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I am working on my Statics homework, and I came across a problem that I didn't even know where to begin.

I want to know how to do part (a) after that I can figure out the other parts. The problem is

A random variable $X$ takes values between 0 and $\infty$ with a cumulative distribution function $F_X(x) = A + Be^{-x}$ for $0 \le x \le \infty$

(a) Find the values of A and B and sketch the cumulative distribution function.

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    $\begingroup$ Hint. What must be true of $F(0)$ and the limit of $F(x)$ as $x$ grows? $\endgroup$ – Ethan Bolker Sep 16 '17 at 2:01
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    $\begingroup$ Now for $F(0)$ would we integrate the function then plug in 0? $\endgroup$ – Bartholomew Allen Sep 16 '17 at 2:09
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    $\begingroup$ For $F(0)$, just plug in $0$. But the result must be zero. Why? $\endgroup$ – quasi Sep 16 '17 at 2:14
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    $\begingroup$ You mean you got $A=-B$, right? (since $F(0) = A+B$) $\endgroup$ – quasi Sep 16 '17 at 2:18
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    $\begingroup$ $F$ is not the probability density function -- it's the cumulative distribution function. It's already "integrated", so to speak. Hence, as $x$ approaches infinity, $F(x)$ must approach _______? $\endgroup$ – quasi Sep 16 '17 at 2:22
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Since the random variable does not take values before $0$, $$ F(0) = 0 $$ Since the random variable can take on any nonnegative value, we must have: $$ \lim_{x \to \infty} F(x) = 1 $$ Using these facts we see that:

$$\begin{align}F(0) &= 0 & \lim_{x \to \infty} F(x) &= 1 \\ A + Be^0 &= 0 & \lim_{x \to \infty} \left[ A + Be^{-x} \right] &= 1 \\ A + B &= 0 & A &= 1 \\ 1 + B &= 0 \\ B &= -1\end{align}$$

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  • $\begingroup$ Thank you for the explaination, can you elaborate more on how the limit = 1? Please and thank you! $\endgroup$ – Bartholomew Allen Sep 16 '17 at 2:46
  • $\begingroup$ The cumulative distribution function $F_X(x)$ gives you the probability that $X$ is at most $x$. The CDF is always a nondecreasing function because of this definition, and the probability that $X$ is at most $x$ as $x$ increases must be $1$, since as we increase $x$, more and more probability mass is counted - the limit can be seen as the probability $X$ is just any value since the maximum is taken to infinity. This applies to all CDFs $\endgroup$ – Manuel Guillen Sep 16 '17 at 2:50

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