3
$\begingroup$

Given a uniform probability distributed random variable $X$ between the interval $[0,100]$. If I construct a new random variable $Y$ as $Y = X \mod N$, with a given $N$. Will $Y$ still be uniformly distributed?

I would say yes, the only thing that will change is the interval, which will be from $[0,N]$. Am I correct?

$\endgroup$
1
$\begingroup$

Let $m$ be the integer with $mN\leq 100<(m+1)N$.

If $mN=100$ then the distribution of $Y=X\pmod N$ is indeed uniform on $[0,N]$. That is because for $0\leq t\leq N$, we have $$P[Y\leq t]=P[X\leq t]+P[N\leq X<N+t]+...+P[(m-1)N\leq X<(m-1)N+t]=\dfrac{t}{100}+\dfrac{t}{100}+...+\dfrac{t}{100}(m\text{ times})=\dfrac{mt}{100}=\dfrac{t}{N}$$ since $N=\dfrac{100}{m}$

If however $mN<100<(m+1)N$ then we have two cases depending on whether $mN+t>100$ or $mN+t\leq 100$.

In the former, $P[mN\leq X<mN+t]=\dfrac{100-mN}{100}$. Therefore, like the previous sum, $P[Y\leq t]=P[X\leq t]+P[N\leq X<N+t]+...+P[(m-1)N\leq X<(m-1)N+t]+P[mN\leq X<mN+t]=\dfrac{mt}{100}+\dfrac{100-mN}{100}=1-\dfrac{m(N-t)}{100}$.

In the latter, however, $P[mN\leq X<mN+t]=\dfrac{t}{100}$ and so $P[Y\leq t]=\dfrac{(m+1)t}{100}$ following above argument.

$\endgroup$
1
$\begingroup$

No. For example, if $N = 3$, $Y \in [0,1)$ if $X \in [0,1) \cup [3, 4) \cup \ldots \cup [99,100)$ (probability $34/100$), but $Y \in [1,2)$ if $X \in [1,2) \cup [4,5) \cup \ldots \cup [97,98)$ (only $33/100$).

$\endgroup$
  • $\begingroup$ Thank you for your reply, I was wondering how the $\mod$ then would affect the distribution of $Y$. Could you maybe elaborate on that aspect? $\endgroup$ – Snowflake Dec 26 '16 at 1:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.