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Let $X \sim \mathcal{U}[a, b]$ be a uniformly distributed random variable with $a < b$ and $a, b \in \mathbb{R}$.

How is $Y = c \cdot X$ with $c > 0$ distributed?

My thoughts

The probability density function of $X$ is

$f_X(x) = \begin{cases}\frac{1}{b-a} &\text{for } x \in [a, b]\\0&\text{otherwise}\end{cases}$

Intuitively, I guess the probability density function of $Y$ is

$f_Y(x) = \begin{cases}\frac{1}{c(b-a)} &\text{for } x \in [c \cdot a, c \cdot b]\\0&\text{otherwise}\end{cases}$

Hence $Y \sim \mathcal{U}[ac, bc]$. However, I don't know how to show this properly.

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You can build the cdf of $Y$. $$F_y(y)=P(Y \leq y)=P(cX\leq y)=P(X \leq \frac{y}{c})=F_X(\frac{y}{c})$$

Let $f_Y$ denote the density function of $Y$ ,it is defined as $$f_Y(y)=F'_Y(y)=(F_X(\frac{y}{c}))'=\frac{1}{c}f_X(\frac{y}{c})$$

which is the result you were looking for.

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