# cumulative distribution function question

• Oct 8th 2010, 12:44 PM
Sneaky
cumulative distribution function question
consider rolling a die.
S= {1,2,3,4,5,6}
P(s)=1/6 for all s in S
X= number on die so that X(s)=s for all s in S
Y= X^2
compute the cumulative distribution function Fy(y) = P(Y<=y), for all y in the set of real numbers.

My guess
for Y=1 i get
P(-inf<y<=1)=P(Y<=1)-P(Y<-inf)=Fx(1)-Fx(-inf)
= Fx(1)-0
= Fx(1)

Is this all I have to do for Y=1, or do I have to integrate, or is there anything wrong?
• Oct 8th 2010, 01:09 PM
harish21
you have been told that $\displaystyle Y = X^2$

so,

$\displaystyle F_{Y}(y) = P(Y \leq y) = P(X^2 \leq y)$

continue further to find the cdf of Y....
• Oct 8th 2010, 01:13 PM
Sneaky
i go further but i am stuck at
1-P(-x<y)-P(y<x)
• Oct 8th 2010, 01:26 PM
Plato
$\displaystyle P(Y \leqslant x) = F_Y(x) = \left\{ {\begin{array}{rl} {0,} & {x < 1} \\ {\frac{1} {6},} & {1 \leqslant x < 4} \\ {\frac{2} {6},} & {4 \leqslant x < 9} \\ { \vdots ,} & \vdots \\ {1,} & {36 \leqslant x} \\ \end{array} } \right.$

You fill in the blanks.
• Oct 8th 2010, 01:48 PM
Sneaky
ok i understand now.