Density Function of Uniform Distribution?

Question

I had this problem from my last question:

Given that: $X \sim \text{Uni}(a,b)$.

I know that the density function is:

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

Then How can this help me to find the density function of : $-X$?

Note: what is the right way to write the density function of : $-X$? will it be $f_{-X}(-x)$ or $f_{X}(-x)$ or $f_{-X}(x)$?

Answer 1

Just consider the cumulative distribution function: $$F_X(x)=\int_a^x f_X(x) \quad dx=P(X<x) $$ Then if $Y=-X$, by definition we have: $$ F_Y(y)=P(Y\leq y)=P(-X\leq y)=P(y\geq -X)=1-F_X(-x) $$

So, again by definition: $f_Y(y)=F_Y'(y)=(-F_X(-x) )' = f_X(-x) $.

There you go. And the right way to write the density function of $-X$
would be $f_{-X}(-x)$, or more commonly: $f_Y(y)$.