty}F(x)=0$
Now, in this case, since $X$ takes only positive values, the lower limit value of $0$ is already attained at $x=0$ (and possibly even before, but certainly for $x=0$). Now, substituting the given $F$ in the equations above yields $$1=\lim_{x\to \infty}F(x)=\lim_{x\to \infty}\frac{A+Bx}{9+8x}=\lim_{x\to \infty}\frac{\frac{A}{x}+B}{\frac{9}{x}+8}\frac{\not x}{\not x}=\frac{B}{8}$$ and $$0=F(0)=\frac{A+B\cdot0}{9+8\cdot 0}=\frac{A}{9}$$ Putting these together you obtain that $$\begin{cases}\frac{B}{8}=1\\\frac{A}{9}=0\end{cases} \implies \begin{cases}B=8\\[0.2cm]A=0\end{cases} \implies F(x)=\frac{8x}{9+8x}$$ for $x\ge 0$. Thus $$P(X>2)=1-P(X\le 2)=1-F(2)=\frac{8(2)}{9+8(2)}=\frac{16}{25}=0.6400$$
• @Sid You are welcome! This is only a way to calculate the limit, it has nothing to do with $x$ going to $\infty$. . I just factored out $x$ from numerator and denominator. You can also do it with L' Hopital (but I think it is pretty much straightforward). – Jimmy R. Dec 4 '14 at 19:09