I have this problem where I need to graph the CDF, for that I need to find the constant $c$. The formula below is a PDF:

$f(x) = c(x^2+1)\space\space\space if\space X \in [0,1];\space$ otherwise $0$

My attempt: $P(0\leq X \leq 1) = \int_0^1c*(x^2+1) = 1 \Rightarrow cx+\frac {cx^3}3 = 1 \Rightarrow c = 3/4$

The problem here is that when we put $x=1$ we get $f(1) = \frac 32 > 1$, which is wrong as probability cannot be higher than $1$. I need to construct the CDF, but because of that, I cannot do it.


The CDF cannot exceed $1$, but the PDF can. Your CDF is $\frac{3}{4}x+\frac{1}{4}x^3$ on $[0,\,1]$, just as you calculated.

  • $\begingroup$ Then how can you find the CDF from the PDF? $\endgroup$ – Tigran Minasyan Mar 21 at 20:12
  • $\begingroup$ @TigranMinasyan You basically did it already; see my edit. $\endgroup$ – J.G. Mar 21 at 20:13
  • 1
    $\begingroup$ Indeed. The CDF is $F(x)~{=\mathbf 1_{0\leq x\lt 1}~\int_0^x c~(s^2+1)~\mathrm d s+\mathbf 1_{1\leq x}\\= \tfrac 34(\tfrac 13x^3+x)~\mathbf 1_{0\leq x\lt 1}+\mathbf 1_{1\leq x}}$ $\endgroup$ – Graham Kemp Mar 21 at 21:47

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