I just got really unsure, can someone confirm/rectify? I have the CDF defined as $F(x)= \begin{cases}0, &\text{if}~x < 0,\\ 4x^2 &\text{if}~ 0 \leq x < \frac{1}{4} \\ 1-\frac{4}{3}(1-x)^2 & \text{if}~ \frac{1}{4} \leq x < 1, \\ 1, &\text{if}~ x \geq 1. \end{cases} $

I want to find the PDF, is it simply $f(x)=8x \mathbb{1}_{0\leq x< \frac{1}{4}} + \frac{8}{3}(1-x) \mathbb{1}_{\frac{1}{4} \leq x < 1}$ ?


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    $\begingroup$ You might want to be careful with the limits. CDFs are right continuous so I think the limit should be <1/4 instead of ≤ 1/4. $\endgroup$ – Moss Murderer Dec 20 '17 at 21:33
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    $\begingroup$ @MossMurderer $4(1/4)^2=1/4=1-4/3(3/4)^2$ so it doesn't matter in this case $\endgroup$ – user179309 Dec 20 '17 at 21:35
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    $\begingroup$ Your CDF is incompletely defined. The CDF $F(x)$ is the probability that the random variable $X$ is no larger than $x$. Thus, the CDF is defined for all $x$, is a nondecreasing function of $x$, and it must approach value $1$ (or have value $1$) as $x$ tends to $\infty$. In this instance, the CDF has value $1$ at $x=1$ and it continues to have this value for all $x>1$ too. And yes, the pdf is the derivative of the CDF and it is (almost) what you find it to be; you don't need to worry about discontinuities in $F(x)$. $\endgroup$ – Dilip Sarwate Dec 21 '17 at 3:41
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    $\begingroup$ Depending on taste, one could also restrict the range of $x$ to something different than $\mathbb{R}$. For instance an interval $\lbrace x \in \mathbb{R} \;\vert\; 0 \leq x \leq 1 \rbrace$. $\endgroup$ – Sextus Empiricus Dec 21 '17 at 8:43
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    $\begingroup$ @MartijnWeterings "Taste" sometimes is not rigorous, if not incorrect. Every distribution function of a random variable must be defined at every point in $\mathbb{R}^1$ --- it's not about the "range of $x$", but the domain of $F$. $\endgroup$ – Zhanxiong Dec 22 '17 at 18:30

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