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If we multiply a singular continuous random variable by a positive constant, is the resulting random variable also singular continuous?

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Sorry, what do you mean by "singular"? –  David Mitra Mar 8 '12 at 17:20
    
$\mu$ is singular continuous if $\mu\{x\}=0$ for all $x\in\mathbb{R}$, but there is $S\subseteq\mathbb{R}$ with $\lambda(S)=0$ and $\mu(S^{c})=0$. –  Vahid Mar 9 '12 at 18:42

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Yes.

Let $X$ be a singular continuous random variable with CDF $F$. Then $F$ is a singular continuous function. Fix $a\in\mathbb{R}$, with $a>0$ for simplicity. Denote the CDF of $aX$ by $F_a$. For $t\in\mathbb{R}$,

$$F_a(t)=\mathbb{P}[a X\leq t]=\mathbb{P}\left[X\leq \frac{t}{a}\right] = F(t/a)$$

So $F_a$ is continuous.

The derivative of $F$ vanishes almost everywhere, so the same holds for $F_a$, by a little analysis. Hence $F_a$ is singular.

So $aX$ is a singular continuous random variable.

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