# Finding the cdf from pdf

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.

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

The pdf is $12 \cdot y^2 (1-y)$. You have the wrong proportionality constant.
• I am not a smart man. So is the correct cdf $(4-3x)x^3$? – Curious Student Oct 8 '17 at 20:07
• 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 $\int_{-1}^1 = 8$ – gammatester Oct 8 '17 at 20:10
• @gammatester I think the support is [0, 1] as indicated by the indicator function in $f(y)$. – Srikant Oct 8 '17 at 20:12