# Cumulative distribution function on uniform distribution

For $$l>0$$, $$X$$ is a random variable according to the uniform distribution on the interval $$[0, l].$$

At this time, the probability density function $$f(x)$$ of $$X$$ is

$$f_X(x)= \cases{0 & if x<0\\ \frac{1}{l} & if 0 \le x \le l\\ 0 & if x > l}$$

The length of the shorter interval obtained by dividing the following interval $$[0,l]$$ into 2 at random is expressed by a random variable $$Y$$.

The length of the longer interval obtained by randomly dividing interval $$[0,l]$$ into two parts is expressed by a random variable $$Z$$.

Find the distribution function $$F_y$$ of Y. Find the distribution function $$F_Z$$ of Z.

$$f_Y(x)= \cases{0 & if x<0\\ \frac{1}{l/2} & if 0 \le x <\frac{l}{2}\\ 0 & if x \le l}$$

$$f_Z(x)= \cases{0 & if x<0\\ \frac{1}{l/2} & if 0 \le x <1/2\\ \frac{4}{(2-l)l} & if l/2 \le x <1\\ 0 & if x \le l}$$

I'm not sure, am I correct? And after this, do I just need to integrate it to find $$F_Y$$?

• I don't know how to interpret your function $f$. It seems to be a function of $x$ and $\ell$. On a closer look, however, it is not a function because if $x=0$ and $\ell=1$ its value is 1 (according to the second line) and 0 (according to the third line). Commented Jun 11, 2019 at 14:40
• Your $f(x)$ does not make sense if $X\sim U(0,1)$. And you have $Y=\min (X,1-X), Z=\max(X,1-X)$. Commented Jun 11, 2019 at 16:17
• @GerhardS.@StubbornAtom im sorry i have so many typo, i fixed it , i also dont quite understand what partition into two in uniform distribution Commented Jun 11, 2019 at 16:38
• What is the use of $l$? Commented Jun 11, 2019 at 18:14
• @StubbornAtom i edit my post again, l is for interval $[0,l]$ Commented Jun 11, 2019 at 19:30

You are not quite correct, but not far away.

Apart from your typographical confusion between $$1$$ and $$l$$, you have some inequalities the wrong way round and some major issues in the density function for $$Z$$: it cannot be less than $$l/2$$ since it is the longer of the two pieces, and in fact it has a uniform distribution between $$l/2$$ and $$l$$. I think you should have something like

$$f_Y(x)= \cases{0 & if x<0\\ \frac{1}{l/2} & if 0 \le x \le \frac{l}{2}\\ 0 & if x \gt l/2}$$

$$f_Z(x)= \cases{0 & if x<0\\ 0 & if 0 \le x \le l/2\\ \frac{1}{l/2} & if l/2 \lt x \le l\\ 0 & if x \gt l}$$

Then, as you suggest, you should integrate to get the cumulative distribution functions

• thankyou so much but why in the $f_Z$ for $0<=x<=l/2$ must be 0? Commented Jun 12, 2019 at 1:48
• @devss "The length of the longer interval obtained by randomly dividing interval $[0,l]$ into two parts is expressed by a random variable $Z$" so $Z$ is the longer of the two parts and must be at least $l/2$ Commented Jun 12, 2019 at 7:30
• but the two of the partition are completely different part? so for cumulative we dont need to sum the short and large partition? Commented Jun 12, 2019 at 8:12
• @devss The distribution of $Z$ is the same as the distribution of $l-Y$ from their definitions. Similarly $0 \le Y \le l/2$ and $l/2 \le Z \le l$ Commented Jun 12, 2019 at 13:33