0

Let $X$ be a uniform r.v. defined on the interval $[0,12]$. Show the density function of $Y= \lfloor X \rfloor +1 .$

My attempt :

$F_Y(y) = P(Y\leq y) = P( \lfloor X \rfloor +1 \leq y) = P(X<y) = F_X(y^-) = F_X(y)$,

using the inequality $\lfloor X \rfloor \leq X <\lfloor X \rfloor +1.$

Differentiating, we would obtain:

$f_X(y) = \dfrac{d}{dy}F_Y(y) = f_X(y)\dfrac{dy}{dy} = f_Y(y) = \dfrac{1}{12} \mbox{ for }1<y<13$

Is this actually correct? Is the interval for $Y$ correct also?

user2345678
  • 2,885

1 Answers1

0

$Y$ is a discrete random variable that takes values in $\{1,2,\ldots,13\}$.

For such $y \in \{1,\ldots,12\}$ we have $P(Y=y) = P(y \le X < y+1) = \frac{1}{12}$, and we have $P(Y=13)=P(X=12)=0$.

angryavian
  • 89,882