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Let $R$ be a random variable with a $U(0,1)$ distribution, $k$ be a nonzero integer constant, and $c$ be a real constant. Let $$S\equiv kR + c \ \mod 1$, \ \text{with} \ 0\leq S \leq 1$$ Show that $S$ has a $U(0,1)$ distribution.

Attempted proof - Let $R\sim U(0,1)$ and consider $c = 0$. So we have $S = kR$. So we have $$f_{kR}(t) = \begin{cases} 1 \ &\text{if} \ t\in (0,1)\\ 0 \ &\text{otherwise} \end{cases}$$ I am not sure if this works, I am being thrown off by the constant $k$ any suggestions on solving this problem are greatly appreciated.

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    How do you conclude the "so we have" part? I mean, it's true, but I don't think you've properly established it. I think setting $c = 0$ first is a viable approach, but when treating $S$ as a pure integer multiple of $R$, you should consider the probability that $S \bmod 1 \leq t$ for some $t \in (0, 1)$. For what values of $R$ does that happen? Show that the probability that $R$ satisfies these conditions is equal to $t$. – 2017-02-21
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    @BrianTung Could you provide an answer? I am a bit lost still. – 2017-02-21

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Basic approach. Suppose $R \sim U(0, 1)$ and $k = 2$. Then $S = 2R \bmod 1$, with $0 \leq 2R \leq 2$.

We then consider the CDF $F_S(t) \equiv P(S \leq t)$, for $0 \leq t \leq 1$. Based on the above, we can write

\begin{align} F_S(t) & = P(0 \leq S \leq t) \\ & = P(0 \leq 2R \leq t \textbf{ or } 1 \leq 2R \leq 1+t) \\ & = P(0 \leq 2R \leq t) + P(1 \leq 2R \leq 1+t) \\ & = P\left(0 \leq R \leq \frac{t}{2}\right) + P\left(\frac{1}{2} \leq R \leq \frac{1+t}{2}\right) \\ & = F_R\left(\frac{t}{2}\right)-F_R(0) + F_R\left(\frac{1+t}{2}\right)-F_R\left(\frac{1}{2}\right) \\ & = \frac{t}{2}-0+\frac{1+t}{2}-\frac{1}{2} \\ & = \frac{t}{2}+\frac{t}{2} = t \end{align}

This establishes that $S$ is uniform on $(0, 1)$ for $k = 2$ and $c = 0$. Now generalize.

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    Just curious, but how could we generalize without assuming some value for $k$? – 2017-02-23
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    Also, do you know how to answer this one:http://math.stackexchange.com/questions/2157289/linear-congruential-generator-proof-by-induction – 2017-02-23
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    If you assume $c = 0$, then you just have to sum up over $k$ intervals: between $0$ and $t/k$, then between $1/k$ and $(1+t)/k$, then between $2/k$ and $(2+t)/k$, etc. – 2017-02-24
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    Do you think you could show the generalization, I am not seeing able to get the result desired – 2017-02-24
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    I try the same method you do but when I generalize it I get $\frac{2t - k}{k}$ – 2017-02-24
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    Look at the intervals I listed. Aren't all of them of length $t/k$? And if there are $k$ of them, isn't their total combined length equal to $t$? – 2017-02-24
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    Yes, but if we assume $c=0$ then $S = kR \mod 1$, with $0\leq kR\leq k$. Then $$\mathbb{P}(0\leq s \leq t) = \mathbb{P}(0\leq kR\leq t \ \text{or} \ k\leq kR \leq 1 + t)$$ and so on, so I don't see how we get $t$ at the end. – 2017-02-24
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    Your expression is incorrect. There should be $k$ terms or'd together, each covering an interval of length $t$. Divide them all by $k$, and you get $t$ back again. – 2017-02-24
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    Could you clarify the correct expression? – 2017-02-24