Find the following distribution?

Question

I have been given the following problem:

The probability density function of a random variable X is given by:

• $f(x;θ) = \dfrac{2(θ−x)}{θ^2}$, if $0< x<θ$, $0$ otherwise*

Find the distribution of $U = X/θ$ and specify its domain where the pdf is non-zero.

In order to solve this do I need to find the estimator $θ$ and divide $2(θ−x)/θ^2$ by the estimator?

From what I have read I can find the estimator by finding the partial derivative of $2(θ−x)/θ^2$ with respect to $θ$ and setting it to $0$?

Answer 1

If $0\le u\le 1$ then $$ \Pr(U\le u) = \Pr\left( \frac X \theta \le u\right) = \Pr(X\le \theta u) = \int_0^{\theta u} \frac{2(\theta - x)}{\theta^2} \, dx = 1 - (1-u)^2. $$ Hence $$ f_U(u) = 2(1-u) \text{ if }0\le u\le 1 $$ and $=0$ if $u>1$ or $u<0$.

Answer 2

No. It's much simpler than that. If $X$ has PDF $f_X(x;\theta)$, then $U = X/\theta$ is just a scale transformation of $X$, so its PDF is directly given by $$f_U(u;\theta) = \theta f_X(\theta u;\theta).$$ The idea is to calculate this and simplify it accordingly.