I have an equation $Y = 5 + 3\times X$ and I assume that $X$ is a random variable taking values from a uniform distribution. Can I consider that also $Y$ is a random variable which takes values from a uniform distribution also but in different space?
Thank you in advance
If X is uniform on (0,1) then Y=5+3X is uniform on (5,8). More generally, for every nonzero b, Z=a+bX is uniform on (a,a+b).
Not an expert, but I think $Y$ will be distributed uniformly as well.
Thank you for pointing out the error Did!
The uniform (continuous) distribution is determined by its cumulative distrubition function.
$$P(X \leq x) = F_X(x)= \begin{cases} 0 &\text{if } x < a\\ \frac{x-a}{b-a} &\text{if } x \in [a,b[ \\ 1 &\text{if } x \geq b\end{cases}$$
Looking at $F_Y(y)$:
$$F_Y(y) = P(Y\leq y) = P(5+3X \leq y) = P\left(X \leq \frac{y-5}{3}\right)$$
You can find the boundaries where this makes sense using: $$\frac{y-5}{3} \geq a \qquad \frac{y-5}{3} < b$$
Which concludes towards: $$P(Y \leq y) = F_Y(y)= \begin{cases} 0 &\text{if } y < 3a+5\\ \frac{\frac{y-5}{3}-a}{b-a} &\text{if } x \in [3a+5,3b+5[ \\ 1 &\text{if } x \geq 3b+5\end{cases}$$
($X$ will not be discrete, the overal idea was fine though)
If $X \stackrel{d}{=} U(a,b)$ then
$$P(X =x) = \begin{cases} \frac{1}{b-a} \quad &\text{ if } x \in [a,b] \\ 0 &\text{ if not}\end{cases}$$
Looking at $P(Y=y)$ which determines the distribution you get:
$$P(Y=y) = P(5+3X = y) = P\left(X = \frac{y-5}{3}\right)$$
Which means $Y$ will also be distributed uniformly.
