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I am doing a problem from my textbook, and I have a question about how they write the answer. The problem is the following

If $\begin{align*} f(x)= \begin{cases} 3(1-x)^2, 0 < x < 1\\ 0, \text{otherwise}\\ \end{cases} \end{align*}$

find the cdf $F(x)$. First, I integrated to find that $F(x) = C-(1-x)^3$. I know that $1 = F(1) = C-(1-1)^3$ which means that $C=1$. Thus, I write my answer as

$\begin{align*} F(x)= \begin{cases} 0, x \leq 0\\ 1-(1-x)^3, 0 < x < 1\\ 1, x \geq 1\\ \end{cases} \end{align*}$

However, the book's answer has different bounds. They write

$\begin{align*} F(x)= \begin{cases} 0, x < 0\\ 1-(1-x)^3, 0 \leq x < 1\\ 1, x \geq 1\\ \end{cases} \end{align*}$

Can anyone explain why they include $0$ on the second case? Since it is a continuous PDF, is it ireelevant? Thanks.

mXdX
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    Yes you can. You can include equality in either of the first two cases. However, note that the last value of the CDF should be $1$ for $x\geq 1$, and not $0$ as you have written. A CDF is always non-decreasing. – P. N. Karthik Sep 23 '19 at 05:25

1 Answers1

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At first in last case of $F(x)$, that $x \geqslant 1$, you should write $F(x)=1$ instead of $0$!

But about your question, since CDF is right continuous function by definition, you should write bounds as $m \leqslant x < M$ except of $m \longmapsto -\infty$.

Ali Ashja'
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