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Let X be a random variable distributed as $U[0, K]$ for an integer K. Find the density function of $Y = f (x) = x- [x]$, where [x] denotes the integer part of the real number x.

I think that [X] represents a discrete uniform but this by definition would be with values at x = 1,2, ... k, and its density would be 1/k which is the same as that of the continuous uniform.

Now, I understand that the distribution [X] is dependent on X, then I could not assume independence to use some kind of transformation because I do not know the joint.

I appreciate if you can give me some other way that I have not considered, thank you very much.

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So, set $\{x\}=x-\lfloor x\rfloor$ to be the fractional part of $x$. Check that $\{x\} \in [0,1)$. Hence, $f_X(x)$ is defined for $x\in [0,1)$. Let's compute the CDF. Fix a $c\in[0,1)$, and study $\mathbb{P}(\{x\}\leq c)$. Observe that, $$ \{\{x\} \leq c\}=\bigcup_{k=0}^{K-1}\{x\in [k,k+c)\}, $$ hence $\mathbb{P}(\{x\}\leq c)=K\cdot \frac{c}{K}=c$. Hence, it turns out that, $\{x\}$ is uniform on $[0,1)$ (and also, $\lfloor x\rfloor$ is uniform on $\{0,\dots,K-1\}$.

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  • $\begingroup$ Thanks, I understood. $\endgroup$ – Cristian Perdomo Dec 31 '18 at 0:26

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