# Math Help - Functions of a random variable question

1. ## Functions of a random variable question

I have this question i cant nut out

f(x)=3/32(4-x^2) when -2<x<2
hence F(x)=0 when -2>x
(3/8)x-(x^3)/32+1/2 when -2<x<2
1 when 2<x

let Y=(X^2)+1, find the CDF of Y

Heres my (probably wrong method or something) working out:
Fy(y)=P(Y<y)=P((x^2+1)<y)=P(X<sqrt(Y-1))=Fx(sqrt(y-1))

therefore Fy(y)=3/8(sqrt(y-1))^(1/2)-((sqrt(y-1))^3)/32+1/2
this isn't a proper CDF when i checked graphing it...

Any help would be appreciated

Edit: yeah, my mistake, i want the CDF of Y

2. I assume you want the CDF of Y, not X.

3. Originally Posted by skirk34
I have this question i cant nut out

f(x)=3/32(4-x^2) I(-2,2)(x)
hence F(x)=(3/8)x-(x^3)/32+1/2

let Y=(X^2)+1, find the CDF of Y

Heres my (probably wrong method or something) working out:
Fy(y)=P(Y<y)=P((x^2+1)<y)=P(X<sqrt(Y-1))=Fx(sqrt(y-1))

therefore Fy(y)=3/8(sqrt(y-1))^(1/2)-((sqrt(y-1))^3)/32+1/2
this isn't a proper CDF when i checked graphing it...

Any help would be appreciated

Edit: yeah, my mistake, i want the CDF of Y
What does the stuff in red mean?

4. It is an indicator function (i think thats its name), it equals one when -2<x<2 and 0 otherwise

5. Originally Posted by skirk34
It is an indicator function (i think thats its name), it equals one when -2<x<2 and 0 otherwise
$F_y (y) = \Pr(Y \leq y) = \Pr(X^2 + 1 \leq y) = \Pr(-\sqrt{y - 1} \leq X \leq \sqrt{y - 1})$ for $1 \leq y \leq 5$

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$= \frac{(13 - y) \sqrt{y - 1}}{16}$ for $1 \leq y \leq 5$.

Note that $F_y (y) = 0$ for $y \leq 1$ and $F_y (y) = 1$ for $y \geq 5$.