I was reading up on finding the cdf from a pdf, and noticed that my cdf did not satisfy the condition of $F_Y{(−1)}=0$ and $F_Y{(1)} = 1$. Given the equation $f_Y(y) = cy^2(1-y)1_{[0,1]}(y)$ I calculated the cdf to be $\frac{1}{12}(4-3x)x^3$ by integrating the pdf: $\int_{0}^{x}\frac{1}{12}y^2(1-y)dy$. However this integral does not satisfy the above conditions, so I was hoping to see if someone could point out what I did wrong.

  • $\begingroup$ "$F_Y{(−1)}=0$" should be $F_Y{(0)}=0$ because the support of the pdf is $[0,1]$. $\endgroup$ – Jean Marie Oct 8 '17 at 20:06
  • $\begingroup$ @JeanMarie Ah i did not know that. So the cdf should be 0 for the endpoints of the range? $\endgroup$ – Curious Student Oct 8 '17 at 20:08
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    $\begingroup$ only for the leftmost one. for the rightmost one it should be 1. $\endgroup$ – mathreadler Oct 8 '17 at 20:10
  • $\begingroup$ @jean-marie: Then there would be no need for $1_{[0,1]}$ $\endgroup$ – gammatester Oct 8 '17 at 20:12
  • $\begingroup$ @gammatester No : $1_{[0,1]}$ should still be necessary for the pdf. $\endgroup$ – Jean Marie Oct 8 '17 at 20:14

The pdf is $12 \cdot y^2 (1-y)$. You have the wrong proportionality constant.

  • $\begingroup$ I am not a smart man. So is the correct cdf $(4-3x)x^3$? $\endgroup$ – Curious Student Oct 8 '17 at 20:07
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    $\begingroup$ yes, that is correct. $\endgroup$ – Srikant Oct 8 '17 at 20:08
  • $\begingroup$ Looks like the support is $[-1,1],$ then the PDF is is $f_Y(y) = 12y^2(1-y)1_{[0,1]}(y).$ Your function has $\int_{-1}^1 = 8$ $\endgroup$ – gammatester Oct 8 '17 at 20:10
  • $\begingroup$ @gammatester I think the support is [0, 1] as indicated by the indicator function in $f(y)$. $\endgroup$ – Srikant Oct 8 '17 at 20:12

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