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Given probability density function

$$f(x) = \begin{cases} 3x^2 &\mbox{for } 0 < x \le1 \\ 0 & \mbox{otherwise } \end{cases}$$

Show that the cumulative probability density function $F$ corresponding to $f$ is given by $$ F(x) = \begin{cases} 0 &\mbox{for } x<0 \ \\ x^3 & \mbox{for }0 \leq x \le1 \\1 & \mbox{for }x>1 \end{cases} $$

I know that the probability density function of a random variable $X$ is given by $$f(x)=f_X(x)=\frac{d}{dx}F(x) $$ , but that's about it. Help appreciated, thanks!

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  • $\begingroup$ Differentiate $F$ or integrate $f$ and see what happens. $\endgroup$ – user251257 Nov 2 '17 at 9:03
  • $\begingroup$ @user251257 I understand that $\int 3x^2dx =x^3+C$, but here there are lots of 'pieces' and boundaries which i don't understand. $\endgroup$ – Programmer Nov 2 '17 at 9:07
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    $\begingroup$ Terminology correction: it is called the cumulative distribution function, not density. $\endgroup$ – Nap D. Lover Nov 2 '17 at 11:49
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The CDF at $x$ is just given by the integration of the PDF from $-\infty$ to $x$. If you want justification, you can look at this answer.

Now, with this information, it is pretty easy to divide $F(x)$ and perform piecewise integration to get your answer.

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