# Distribution of ceiling function and absolute value of random variable

Given a distribution function $f_X$, where $X$ is some random variable. I want to get the distribution functions of $|X|$ and $\lceil X \rceil$( the last one may only have an easy form if $X$ is exponentially distributed, so it would be sufficient to understand how to construct this function if X is exponentially distributed).

The problem is that I do not see how to construct this distribution function.

Any kind of help is highly appreciated.

• What's holding you back from finding $P(|X|\leq x)$ and $P(\lceil X\rceil= x)$? – Stefan Hansen Mar 5 '14 at 19:47
• Have you realized that $\lceil X\rceil$ is an integer-valued (discrete) random variable? Can you find all values of $X$ for which $\lceil X\rceil$ has value $11$, say? Can you calculate the probability that $X$ takes on value in the set you found? Congratulations! You just found $P\{\lceil X\rceil = 11\}$. Lather, rinse, repeat for other values of $11$. – Dilip Sarwate Mar 5 '14 at 19:49
• @StefanHansen yeah, well $P(|X| \le x) = P(X \le x \wedge X \ge -x)$ and then? $P(\lceil X \rceil = 11 ) = P(X \in (10,11])$ – user66906 Mar 5 '14 at 19:54
• @Lipschitz: $P(|X|>x)=P(X>x)+P(X<-x)$. – Stefan Hansen Mar 5 '14 at 19:55
• @StefanHansen ah, so we have $P(|X| \le x) = 1-P(X>x)+ P(X \le -x) = P(X\le x) + P(X\le -x)$ and therefore the distribution function is given by $f_{|X|}(x)= f_X(x)+f_X(-x)$, right? – user66906 Mar 5 '14 at 19:59

$\lceil X\rceil\geq k\iff X>k-1$ so: $$P\left\{ \lceil X\rceil\geq k\right\} =P\left\{ X>k-1\right\}$$
• In the exponential case $P\left\{ X>k-1\right\} =e^{-\lambda\left(k-1\right)}$ leading to $P\left\{ \lceil X\rceil=k\right\} =P\left\{ \lceil X\rceil\geq k\right\} -P\left\{ \lceil X\rceil\geq k+1\right\} =e^{-\lambda\left(k-1\right)}-e^{-\lambda k}$ – drhab Mar 6 '14 at 8:15