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If $f(y)=3(1-2y+y^2)$ where $0<y<1$

1)Determine cumulative distribution function (cfd) of random variable $Y$?

2) Find $P(0.5<Y<1.5)$

for Q1 I found the cfd: $3(y-y^2+y^3/3)$ but I'm struggling with Q2 anyone can help since $y$ can't be more than $1$ and here he want it to be $(0.5<y<1.5)$!!

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  • $\begingroup$ sorry I don't know what happened my question is missing some parts! Q2 is find P(0.5<Y<1.5) $\endgroup$ – Alyah Oct 21 '16 at 6:52
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I suspect that $f(y)=3(1-2y+y^2)$ for $y\in(0,1)$.

But that is not a complete PDF because it does not mention what values it takes on $\mathbb R-(0,1)$.

I bet that the right addition is: $f(y)=0$ on $\mathbb R-(0,1)$.

Determining the CDF and denoting it by $F$ we then find that

  • $F(y)=0$ if $y\leq0$
  • $F(y)=3y-3y^2+y^3$ if $y\in(0,1)$ (as you found out yourself)
  • $F(y)=1$ if $y\geq1$

Equipped with this can you find $\Pr(0.5<Y<1.5)$ yourself?

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  • $\begingroup$ The final result =0,125 right??? $\endgroup$ – Alyah Oct 21 '16 at 7:55
  • $\begingroup$ It is $F(1.5)-F(0.5)=1-F(0.5)=\dots$. That's all I am saying :). I hate calculations like this, and never trust myself in it. $\endgroup$ – drhab Oct 21 '16 at 7:58
  • $\begingroup$ The most important lesson for you is: look at PDF's and CDF's as functions on $\mathbb R$ and not just on a part of $\mathbb R$. $\endgroup$ – drhab Oct 21 '16 at 8:04
  • $\begingroup$ Ok.. Thanks for your help ^^ $\endgroup$ – Alyah Oct 21 '16 at 10:47

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