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Let $F(x) = A(1 - \dfrac{1}{e^{x-1}})$ where $1 < x < \infty$ and $0 \leq F(x) \leq 1$
The question is asking to solve for $A$. My idea is that, $y = (1 - \dfrac{1}{e^{x-1}}) \in (0, 1)$ which implies $0 \leq Ay \leq 1$ Thus $0 \leq A \leq \dfrac{1}{y}$. Does this approach make sense?

3 Answers 3

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Try this: $ f(x)=F'(x)=Ae^{1-x}\\ \int_{1}^{\infty}f(x)=1 \Leftrightarrow A=\frac{1}{\int_{1}^{\infty}e^{1-x}dx}=1 $

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Set $A=1$. Then $F(1) = 0$ and $F(\infty) = 1$, which solves your problem, no?

I cannot really follow your approach. What you'd usually do is start with a factor (i.e. $A$) of $1$, and check that $F(a) = 0$ where $a$ is the starting point of the domain of $F$. If not, a factor won't help, you'll need to substract $F(0)$. If it holds, then you find $\alpha = F(\infty)$ (actually $\lim_{x\to\infty}F(x)$). If it's not already $1$, you divide $F$ by $\alpha$. In other words, if $F$ is a monotone function on some (sensible) set $A$, you get a cumulative distribution function $F'$ by setting $ F'(x) = \frac{F(x) - F(\text{min }A)}{F(\text{max } A) - F(\text{min }A)} $ You can easily check that $F'$ then always satisfies $F'(\text{min }A) = 0$, $F'(\text{max }A) = 1$.

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I think it's enough if you just make $F(\infty) - F(1) = 1$ and then compute $A$.

I mean, in order to have a distribution, it has to be satisfied (among other things) that $\int_X f(x) dx =1,$ where $X$ is the set where $x$ takes its values and $F'(x) = f(x)$ is the density function associated to the distribution $F$.

But that integral above is, applying Fudamental Calculus Theorem, just evaluating $F$ on the bounds, and since you already have $F$, you are set.